Chapter 7: Eigenvalues and Eigenvectors
1
SECTION F Applications of Eigenvalues and Eigenvectors to Quadratic Forms
By the end of this section you will be able to
 find a matrix given a quadratic form
 convert a quadratic into diagonal form
 decide whether a matrix is positive, negative or indefinite
You might be surprised that quadratic forms which involve expressions of degree 2 are
covered in linear algebra because in essence linear algebra is the study of linear equations.
In this section you will find that linear algebra is a useful tool to describe quadratic forms.
You will need to recall your work from section D regarding the diagonalization of symmetric
matrices.
F1 Introduction to Quadratic Forms
What does the term quadratic mean?
It is an expression of second degree. For example a quadratic polynomial is a polynomial
such as ax 2  bx  c . An expression of the form
(7.35)
[a, b and h are real numbers]
ax 2  2hxy  by 2
is called a quadratic form in two variables x and y. Of course we can have a quadratic form
of more than two variables as well. In this section we place quadratic forms into matrix form
as the examples below show. Putting a quadratic into matrix form means we can use matrices
to solve quadratic equations.
Quadratic forms are also used to find where a function of more than one variable is a
maximum or minimum.
Example 37
 2 4 
 x
T
Let A  
 and x    . Determine x Ax .
4
3
y


 
Solution
What is xT equal to?
Well xT is the vector x transposed. That is xT   x y  . Thus we have
 2 4   x 
y
 
 4 3 y 
 2 x  4 y 
  x y

 4x  3y 
xT Ax   x
  x  2 x  4 y   y  4 x  3 y 
 2 x 2  4 xy  4 xy  3 y 2  2 x 2  8 xy  3 y 2


8 xy
Note that x Ax  2 x  8 xy  3 y 2 is an example of a polynomial in quadratic
form.
The matrix A is called the matrix of quadratic form.
The following is not a quadratic form (of degree 2) in two variables x and y:
x2  x2 y  y 2
Why not?
Because the middle term x 2 y   xxy  is cubic (of degree 3) in nature.
T
2
Chapter 7: Eigenvalues and Eigenvectors
2
Quadratic form is given by xT Ax where A is a symmetric matrix. We carry out the following
operations:
gives quadratic form xT Ax
Figure 1
Quadratic forms are used in physics and engineering. For example the kinetic energy, ke, of a
rigid body is given by
1
ke   T A
2
where ω is the angular velocity vector and A is a 3 by 3 symmetric matrix.
Another application of quadratic form is finding the length Ax of the vector Ax :
Ax   Ax  Ax  xT  AT A  x
2
T
Matrix A may not be symmetric so we use the symmetric matrix AT A to find the length.
We can also use quadratic forms to find the norm of a matrix. The norm of a matrix A tells
us how much the matrix A magnifies a given vector x when applied to it.
For most applications the norm of any matrix A denoted by A is defined as
A  max  xT  AT A  x 
2
provided x  1
[In Matlab the command norm(A) evaluates this norm of the matrix A.]
The function max is the maximum value of the quadratic form xT  AT A  x for all unit
vectors x.
Example 38
 3 5 
Find the quadratic form xT Ax of the matrix A  
.
 5 4 
Solution
 x
Let x    then xT   x y  and we have
 y
 3 5  x 
xT Ax   x y  
 
 5 4  y 
 x

 3 x  5 y 
y

 5 x  4 y 
 x  3x  5 y   y  5x  4 y 
We have xT Ax  3 x 2
 3x 2  10 xy  4 y 2
 10 xy  4 y 2 .
What do you notice about the matrix A in both the above examples?
 2 4 
 3 5 
We have A  
 and A  
 . In both cases the matrix is symmetrical, that is
 4 3
 5 4 
the transpose of A is A; AT  A .
In the next example we derive the general quadratic form.
Example 39
a h
 x
T
Let A  
 where a, b and h are real numbers and x    . Determine x Ax .
h
b
y


 
Solution
Chapter 7: Eigenvalues and Eigenvectors
Note that A is a symmetric matrix. We have xT   x
3
y  therefore
 a h x 
y
 
 h b y 
 ax  hy 
  x y

 hx  by 
xT Ax   x
  x  ax  hy   y  hx  by 
 ax 2  2hxy  by 2
What do you notice about ax 2  2hxy  by 2 ?
This is the above quadratic form given in (7.35). Thus we conclude that the quadratic form:
a h
 x
 x where x   
a x 2  2 h xy  b y 2  xT 
h b
 y


We can extend quadratic form to any number of variables. We write a quadratic form in n
variables x1 , x2 , x3 ,  and xn by using an n by n matrix as the following definition states:
Definition (7.36). Let A be an n by n symmetric matrix. The function f given by
 x1 
 
f  x   xT Ax where x    
x 
 n
is called the quadratic form in n variables x1 , x2 , x3 ,  and xn .
Example 40
1 4 7


Find the quadratic form xT Ax for the symmetric matrix A   4 2 5  .
7 5 3


Solution
Let xT   x y z  then we have
1 4 7 x 

 
x Ax   x y z   4 2 5   y 
7 5 3 z 

 
 x  4 y  7z 


  x y z   4 x  2 y  5z 
 7 x  5 y  3z 


T
  x  x  4 y  7 z   y  4 x  2 y  5 z   z  7 x  5 y  3 z 
 x 2  4 xy  7 xz  4 xy  2 y 2  5 yz  7 xz  5 yz  3 z 2
 x 2  8 xy  14 xz  2 y 2  10 yz  3 z 2
Collecting like terms
Note that xT Ax  x 2  8 xy  14 xz  2 y 2  10 yz  3 z 2 is a quadratic
form in 3 variables x, y and z.
Next we derive the general quadratic form in three variables.
Chapter 7: Eigenvalues and Eigenvectors
Example 41
a

Let A   d
e

Solution
We have xT
4
e
 x

f  and x   y  . Determine xT Ax .
z
c 
 
d
b
f
 x
y
z  . Therefore
a

z d
e

e  x 
 
x Ax   x y
b f  y 
f c   z 
 ax  dy  ez 


  x y z   dx  by  fz 
 ex  fy  cz 


T
d
  x  ax  dy  ez   y  dx  by  fz   z  ex  fy  cz 
 ax 2  dyx  ezx  dxy  by 2  fzy  exz  fyz  cz 2
Collecting like terms 
 ax 2  2dyx  2ezx  by 2  2 fzy  cz 2
Thus the general quadratic form in three variables x, y and z is given by
ax 2  2d  yx   2e  zx   by 2  2 f  zy   cz 2
a

where A   d
e

(7.37) ax 2
d
b
f
 xT Ax
e
 x

f  is a symmetric matrix and x   y  . This is equivalent to
z
c 
 
 2 d  xy   2 e  zx   by 2
 2 f
 zy 
 cz 2
a

 xT  d

 e

d
b
f
e 

f x

c 
F2 Finding the Matrix given the Quadratic Form
The other way round of finding the matrix given the quadratic form is more difficult.
Examining the matrix in the above formula (7.37), we have the entries on the leading
diagonal give the coefficients of x 2 , y 2 and z 2 respectively. The d entry of the matrix in the
above (7.37) is half the xy coefficient because the xy coefficient is 2d . Similarly the e entry
is half the zx coefficient and the f entry is half the zy coefficient.
The coefficients of the single variable like x 2 , y 2 and z 2 lie on the leading diagonal.
Halving the coefficients of mixed variables such as xy, xz and yz gives the entries to the
right and left of the leading diagonal.
Example 41
Write the quadratic form 2 x 2  7 yx  3 zx  8 y 2  2 zy  3 z 2 as xT Ax .
Solution
7
3
2
Using the above (7.37) with a  2, d  , e  , b  8, f    1 and c  3 :
2
2
2
2 x  7  yx   3  zx   8 y  2  zy   3 z   x
2
2
2
y
7 / 2 3 / 2 x 
 2

 
z   7 / 2 8
1   y 
 3 / 2 1
3   z 

Chapter 7: Eigenvalues and Eigenvectors
5
Example 42
Write the quadratic form 2 x 2  7 xy  8 y 2 as xT Ax .
Solution
Since we have 2 variables x and y so the matrix is of size 2 by 2. The entries on the leading
diagonal are the coefficients of x 2 and y 2 respectively. Both entries along the other diagonal
is 7 / 2 because the xy coefficient is 7. Thus we have
 2 7 / 2   x 
2 x 2  7 xy  8 y 2   x y  
 
 7 / 2 8   y 
Example 43
Convert the quadratic form x 2  y 2  z 2 to matrix form xT Ax .
Solution
What size matrix will we have this time?
We have three variables x, y and z therefore the matrix size is 3 by 3. The coefficients of
x 2 , y 2 and z 2 is 1 therefore all the leading diagonal entries are 1. There are no mixed xy, xz
and yz terms therefore the remaining entries in the matrix are zero. We have
1 0 0 x 
 x

 
 
2
2
2
x  y  z   x y z0 1 0 y    x y zI y 
0 0 1 z 
z

 
 
The 3 by 3 identity matrix I is the matrix of the quadratic form x 2  y 2  z 2 or we can say
xT Ix  xT x  x 2  y 2  z 2
This quadratic form xT Ix gives the length or norm of the vector x.
F3 Converting to Diagonal form
In this subsection we change the quadratic form in variables x1 , x2 , x3 ,  and xn into new
variables y1 , y2 , y3 ,  and yn . Why?
We will find it simpler to deal with these new variables because we convert the given
quadratic form into a more easy to use format called diagonal form.
Remember we discussed in section D that it is easier to deal with a diagonal matrix. We
convert a given quadratic form xT Ax into a diagonal format y T Dy where D is a diagonal
matrix. This y T Dy is called diagonal form.
What is the point of converting the quadratic form xT Ax to y T Dy ?
Well y T Dy contains no mixed or cross product terms, that is terms of the sort y1 y2 , y1 y3 ,
y2 y3 … are not part of y T Dy . Thus we can show that y T Dy only contains terms of the form
y12 , y2 2 , y32 ,  and yn 2 . Actually y T Dy is a sum of squares.
The next example illustrates this.
Example 44
Write the quadratic form 10 x12  8 x1 x2  4 x2 2 as xT Ax .
Find the orthogonal matrix Q which diagonalizes A and the diagonal matrix D such that
y 
QT AQ  D . Also determine the diagonal form f  y   y T Dy where y   1  .
 y2 
Solution
Chapter 7: Eigenvalues and Eigenvectors
6
What are the leading diagonal entries in the matrix A of quadratic form 10 x12  8 x1 x2  4 x2 2 ?
10 and 4 because these are the coefficients of x12 and x2 2 respectively. What are the other
entries in the matrix A?
4 because half of the cross product x1 x2 coefficient  8  is equal to 4 . Thus we have
10 x12
 8 x1 x2
 4 x2 2

 x1
 10 4   x1 
x2  
 
4   x2 
 4
 10 4 
The matrix A  
 is a symmetric matrix which means we can orthogonally
 4 4 
diagonalize this matrix by finding the eigenvalues and corresponding normalised
eigenvectors. This was covered in section D and we will assume that you can verify the
following eigenvalues and eigenvectors belonging to these eigenvalues in your own time:
1 1 
1  2
1  2, u 
  and 2  12, v 
 
5  2
5  1 
What is the eigenvector matrix Q equal to?
2
1 1
Q  u v  
 Remember Q   u v  


5  2 1 
What is the eigenvalue (diagonal) matrix D equal to?
The matrix with the leading diagonal entries being the eigenvalues of matrix A:
2 0 
 Because 2 and 12 are
D



 0 12 
 the eigenvalues of A 
How do we find f  y   y T Dy ?
f  y   y T Dy 
 y1
 2 y1 
 2 0   y1 
y2  
  y    y1 y2   12 y 
 0 12   2 

2
2
 2 y1  12 y2 2
What is the advantage of converting the given quadratic form into diagonal form?
Well if we examine the above example we converted from the given quadratic to a sum of
squares:
10 x12  8 x1 x2  4 x2 2
2 y12  12 y2 2
This new diagonal form is easier to graph as will be shown in the next section. Also the
solution to the equation 10 x12  8 x1 x2  4 x2 2  c is equivalent to solving the easier equation
2 y12  12 y2 2  c
In general a complicated problem can been reduced to something more manageable by using
diagonal form.
Next we check the above result is correct by going from the y back to the x variable.
Example 45
Check that f  y   2 y12  12 y2 2  10 x12  8 x1 x2  4 x2 2 .
Solution
From Proposition (7.38) we have Qy  x . Taking inverse Q of both sides of Qy  x gives
y  Q 1x  QT x
The new variables  y1 and y2 
 Because Q is orthogonal therefore Q 1  QT 
in vector y are obtained from the old variables  x1 and x2 
Chapter 7: Eigenvalues and Eigenvectors
7
in vector x by applying the matrix QT to vector x, that is y  QT x .
 x1 
y 
Substituting x    and y   1  into y  QT x gives
 y2 
 x2 

2 
1 1
 Because Q 


5  2 1  

2   x1  1  x1  2 x2   y1 
1 1


 

  
5  2 1   x2 
5  2 x1  x2   y2 
1
1
From this we have the new variables y1 
 x1  2 x2  and y2   2 x1  x2  in terms of
5
5
the old variables, x1 and x2 .
T
2   x1 
 y1  1  1
 

  
5  2 1   x2 
 y2 
Evaluating f  y  by using the given result we have
f  y   2 y12  12 y2 2
2
2
 1

 1

 2
 x1  2 x2    12   2 x1  x2 
 5

 5

2
12
  x12  4 x1 x2  4 x2 2    4 x12  4 x1 x2  x2 2 
5
5
2
  x12  4 x1 x2  4 x2 2  6  4 x12  4 x1 x2  x2 2  
5
Substituting
y1 and y2 
 Expanding 

2
12
 2 
Taking out 5 because 5  6  5  
 

2
 25 x12  20 x1 x2  10 x2 2 
5
2
2
2
2

 2 5 x1  4 x1 x2  2 x2   10 x1  8 x1 x2  4 x2

Taking out 5
We have
xT Ax  10 x12
 8 x1 x2
 4 x2 2
 2 y12
 12 y2 2
 y T Dy
We have xT Ax  y T Dy which means we have converted a given quadratic into diagonal
form which is easier to deal with because we only have sum of squares 2 y12  12 y2 2 and
no cross products.
Proposition (7.38). Let x  Qy where Q is an orthogonal matrix which diagonalizes a
symmetric matrix A, and D be the diagonal eigenvalue matrix. Then any given quadratic
form can be written in diagonal form as:
xT Ax  y T Dy
Proof.
Consider the quadratic form xT Ax :
xT Ax   Qy  A  Qy 
T
  y T QT  A  Qy 
 y T  QT AQ  y
 y T Dy
Substituting x  Qy 
 Using  XY T  YT XT 


 By (7.24) we have QT AQ  D 
■
Actually we can state and prove the following more useful result called ‘The Principle Axes
Theorem’.
Chapter 7: Eigenvalues and Eigenvectors
8
The Principle Axes Theorem (7.39). A quadratic form in n variables xT Ax is given by
xT Ax  y T Dy  1 y12  2 y2 2    n yn 2
 y1 
 
where 1 , 2 ,  and n are the eigenvalues of the matrix A and y     .
y 
 n
Proof.
By the above Proposition (7.38) we have xT Ax  y T Dy where D is a diagonal matrix
with the eigenvalues  of matrix A as entries on the leading diagonal.
xT Ax  y T Dy
 1 0

0 2
  y1 y2  yn  
0 

0 0
 1 y1 


  y1 y2  yn    
 y 
 n n
2
2
 1 y1  2 y2  

0   y1 
 
0 0   y2 
    
 
 n   yn 
 n yn 2
■
Remember the advantage of writing
xT Ax  1 y12
 2 y2 2
   n yn 2
is that there are no cross product terms such as y1 y2 , y1 y3 , y2 y3 , … .
In a nutshell the Principle Axes Theorem says we can write any quadratic form as sum of
squares with the coefficients given by the eigenvalues.
Example 46
Transform  x 2  2 xy  y 2  z 2 into diagonal form.
Solution
First we need to find the matrix A such that  x 2  2 xy  y 2  z 2  xT Ax .
The entries on the leading diagonal are the coefficients of x 2 , y 2 and z 2 which are 1 in
each case. The only other non-zero entry is half the xy coefficient (2) which is 1. Thus
x
2
 2 xy 
y
2

z
2

 1 1 0 


x Ax where A   1 1 0 
 0 0 1 


T
What do we need to find so that the quadratic form has no cross-product terms?
The eigenvalues of the symmetric matrix A.
Verify that the eigenvalues of matrix A are 1  0, 2  1 and 3  2 .
Let the new variables be x, y and z  . Therefore using the above Principle Axes Theorem
(7.39) we have
2
2
2
 x 2  2 xy  y 2  z 2  1  x   2  y   3  z  
 0  x    1 y    2  z  
2
   y   2  z  
2
2
2
2
Chapter 7: Eigenvalues and Eigenvectors
9
In the above example we converted:
  y   2  z  
 x 2  2 xy  y 2  z 2
It is simpler to use the sum of squares on the Right Hand Side.
Note that to write a given quadratic form without any cross-products we only need to find the
eigenvalues of the matrix.
In order to find the relationship between x1 , x2 , x3 and the new variables y1 , y2 , y3 we
2
2
need to find the eigenvector matrix Q because y  QT x .
If x is a non-zero vector then y  QT x is also non-zero because Q is an orthogonal matrix.
F4 Definite Matrices
In this subsection we discuss positive, negative and indefinite matrices.
A symmetric matrix A is positive definite if xT Ax is a positive real number for all non-zero
vectors x. The formal definition is:
Definition (7.40). A symmetric matrix A is positive definite  the quadratic form xT Ax is
positive for all non-zero vectors x. A positive definite matrix A is denoted by A  0 .
 10 4 
The matrix A  
 of Example 44 is positive definite because
 4 4 
(*)
xT Ax  y T Dy  2 y12  12 y2 2
It is easier to check a matrix is positive definite (or negative definite) in diagonal form. For
example in (*) we can see straight away that the Right Hand Side is a positive real number so
matrix A is positive definite, A  0 .
The identity matrix I is also positive definite because
xT Ix 
 x1 
2
 x2 

2
  
 xn 
2
or I  0
Remember this quadratic form xT Ix gives the length of the vector x in  n .
Proposition (7.41). A symmetric matrix A is positive definite  all the eigenvalues of A are
positive.
What does this statement mean?
Means we can check a given matrix is positive definite by examining the polarity of the
 1 4 5 


eigenvalues. For example the matrix A   0 2 6  is positive definite because we have
 0 0 3


a triangular matrix so the eigenvalues are the entries on the leading diagonal 1, 2 and 3 which
are all positive. Hence A  0 .
How do we prove this result?
We have  in the statement therefore we need to prove the result both ways,  and  .
Proof.
(  ). Let x be an arbitrary non-zero vector and all the eigenvalues  of matrix A be
positive. Let y be the vector of new variables so by the Parallel Axes Theorem (7.39):
xT Ax  y T Dy


1 y12
 2 y2 2
   n yn 2
By (7.39)
0
 Because all the  's are positive
Chapter 7: Eigenvalues and Eigenvectors
10
Thus matrix A is positive definite.
(  ). Assume matrix A is positive definite then by the above definition (7.40):
(7.40). A symmetric matrix A is positive definite  xT Ax is positive (x  O ).
This means for all non-zero vectors x:
xT Ax  1 y12  2 y2 2    n yn 2  0 [Positive]
Required to prove all the eigenvalues are positive. We prove this by contradiction.
Suppose one of the eigenvalues is negative or zero, that is k  0 [Negative or zero] for
 v1 
 
some k between 1 and n. Let x  v k     where v k is an eigenvector belonging to the
v 
 n
negative eigenvalue k of matrix A. Then
xT Ax   v k  Av k
T
  v k  k v k
T
Substituting x  v k 
 Because v k is an eigenvector;
Av k  k v k 
 k  v k  v k
T
 k  v1 v2
 v1 
 
 vn    
v 
 n
 k v12  v2 2  v32    vn 2 
We are assuming xT Ax  0 [Positive] for all non-zero vectors x therefore
T
xT Ax   v k  Av k
 k v12  v2 2  v32    vn 2   0  Positive
This implies k  0 [Positive]. We have a contradiction because k  0 and k  0 . This
means our supposition k  0 [Negative or zero] is incorrect so k  0 . Hence all our
eigenvalues are positive.
■
The identity matrix I is positive definite because all the eigenvalues are 1 [Positive].
Also if symmetric matrices A and B are of the same size and A  B is positive definite,
A  B  0 , then A  B . We can use this positive definiteness to compare matrices.
What do you think the term negative definite matrix means?
A is negative definite if xT Ax is a negative real number for all non-zero vectors x.
Definition (7.42). A symmetric matrix A is negative definite  the quadratic form xT Ax is
negative for all non-zero vectors x. We denote this by A  0 .
The matrix A of the above Example 46 is negative definite because
2
2
xT Ax    y   2  z   [Negative]
In diagonal form it is easy to notice that the quadratic is a negative real number.
Proposition (7.43). A symmetric matrix A is negative definite  all the eigenvalues of A
are negative.
Proof. Very similar to proof of Proposition (7.41). See Exercise 7(f).
Chapter 7: Eigenvalues and Eigenvectors
11
 1 2 
The matrix A  
 is negative definite because both the eigenvalues are negative;
 0 3 
1 and  3 so by (7.43) we have A  0 .
Definition (7.44). A symmetric matrix A is indefinite  the quadratic form xT Ax takes on
both positive and negative values.
In general we say a quadratic form is positive definite, negative definite or indefinite if its
matrix is positive definite, negative definite or indefinite respectively.
The following quadratic form is popular in advanced physics:
f ( x, y , z , t )  x 2  y 2  z 2  t 2
This is an indefinite quadratic form because f can take both positive and negative values.
Example 47
Consider the quadratic form f  x   2 x 2  4 yx  4 zx  y 2  8 zy  z 2 . Decide whether this
quadratic form is positive, negative or indefinite.
Solution
Putting the given quadratic form f  x   2 x 2  4 yx  4 zx  y 2  8 zy  z 2 into xT Ax gives:
2 x 2  4 yx  4 zx  y 2  8 zy  z 2   x
y
 2 2 2   x 

 
z   2 1 4   y 
 2
4 1   z 

 2 2 2 


You can verify that the matrix A   2 1 4  has the eigenvalues 1  3, 2  3 and
 2
4 1 

3  6 . Applying the Principle Axes Theorem (7.39) which gives us the sum of squares:
xT Ax  1  x   2  y   3  z  
2
2
2
With 1  3, 2  3 and 3  6 we have
xT Ax  3  x   3  y   6  z  
What type is this quadratic form?
2
2
2
Indefinite because 3  x   3  y   6  z   could be either positive or negative.
2
2
2
Also matrix A is indefinite.
These positive, negative and indefinite matrices are used in optimization problems of finding
maximum and minimum of a function f  x  of more than one variable.
SUMMARY
A quadratic form can be written in matrix form as xT Ax where A is a real symmetric matrix.
We can convert a quadratic form into a diagonal form by xT Ax  y T Dy where
y T Dy  1 y12  2 y2 2    n yn 2 [Sum of squares]
Chapter 7: Eigenvalues and Eigenvectors
12
Exercise 7(f)
1. Determine xT Ax for the following:
1
1 0
 x
(a) A  
(b) A  
, x   
0
0 1
 y
1 2
 x
 1
(c) A  
(d) A  
, x   
2 1
 y
 4
1/ 2 2 
 x
 1
(e) A  
(f) A  
, x   
 2 1/ 3 
 y
 
0
 x1 
, x   x 
1
 2
4 
 x
, x   
7 
 y

 x
, x   
e
 y
2. Find xT Ax for the following:
1 0 0
 x


 
(a) A   0 1 0  , x   y 
0 0 1
z


 
 1 2 7 
 x


 
(c) A   2 5 8  , x   y 
 7 8 9 
z


 
2 2
 x

 
2 2, x   y 
z
2 2 
 
5 7
 x

 
0 3  , x   y 
z
3 0 
 
2

(b) A   2
2

0

(d) A   5
7

3. For the following quadratic forms determine the symmetric matrix A and write these
quadratic forms as xT Ax .
(a) 9 x 2  2 xy  y 2
(b) x 2  y 2
(c) x 2  y 2
(d) 31x 2  41xy  69 y 2
(e) x 2  xy
(f)  x 2
(g) xy
(h) xy 2
4.
(a)
(c)
(e)
(g)
Express the following quadratic forms into
x 2  2 xy  4 xz  2 y 2  10 yz  5 z 2
3 x 2  6 xy  12 xz  7 y 2  18 yz  z 2
 x2  y 2  z 2
yz  xz
xT Ax :
(b) 2 x 2  6 xy  32 y 2  24 yz  z 2
(d) 3 x 2  4 xz  7 y 2  5 z 2
(f) 2x 2  xy  xz
(h) 3 x 2  5 xy  7 xz  3 yz  2 z 2
5. Transform the following quadratic forms into diagonal form:
(a) 2 x 2  4 xy  y 2
(b) 2 x 2  2 xy  2 y 2 (c) 5 x 2  24 xy  5 y 2 (d) x 2  y 2
(e) 2xy
(f) x 2  2 xy
6. Transform the following into diagonal form:
(a) 3 x 2  8 xy  3 y 2  5 z 2
(b) 2 x 2  4 xy  4 xz  2 y 2  4 yz  2 z 2
(c) 2 x 2  4 xy  4 xz  y 2  8 yz  z 2
(d)  x 2  y 2  z 2
(e)  x 2  2 y 2  2 yz  2 z 2
(f) 2xy  xz
A quadratic form xT Ax is called positive semi-definite  xT Ax is positive or zero for
all x. (   0 must be an eigenvalue of matrix A for semi-definite.)
Chapter 7: Eigenvalues and Eigenvectors
13
A quadratic form xT Ax is called negative semi-definite  xT Ax is negative or zero
for all x.
7. Determine which of the quadratic forms of question 5 and 6 are positive definite,
negative definite, indefinite, positive semi-definite or negative semi-definite.
8. Determine which of the following matrices are positive definite, negative definite,
indefinite, positive semi-definite or negative semi-definite:
 1 0 0
 25 0 0 
1 4
1 4 




(a) A  
(c) A   3 2 0  (d) A   0 2 0 
 (b) A  

4 0
 4 16 
 3 5 3 
 0 0 3




The norm A of a symmetric matrix A is given by the absolute maximum eigenvalue of
A. For non-symmetric matrices A the norm is given by the absolute singular value of the
matrix A.
Find the matrix norm A in each the above cases.
[Use Matlab to find the norm of part (c).]
9. Prove Proposition (7.43).
a b
10. Show that the symmetric matrix A  
 is positive definite  a  b  0 .
b a
11. Let I be a n by n identity matrix and x be n by 1 column vector. Show that the
quadratic form xT Ix satisfies the following:
2
(i) xT Ix  xT x  x
(ii) xT Ix is positive definite
12. Let A be a positive definite matrix. Prove that
(a) A m where m   is positive definite
(b) A 1 is positive definite
13. Let A be a negative definite matrix. Prove that A 2 is positive definite matrix.
14. Let A be a negative definite matrix of size n by n. Prove that
(a) det  A   0 provided n is odd
(b) tr  A   0 where tr is the trace of the matrix
15. Prove that if A is positive definite matrix then AT is also a positive definite matrix.
16. Prove that if A is an invertible matrix then AT A is a positive definite matrix. [Hint:
2
vT v  v ].
17. (i) If x  Qy where Q is an orthogonal matrix then show that y  x .
(ii) Let f  x   xT Ax be a quadratic form such that the eigenvalues of A satisfy
1  2  3    n . Prove that if x  1 then 1  f  x   n .
18. Prove that a symmetric matrix which is positive or negative definite is invertible.
Chapter 7: Eigenvalues and Eigenvectors
14
Brief Solutions to Exercise 7(f)
1. (a) x 2  y 2
(b) x12  x2 2 (c) x 2  4 xy  y 2
(d)  x 2  8 xy  7 y 2
1
1
(e) x 2  4 xy  y 2
(f)  x 2  2 xy  ey 2
2
3
2. (a) x 2  y 2  z 2
(b) 2 x 2  4 xy  4 xz  2 y 2  4 zy  2 z 2
(c)  x 2  4 xy  14 xz  5 y 2  16 zy  9 z 2
(d) 10 xy  14 xz  6 zy
 9 1  x 
 x
1 0 x 
3. (a)  x y  
(b)  x y  I  
(c)  x y  
 
 
 1 1  y 
 y
 0 1   y 
 31 41/ 2   x 
 1 1/ 2   x 
(d)  x y  
(e)  x y  
 
 
69   y 
0  y 
 41/ 2
1/ 2
 1 0   x 
 0 1/ 2   x 
(f)  x y  
(g)  x y  
 
   (h) Not in quadratic form.
0  y 
 0 0 y 
1/ 2
1

4. (a)  x y z   1
2

 3

(c)  x y z   3
 6

 1

(e)  x y z   0
 0

 0

(g)  x y z   0
1/ 2

1 2 x 
 
2 5 y 
5 5   z 
3 6   x 
 
7 9   y 
9 1   z 
0 0 x 
 
1 0   y 
0 1   z 
0 1/ 2   x 
 
0 1/ 2   y 
1/ 2
0   z 
5. (a) 3  x   2  y 
2
3
0  x 
 2

 
(b)  x y z   3 32 12   y 
 0 12
1   z 

2 x 
3 0

 
0 y 
(d)  x y z   0 7
 2 0 5   z 

 
 2 1/ 2 1/ 2   x 

 
0
0  y 
(f)  x y z   1/ 2
 1/ 2
0
0   z 

5 / 2 7 / 2   x 
 3

 
0
3 / 2   y 
(h)  x y z   5 / 2
 7 / 2 3 / 2
2   z 

(b)  x   3  y 
2
2
2
(c) 13  x   13  y 
2
2
1  5   x  1  5   y
(f)
2
(d) x  y
2
(e)   x    y 
2
2
6. (a) 5  x   5  y   5  z  
2
2
2
2
(b) 6  z  
2
2
2
2
(c) 6  x   3  y   3  z  
2
5
5
2
2
 y    z  
2
2
7. For question 5 we have (a) indefinite (b) positive definite (c) indefinite
(d) positive definite
(e) indefinite (f) indefinite
For question 6 we have (a) indefinite
(b) positive semi-definite
(c) indefinite
(d) negative definite
(e) indefinite
8. (a) Indefinite, A  4.5311 (b) Positive semi-definite, A  17
(d)  x 2  y 2  z 2
(c) Cannot find definiteness, A  7.25
(e)
(d) Indefinite, A  25
9. Very similar to the proof of Proposition (7.41).
2
Chapter 7: Eigenvalues and Eigenvectors
15
10. Show that the characteristic equation is given by  2  2a  a 2  b 2  0 then by using
the quadratic formula derive   a  b .
2
11. (i) Show that xT Ix  x12  x2 2  x32    xn 2  x .
(ii) Eigenvalues are equal to 1 therefore positive definite.
12. Use Proposition (7.8) (a) which says that the eigenvalues of A m are  m and (7.8) (b)
which says that the eigenvalues of A 1 are  1 .
13. Apply Proposition (7.8) (a) again with  being the eigenvalue of A then  2 is the
eigenvalue of A 2 .
14. Use Proposition (7.9) (a), (b) which is det  A   1    n and tr  A   1    n .
15. By question 16 of Exercise 7B the eigenvalues of A and AT are identical.
2
T
16. Show that xT  AT A  x   Ax   Ax   Ax  0 .
17. (i) Show the following y  y T y   QT x  QT x  xT x  x .
T
2
2
(ii) Use result of part (i) and y  y12  y2 2  y32    yn 2  1 also write f  x  in
diagonal form:
f  x   1 y12  2 y2 2  3 y32    n yn 2
2
Complete Solutions 7(f) 1
Complete Solutions 7(f)
1 0
 x
T
T
1. (a) Substituting A  
 , x    and x   x y  into x Ax gives
0
1
y


 
1 0 x 
xT Ax   x y  
 
0 1 y 
 x
  x y      x2  y 2   x2  y 2
 y
(b) Identical to part (a) with x and y replaced by x1 and x2 . Thus we have x12  x2 2 .
1 2
 x
T
T
(c) Substituting A  
 , x    and x   x y  into x Ax gives
2 1
 y
1 2 x 
xT Ax   x y  
 
 2 1 y 
 x
  x  2 y 2x  y   
 y
  x  2 y  x   2 x  y  y 
 x 2  2 xy  2 xy  y 2  x 2  4 xy  y 2
 1 4 
 x
(d) Clearly these are all very similar. Substituting A  
 , x    and
 4 7 
 y
xT   x
y  into xT Ax gives
 1 4   x 
y
 
 4 7   y 
 x
   x  4 y 4 x  7 y   
 y
xT Ax   x
    x  4 y  x   4 x  7 y  y 
  x 2  4 xy  4 xy  7 y 2   x 2  8 xy  7 y 2
(e) Similarly we have
1/ 2 2   x 
y
 
 2 1/ 3   y 
1  x 
1
  x  2 y 2 x  y   
3  y 
2
xT Ax   x
 1
1  


   x  2 y  x   2 x  y  y 
3  


2
1
1
1
1
 x 2  2 xy  2 xy  y 2  x 2  4 xy  y 2
2
3
2
3
(f) We have
Complete Solutions 7(f) 2
 1    x 
y
 
  e  y 
 x
   x   y  x  ey   
 y
xT Ax   x
   x   y  x   x  ey  y 
  x 2   xy   xy  ey 2   x 2  2 xy  ey 2
2. (a) This time we have a matrix of size 3 by 3.
1 0 0 x 

 
T
x Ax   x y z   0 1 0   y 
0 0 1 z 

 
 x
 
  x y z   y   x2  y 2  z 2
z
 
(b) This time we have 3 by 3 matrix A with all the entries being 2:
 2 2 2 x 

 
T
x Ax   x y z   2 2 2   y 
 2 2 2 z 

 
 x
 
  2x  2 y  2z 2x  2 y  2z 2x  2 y  2z   y 
z
 
  2 x  2 y  2 z  x   2 x  2 y  2 z  y   2 x  2 y  2 z  z 
 2 x 2  2 xy  2 xz  2 xy  2 y 2  2 zy  2 xz  2 yz  2 z 2
 2 x 2  4 xy  4 xz  2 y 2  4 zy  2 z 2
(c) In this case we can take out the minus sign and then evaluate xT Ax :
Complete Solutions 7(f) 3
 1

x Ax   x y z   2
 7

1

  x y z 2
7

T
2 7   x 
 
5  8   y 
8 9   z 
2 7 x 
 
5 8 y 
8 9   z 
 x
 
   x  2 y  7 z 2 x  5 y  8z 7 x  8 y  9 z   y 
z
 
   x  2 y  7 z  x   2 x  5 y  8 z  y   7 x  8 y  9 z  z 
   x 2  2 xy  7 xz  2 xy  5 y 2  8 zy  7 xz  8 yz  9 z 2 
   x 2  4 xy  14 xz  5 y 2  16 zy  9 z 2 
  x 2  4 xy  14 xz  5 y 2  16 zy  9 z 2
(d) Similar to the first three we have
0 5 7 x 

 
T
x Ax   x y z   5 0 3   y 
 7 3 0   z 

 
 x
 
  0 x  5 y  7 z 5 x  0 y  3z 7 x  3 y  0 z   y 
z
 
 5 y  7 z  x  5 x  3 z  y   7 x  3 y  z 
 5 xy  7 xz  5 xy  3 zy  7 xz  3 yz
 10 xy  14 xz  6 zy
3. In each case we need to find the symmetric matrix A.
(a) We are given 9 x 2  2 xy  y 2 and we need to write this as xT Ax . How?
First we determine the symmetric matrix A. What is this matrix A equal to?
Well the entries on the leading diagonal are 9 and 1 because these are the coefficients of x 2
and y 2 respectively. What are the other entries equal to?
 9 1
1 because we take half the xy coefficient which is 2. We have A  
.
 1 1
Thus writing the given quadratic form 9 x 2  2 xy  y 2 as xT Ax gives
 9 1  x 
9 x 2  2 xy  y 2   x y  
 
 1 1  y 
(b) We need to find the symmetric matrix A such that the given quadratic form x 2  y 2 can
be written as xT Ax . What are the entries of the matrix A?
Since we are given x 2  y 2 therefore the leading diagonal entries are 1, 1 and the remaining
entries are 0. Thus we have
Complete Solutions 7(f) 4
1 0 x 
 x
y
    x yI 
0 1 y 
 y
In this case the symmetric matrix A is the identity matrix I.
(c) This is very similar to part (b) because we are given x 2  y 2 which means that the
symmetric matrix A has the leading diagonal entries 1, 1 and the remaining entries 0:
1 0 x 
x2  y 2   x y  
 
 0 1   y 
x2  y 2   x
(d) We are given the quadratic form 31x 2  41xy  69 y 2 and we need to find the symmetric
matrix A such that 31x 2  41xy  69 y 2  xT Ax . What are the entries of the matrix A equal
to?
The leading diagonal entries are 31 and 69 respectively. The remaining entries are half the
41
xy coefficient which is 41 therefore  :
2
 31 41/ 2   x 
31x 2  41xy  69 y 2   x y  
 
69   y 
 41/ 2
(e) We are given the quadratic form x 2  xy . What are the entries of the symmetric matrix
A equal to?
The leading diagonal entries are 1 and 0 because there is no y 2 term. The remaining
entries are half the xy coefficient (1) which means they are ½. Thus we have
 1 1/ 2   x 
x 2  xy   x y  
 
0  y 
1/ 2
(f) We need to write  x 2 in xT Ax form. What are the entries of the matrix A equal to?
The leading diagonal entries are 1 and 0 because there is no y 2 term. Since there is no
xy coefficient therefore the remaining entries are 0. Thus we have
 1 0   x 
 x2   x y  
 
 0 0 y 
(g) How do we write the quadratic form xy as xT Ax ?
Since there are no x 2 and y 2 terms therefore both the leading diagonal entries are 0 and
remaining entries are half the xy coefficient (1) which gives ½.
 0 1/ 2   x 
xy   x y  
 
0  y 
1/ 2
(h) What do you notice about xy 2 ?
It is not a quadratic form because xy 2 is a cubic so we cannot write this in xT Ax form.
4. (a) We are given the quadratic form x 2  2 xy  4 xz  2 y 2  10 yz  5 z 2 which has the
three variables x, y and z. What is the size of the matrix A in xT Ax ?
We have three variables therefore we have a 3 by 3 matrix. What are the entries of this
matrix A equal to?
The leading diagonal entries are the coefficients of x 2 , y 2 and z 2 which are 1, 2 and 5
respectively. What are the other entries in the matrix A equal to?
Half the xy coefficient (2) which gives 1, half the xz coefficient (4) which gives 2 and half
the yz coefficient (10) which gives 5. Thus we have
Complete Solutions 7(f) 5
1 1 2 x 

 
x  2 xy  4 xz  2 y  10 yz  5 z   x y z   1 2 5   y 
 2 5 5 z 

 
2
2
2
(b) We are given the quadratic form 2 x  6 xy  32 y  24 yz  z which we need to
convert into the form xT Ax . What is the matrix A equal to?
The leading diagonal entries are the coefficients of x 2 , y 2 and z 2 which are 2 , 32 and
1 respectively. What are the other entries in the matrix A equal to?
Half the xy coefficient (6) which gives 3 and half the yz coefficient ( 24 ) which gives
12 . Note that there is no xz coefficient which means this entry is 0. Thus we have
3
0  x 
 2

 
2
2
2
2 x  6 xy  32 y  24 yz  z   x y z   3 32 12   y 
 0 12
1   z 

(c) We have 3 x 2  6 xy  12 xz  7 y 2  18 yz  z 2 and we need to convert this into xT Ax
form. How?
We find the symmetric matrix A. The leading diagonal entries are 3,  7 and 1 . The
remaining entries are half the cross-product terms. We have
 3 3 6   x 

 
2
2
2
3 x  6 xy  12 xz  7 y  18 yz  z   x y z   3 7 9   y 
 6 9 1   z 

 
2
2
2
(d) We are given the quadratic form 3 x  4 xz  7 y  5 z and we need to convert this into
xT Ax . What is the symmetric matrix A equal to?
The leading diagonal entries are 3, 7 and 5 . The only other non-zero entry in the matrix is
half the xz coefficient (4) which gives 2. We have
2 x 
3 0

 
2
2
2
3 x  4 xz  7 y  5 z   x y z   0 7
0 y 
 2 0 5   z 

 
2
2
2
(e) For the quadratic form  x  y  z the symmetric matrix A only has non-zero entries
on the leading diagonal. Why?
Because there no cross-product terms such as xy, xz and yz. What are the leading diagonal
entries?
All are 1 . We have
 1 0 0   x 

 
 x 2  y 2  z 2   x y z   0 1 0   y 
 0 0 1   z 

 
2
(f) We are given the quadratic form 2x  xy  xz . What are the entries of the symmetric
matrix A equal to?
The only non-zero entry on the leading diagonal is 2 (which is the first entry on the
diagonal) because we have no y 2 or z 2 . The other entries are given by finding half the xy,
xz and yz coefficients which are 1/ 2 , 1/ 2 and 0 respectively. We have
2
2
2
Complete Solutions 7(f) 6
 2 1/ 2 1/ 2   x 

 
2 x  xy  xz   x y z   1/ 2
0
0  y 
 1/ 2
0
0   z 

(g) We can write yz  xz  xz  yz . What are the entries of the symmetric matrix A equal
to?
We have a 3 by 3 matrix because we have 3 variables x, y, z . The leading diagonal
entries are all zero because we have no x 2 , y 2 or z 2 . The remaining entries are half xy, xz
and yz coefficients. These are 0, ½ and ½. We have
0 1/ 2   x 
 0

 
yz  xz  xz  yz   x y z   0
0 1/ 2   y 
1/ 2 1/ 2
0   z 

(h) What is the symmetric matrix A for the given quadratic form
3 x 2  5 xy  7 xz  3 yz  2 z 2 ?
The leading diagonal entries are 3, 0 and 2 respectively. The remaining entries are half xy,
xz and yz coefficients. These are 5 / 2 , 7 / 2 and 3 / 2 respectively. We have
5 / 2 7 / 2   x 
 3

 
3 x 2  5 xy  7 xz  3 yz  2 z 2   x y z   5 / 2
0
3 / 2   y 
 7 / 2 3 / 2
2   z 

2
5. (a) We are given the quadratic form 2 x 2  4 xy  y 2 . First we find the symmetric matrix
A which represents this quadratic form in xT Ax . What is A equal to?
The leading diagonal entries are 2 and 1. The remaining entry is half the xy coefficient
 2 2 
(4) which is 2. The symmetric matrix A  
 and
 2 1
 2 2   x 
xT Ax   x y  
 
 2 1 y 
How do we write the given quadratic form 2 x 2  4 xy  y 2 into one without any crossproduct terms such as xy?
We apply the Principles Axes Theorem (7.39) which says
xT Ax  1 y12  2 y2 2    n yn 2
where  ’s are the eigenvalues of the matrix A. In this case we transform from the variables
x and y to x and y respectively. Applying the above theorem (7.39) we have
2 x 2  4 xy  y 2  xT Ax  1  x   2  y 
2
2
where 1 and 2 are the eigenvalues of the matrix A.
What are the eigenvalues of matrix A?
You can verify in your own time that 1  3 and 2  2 . Therefore substituting these
1  3 and 2  2 into 2 x 2  4 xy  y 2  1  x   2  y  gives
2
2
2 x 2  4 xy  y 2   3 x   2  y 
2
 3  x   2  y 
2
2
2
Complete Solutions 7(f) 7
(b) We are given the quadratic form 2 x 2  2 xy  2 y 2 . First we find the symmetric matrix A
which represents this quadratic form in xT Ax . What is A equal to?
The leading diagonal entries are 2 and 2. The remaining entry is half the xy coefficient
 2 1 
( 2 ) which is 1 . The symmetric matrix A  
 and
 1 2 
 2 1   x 
2 x 2  2 xy  2 y 2  xT Ax   x y  
 
 1 2   y 
How do we write the given quadratic form 2 x 2  2 xy  2 y 2 into one without any crossproduct terms such as xy?
We apply the Principles Axes Theorem (7.39) which in this case we transform from the
variables x and y to x and y respectively. Applying (7.39) we have
2 x 2  2 xy  2 y 2  xT Ax  1  x   2  y 
What are the eigenvalues of matrix A?
You can verify in your own time that 1  1 and 2  3 . Therefore substituting these
2
2
1  1 and 2  3 into 2 x 2  2 xy  2 y 2  1  x   2  y  gives
2
2
2 x 2  2 xy  2 y 2  1 x   3  y 
2
  x   3  y  
2
2
2
(c) We are given the quadratic form 5 x 2  24 xy  5 y 2 . What is the symmetric matrix A
equal to?
The leading diagonal entries are 5 and 5 . The remaining entry is half the xy coefficient
12 
5
(24) which is 12. The symmetric matrix A  
 and
12 5 
12   x 
5
5 x 2  24 xy  5 y 2  xT Ax   x y  
 
12 5   y 
How do we write the given quadratic form 5 x 2  24 xy  5 y 2 into one without any crossproduct terms such as xy?
We apply the Principles Axes Theorem (7.39) which in this case we transform from the
variables x and y to x and y respectively. Applying (7.39) we have
5 x 2  24 xy  5 y 2  xT Ax  1  x   2  y 
What are the eigenvalues of matrix A?
You can verify in your own time that 1  13 and 2  13 . Therefore substituting these
2
2
1  13 and 2  13 into 5 x 2  24 xy  5 y 2  1  x   2  y  gives
2
2
5 x 2  24 xy  5 y 2  13  x   13  y 
2
2
(d) We are given the quadratic form x 2  y 2 . What do you notice about this x 2  y 2 ?
It does not contain any cross-product terms which means it is already in diagonal form,
thus x 2  y 2  x 2  y 2 .
(e) We are given the quadratic form 2xy . What is the symmetric matrix A equal to?
Complete Solutions 7(f) 8
The leading diagonal entries are 0 and 0. The remaining entry is half the xy coefficient
 0 1 
( 2 ) which is 1 . The symmetric matrix A  
 and
 1 0 
 0 1   x 
2 xy  xT Ax   x y  
 
 1 0   y 
How do we write the given quadratic form 2xy into one without any cross-product terms
such as xy?
We apply the Principles Axes Theorem (7.39) which in this case we transform from the
variables x and y to x and y respectively. Applying (7.39) we have
2 xy  xT Ax  1  x   2  y 
What are the eigenvalues of matrix A?
You can verify in your own time that 1  1 and 2  1 . Therefore substituting these
2
2
1  1 and 2  1 into 2xy  1  x   2  y  gives
2
2
2 xy  1 x   1 y     x    y 
2
2
2
2
(f) We are given the quadratic form x 2  2 xy . What is the symmetric matrix A equal to?
The leading diagonal entries are 1 and 0. The remaining entry is half the xy coefficient (2)
1 1 
which is 1. The symmetric matrix A  
 (this matrix is also known as the Fibonacci
1 0 
matrix) and
1 1   x 
x 2  2 xy  xT Ax   x y  
 
1 0   y 
How do we write the given quadratic form x 2  2 xy into one without any cross-product
terms such as xy?
We apply the Principles Axes Theorem (7.39) which in this case we transform from the
variables x and y to x and y respectively. Applying (7.39) we have
x 2  2 xy  xT Ax  1  x   2  y 
What are the eigenvalues of matrix A?
1 5
1 5
You can verify in your own time that 1 
. Therefore substituting
and 2 
2
2
1 5
1 5
2
2
these 1 
into x 2  2 xy  1  x   2  y  gives
and 2 
2
2
 1 5 
 1 5 
2
2
x 2  2 xy  
  x   
  y 
 2 
 2 
2
2
1  5   x  1  5   y

2
2
2
6. (a) In this case we are given the quadratic form 3 x 2  8 xy  3 y 2  5 z 2 which means it is
in three variables x, y and z. Hence we need a 3 by 3 matrix A. What is the symmetric
matrix A equal to?
Complete Solutions 7(f) 9
The leading diagonal entries are 3, 3 and 5 . The only other non-zero entry in the matrix
is half the xy coefficient (8) which is 4. Thus we have
 3 4 0 x 

 
2
2
2
T
3 x  8 xy  3 y  5 z  x Ax   x y z   4 3 0   y 
 0 0 5 z 

 
2
2
2
We need to write the given quadratic form 3 x  8 xy  3 y  5 z without any cross-product
terms such as xy, xz and yz. How?
We apply the Principles Axes Theorem (7.39) which in this case we transform from the
variables x, y and z to x , y and z  respectively. Applying (7.39) we have
3 x 2  8 xy  3 y 2  5 z 2  xT Ax  1  x   2  y   3  z  
2
2
2
where 1 , 2 and 3 are the eigenvalues of the matrix A. What are the eigenvalues of A?
You can verify in your own time that 1  5, 2  5 and 3  5 . Substituting these into
the above gives
2
2
2
3 x 2  8 xy  3 y 2  5 z 2  5  x   5  y   5  z  
(b) We are given the quadratic form 2 x 2  4 xy  4 xz  2 y 2  4 yz  2 z 2 . What is the
symmetric matrix A equal to?
The leading diagonal entries are 2, 2 and 2. The remaining entries in the matrix are half the
xy, xz and yz coefficients which are all 4 therefore the matrix entries are 2. Thus we have
 2 2 2 x 

 
2
2
2
T
2 x  4 xy  4 xz  2 y  4 yz  2 z  x Ax   x y z   2 2 2   y 
 2 2 2 z 

 
2
2
We need to write the given quadratic form 2 x  4 xy  4 xz  2 y  4 yz  2 z 2 without any
cross-product terms such as xy, xz and yz. How?
We apply the Principles Axes Theorem (7.39) which in this case we transform from the
variables x, y and z to x , y and z  respectively. Applying (7.39) we have
2 x 2  4 xy  4 xz  2 y 2  4 yz  2 z 2  xT Ax  1  x   2  y   3  z  
2
2
2
where 1 , 2 and 3 are the eigenvalues of the matrix A. What are the eigenvalues of A?
You can verify in your own time that 1  0, 2  0 and 3  6 . Substituting these into the
above gives
2
2
2
2 x 2  4 xy  4 xz  2 y 2  4 yz  2 z 2  0  x   0  y   6  z  
 6  z
2
(c) We are given the quadratic form 2 x 2  4 xy  4 xz  y 2  8 yz  z 2 . What is the symmetric
matrix A equal to?
The leading diagonal entries are 2, 1 and 1 . The remaining entries in the matrix are half
the xy, xz and yz coefficients which are 4, 4 and 8 respectively. Therefore the entries on
the non leading diagonal are 2, 2 and 4 . We have
 2 2 2   x 

 
2 x 2  4 xy  4 xz  y 2  8 yz  z 2  xT Ax   x y z   2 1 4   y 
 2
4 1   z 

Complete Solutions 7(f) 10
We need to write the given quadratic form 2 x 2  4 xy  4 xz  y 2  8 yz  z 2 without any
cross-product terms such as xy, xz and yz. How?
We apply the Principles Axes Theorem (7.39) which in this case we transform from the
variables x, y and z to x , y and z  respectively. Applying (7.39) we have
2 x 2  4 xy  4 xz  y 2  8 yz  z 2  xT Ax  1  x   2  y   3  z  
2
2
2
where 1 , 2 and 3 are the eigenvalues of the matrix A. What are the eigenvalues of A?
You can verify in your own time that 1  6, 2  3 and 3  3 . Substituting these into
the above gives
2
2
2
2 x 2  4 xy  4 xz  y 2  8 yz  z 2  6  x   3  y   3  z  
(d) What do you notice about the given quadratic form  x 2  y 2  z 2 ?
It is already in diagonal form, that is there are no cross product terms which means that
 x2  y 2  z 2   x2  y 2  z 2
(e) First we need to find the symmetric matrix A such that  x 2  2 y 2  2 yz  2 z 2  xT Ax .
What is the matrix A equal to in this case?
The entries on the leading diagonal are the coefficients of x 2 , y 2 and z 2 which are 1 2
and 2 . The only other non-zero entry is half the yz coefficient (2) which is 1. We have
 1 0 0 


2
2
2
T
 x  2 y  2 yz  2 z  x Ax where A   0 2 1 
 0 1 2 


We need to find the eigenvalues of matrix A. Verify that the eigenvalues of matrix A are
1  3, 2  1 and 3  1 .
Let the new quadratic form be in the variables x, y and z  . Therefore using the above
Principle Axes Theorem (7.39) we have
2
2
2
 x 2  2 y 2  2 yz  2 z 2  1  x   2  y   3  z  
 3  x    1 y    1 z  
2
2
 3  x     y     z  
2
2
2
2
(f) We are given the quadratic form 2xy  xz . How do we convert this into diagonal
form?
First we find the symmetric matrix A. What are the diagonal entries in the matrix A equal
to?
There are no x 2 , y 2 or z 2 terms therefore the leading diagonal entries are all zero. What
are the remaining entries in the matrix A equal to?
There are only two non-zero entries which are half the xy and xz coefficients ( 2 and 1
respectively) which are 1 and 1/2. Therefore we have
 0 1 1/ 2   x 

 
T
2 xy  xz  x Ax   x y z   1
0 0  y 
1/ 2 0 0   z 

 
Complete Solutions 7(f) 11
 0 1 1/ 2 


0 0 ?
What are the eigenvalues of the matrix A   1
1/ 2 0 0 


5
5
. Hence we have
and 3  
2
2
2
2
2
2 xy  xT Ax  1  x   2  y   3  z  
You can verify that 1  0, 2 
 0  x  
2
5
5
5
5
2
2
2
2
 y    z     y    z  
2
2
2
2
7. For question 5 we have the following diagonal forms:
2
2
(a) xT Ax  3  x   2  y  can take on both positive and negative values therefore the
quadratic form is indefinite.
2
2
(b) xT Ax   x   3  y  is a quadratic form which can only take on positive values for all
x  O , therefore it is positive definite.
(c) Clearly the quadratic form xT Ax  13  x   13  y  can take on both positive and
negative values therefore it is indefinite quadratic form.
(d) The quadratic form xT Ax  x 2  y 2 can only take on positive values therefore it is
positive definite.
2
2
(e) xT Ax    x    y  is a quadratic form which can take on both positive and negative
values therefore it is an indefinite quadratic form.
2
2
1  5  x   1  5  y  
T
(f) The quadratic form x Ax 
can take on both positive and
2
negative values therefore it is indefinite.
For question 6 we have the following diagonal forms:
2
2
2
(a) 5  x   5  y   5  z   is a quadratic form which can take on both positive and
negative values therefore it is indefinite.
2
(b) The quadratic form 6  z   can have positive or zero values therefore it is positive
semi-definite.
2
2
2
(c) The quadratic form 6  x   3  y   3  z   can take on both positive and negative
values therefore it is indefinite quadratic form.
(d)  x 2  y 2  z 2 can only take on negative values for all x  O therefore the quadratic
form is negative definite.
5
5
2
2
(e) The quadratic form
 y    z  can take on both positive and negative values
2
2
therefore it is an indefinite quadratic form.
2


2


1 4
8. (a) The eigenvalues of A  
 are 1  3.5311,
4 0
indefinite.
2  4.5311 so the matrix A is
Complete Solutions 7(f) 12
1 4 
(b) Since the bottom row is a multiple (4) of the first row in matrix A  
 so two
 4 16 
rows of A are linearly dependent so 0 must be an eigenvalue of matrix A. The other
eigenvalue is 17 so matrix A is positive semi-definite.
 1 0 0


(c) The given matrix A   3 2 0  is not a symmetric matrix so we cannot find
 3 5 3 


whether it is positive, negative or indefinite matrix.
 25 0 0 


(d) The eigenvalues of the given matrix A   0 2 0  are 25, 2 and 3 so the matrix
 0 0 3


is indefinite.
The matrix norm is the maximum absolute value:
(a) A  4.5311 (b) A  17
(d) A  25
(c) The given matrix is not symmetric. We need to find AT A :
9
 19 21


T
A A   21 29 15 
 9 15
9 

The eigenvalues of this matrix AT A are 1  0.1586, 2  4.3219 and 3  52.5195 .
Hence
A  52.5195  7.25
9. The proof is very similar to the proof of Proposition (7.41).
Proposition (7.43). A symmetric matrix A is negative definite  all the eigenvalues of A
are negative.
How do we prove this result?
Since we have an if and only if (  ) therefore we need to prove the result both ways 
and  .
Proof.
(  ). Let x be an arbitrary non-zero vector and all the eigenvalues  of matrix A be
negative. Let y be the vector of new variables which are given by y  QT x where Q is an
orthogonal matrix such that QT AQ  D . Then by (7.39) we have
xT Ax  y T Dy
2
2
2
2

 1 y1  2 y2  3 y3    n yn
By (7.31)
0
 Because all the  's are negative
Thus matrix A is negative definite.
(  ). Assume matrix A is negative definite which means for all non-zero vectors x
xT Ax  1 y12  2 y2 2  3 y32    n yn 2  0
Complete Solutions 7(f) 13
 v1 
 
v
Suppose k  0 [Positive or zero] for some k between 1 and n. Let x  v k   2  where

 
 vn 
v k is an eigenvector belonging to the eigenvalue k of the matrix A. Then
xT Ax   v k  Av k
T
 Because v k
  v k  k v k
T
is e.vector Av k  k v k 
 k  v k  v k
T
 k  v1 v2
 v1 
 
v
 vn   2 

 
 vn 
 k v12  v2 2  v32    vn 2 
Since xT Ax  0 [Negative] for all non-zero vectors x therefore
T
xT Ax   v k  Av k
 k v12  v2 2  v32    vn 2   0  Negative
This gives k  0 . We have a contradiction because k  0 and k  0 . This means our
supposition k  0 is incorrect so k  0 . Hence all our eigenvalues are negative.
■
a b
10. Need to prove that A  
 is positive definite if an only if a  b  0 .
b a
a b
Proof. To prove that A  
 is positive definite we need to show that all the
b a
eigenvalues of A are positive. The eigenvalues  are given by
b 
a 
det  A   I   det 

a
 b
  a     b2
2
  2  2a  a 2  b 2  0
We can solve the quadratic  2  2a  a 2  b 2  0 by using the formula

2a  4a 2  4  a 2  b 2 
2
 ab
This gives the eigenvalues 1  a  b and 2  a  b . If a  b  0 then 1  0 and 2  0
which means that all the eigenvalues are positive.
Thus A is positive definite  a  b  0 .
■
Complete Solutions 7(f) 14
11. We need to show that the quadratic form (i) xT Ix  x
definite.
Proof.
 x1 
 
(i) Let x     then we have
x 
 n
2
and (ii) xT Ix is positive
1
0   x1 

 
x Ix   x1  xn  

  

 
1   xn 
0
 x1 
 
  x1  xn    
x 
 n
T
 x12  x2 2  x32    xn 2  x
2
■
Proof of (ii).
Since all the eigenvalues of the identity matrix I are 1 therefore the matrix I is positive
definite so the quadratic form xT Ix is positive definite.
■
12. We need to prove that if A is a positive definite matrix then
(a) A m where m   is positive definite.
(b) A 1 is positive definite
Proof of (a).
Since the given matrix A is positive definite therefore by Proposition (7.41) all the
eigenvalues  of A are positive. By Proposition (7.8) (a):
(7.8) (a) The eigenvalues of A m are  m
We have the eigenvalues of A m are  m which means that these are also positive therefore
the matrix A m is positive definite.
■
Proof of (b).
Since the given matrix A is positive definite therefore by Proposition (7.41) all the
eigenvalues  of A are positive. By Proposition (7.8) (b):
(7.8) (b) The eigenvalues of A 1 are  1
1
We have the eigenvalues of A 1 are
which means that these are also positive therefore

the matrix A 1 is positive definite.
■
2
13. Required to prove that if A is a negative definite matrix then A is positive definite
matrix.
Proof.
Complete Solutions 7(f) 15
Since the given matrix A is negative definite therefore by Proposition (7.43) all the
eigenvalues  of A are negative. The eigenvalues of A 2 are  2 which means all the
eigenvalues are positive. Hence A 2 is positive definite matrix.
■
14. We need to prove that if A is a negative definite matrix then
(a) det  A   0
(b) tr  A   0 where tr is the trace of the matrix
Proof of (a).
Since the given matrix A is negative definite therefore by Proposition (7.43):
(7.43) Symmetric matrix A is negative definite  all the eigenvalues of A are negative.
All the eigenvalues 1 , 2 , 3 ,  and n of A are negative. By Proposition (7.9) (a):
(7.9) (a) det  A   1  2  3    n
The determinant is given by
det  A   1  2  3    n


Odd n
Since n is odd therefore we have a multiple of odd negative numbers which means that it is
negative so we have our result det  A   0 .
■
Proof of (b).
Since the given matrix A is negative definite therefore by Proposition (7.43):
(7.43) Symmetric matrix A is negative definite  all the eigenvalues of A are negative.
All the eigenvalues 1 , 2 , 3 ,  and n of A are negative. By Proposition (7.9) (b):
(7.9) (b) tr  A   1  2  3    n
The trace of the matrix A is given by
tr  A   1  2  3    n
Since all the  ’s are negative therefore tr  A   0 .
■
15. We need to prove that if A is positive definite matrix then AT is also a positive definite
matrix.
Proof.
Let  be the eigenvalues of the matrix A. Since A is positive definite therefore all the
eigenvalues of A are positive. By question 16 of Exercise 7(b) we know that AT has the
same eigenvalues  therefore all the eigenvalues of AT are positive so AT is a positive
definite matrix.
■
T
16. We need to prove that A A is positive definite provided that A is an invertible matrix.
Proof.
First we need to show that AT A is a symmetric matrix.
A A
T
T
 AT  AT   AT A
T
Thus AT A is a symmetric matrix. Next we examine the quadratic form xT  AT A  x :
xT  AT A  x   Ax 
T
 Ax 
 Ax  0
2
[By Hint]
Complete Solutions 7(f) 16
Thus Ax  0 . Consider the case Ax  0 , then Ax  O and we are given that A is
2
2
invertible so multiplying both sides of Ax  O by A 1 gives x  O . Hence for x  O
2
2
Ax  0 which means that xT  AT A  x  Ax  0 therefore the matrix AT A is positive
definite.
17. (i) We are required to prove that if x  Qy where Q is an orthogonal matrix then
■
y  x .
Proof.
2
Since Q is an orthogonal matrix therefore Q 1  QT . We have y  y T y . What is y equal
to?
We are given that x  Qy so multiplying both sides of this x  Qy by QT
(because Q 1  QT ) gives
T
QT x  Q
Q y  y or y  QT x

I
Substituting this y  Q x into the above y  y T y yields
2
T
y  y T y   QT x  QT x
T
2
T
 xT QQ
x
I
 x Ix  xT x  x
T
We have y  x
2
2
2
and taking the square root of both sides of this gives our result
y  x .
■
(ii) We need to prove if the eigenvalues of A satisfy 1  2  3    n and x  1 then
1  f  x   n .
Proof.
2
2
By the above part (i) we have y  x  1 . This means that y  x  1 and
y  y12  y2 2  y32    yn 2  1
2
(*)
The quadratic form f  x   x Ax in diagonal form is given by
T
f  x   xT Ax  1 y12  2 y2 2  3 y32    n yn 2
(**)
Using the given inequality 1  2  3    n we have
1 1  1  y12  y2 2  y32    yn 2 

1 by (*)
  y  1 y2  1 y32    1 yn 2
2
1 1
2
 1 y12  2 y2 2  3 y32    n yn 2
 f x
 Because 1  2  3    n 
 By (**)
Thus we have our first inequality 1  f  x  . Using (**) we have
Complete Solutions 7(f) 17
f  x   1 y12  2 y2 2  3 y32    n yn 2
 n y12  n y2 2  n y32    n yn 2
 n  y12  y2 2  y32    yn 2   n

 Because 1  2  3    n 
1 by (*)
Hence we have our second inequality, that is f  x   n . Putting the two inequalities
together we have our result 1  f  x   n .
■
18. Need to prove that a symmetric matrix which is positive or negative definite is
invertible.
Proof.
Without loss of generality assume a symmetric matrix A is positive definite. By
Proposition (7.41). A symmetric matrix A is positive definite  all the eigenvalues of A
are positive.
All the eigenvalues of matrix A are positive. By
Proposition (7.7). A square matrix A is invertible (has an inverse)    0 is not an
eigenvalue of the matrix A.
Since   0 is not an eigenvalue of matrix A because all the eigenvalues are positive so
matrix A is invertible.
■