11.2MH1 LINEAR ALGEBRA
EXAMPLES 8: EIGENVALUES AND EIGENVECTORS – SOLUTIONS
/
0.
,
1. 0 E
so E
Suppose u v E
Then A u v
.
Au Av
u
v
u v
E
Au
u
u
so
u v
Also A u
since u v
E
so
u
E
Hence E
is a subspace of n .
is a eigenvalue of A if and only if there exists x 0 such that Ax
x i.e. if and only if
E
0 . Hence, if is not an eigenvalue of A, E
0 and so dimE
0.
2
2.
2
(a) The characteristic equation is det
1 1
1
3
1
3
0.
1
2
ie 2
4 2
2 32
0
3 2 2 5
ie
6
0
2
ie
1
6
0
ie
1
3
2
0
Hence the eigenvalues are
2,
1 and
3.
x y z is an eigenvector corresponding to
2 if and only if
2x 2y 3z
2x i.e. 4x 2y 3z
0
x y z
2y
x 3y z
0
x 3y z
2z
x 3y z
0
has the same solutions as the systems corresponding to the following:
1
4
3 1 0
2 3 0
r2
r2
4r1
i.e. eigenvectors x y z satisfy x
Let z
So
span
1
14
then y
11
14
1
14
x
y
z
and x
1
0
3
14
1 0
1 0
3y
14y
z
z
0
0
11
14
.
and hence the eigenspace corresponding to
1
11
1
14
.
x y z is an eigenvector corresponding to
2x 2y 3z
x i.e. x 2y 3z
x y z
y
x z
x 3y z
z
x 3y 2z
1
1 if and only if
0
0
0
2 is E
2
has the same solutions as the systems corresponding to the following:
1
1
1
2
0
3
3 0
1 0
2 0
r2
r3
r2
r3
1
0
0
r1
r1
i.e. eigenvectors x y z satisfy x
Let z
then y
and x
x
y
z
So
1
1
1
1
1
1
span
2y
2
2
5
3 0
2 0
5 0
3z
y z
r3
2r3
5r2
1
0
0
2
2
0
3 0
2 0
0 0
0
0
.
and hence the eigenspace corresponding to
1 is E 1
.
x y z is an eigenvector corresponding to
3 if and only if
2x 2y 3z
3x i.e.
x 2y 3z
0
x y z
3y
x 2y z
0
x 3y z
3z
x 3y 4z
0
has the same solutions as the systems corresponding to the following:
1
1
1
2
2
3
3 0
1 0
4 0
r2
r3
r2
r3
i.e. eigenvectors x y z satisfy
Let z
So
then y
x
y
z
11
1
14
and x
1
1
1
1
1
1
1
0
0
r1
r1
x
2y
2
4
1
3 0
4 0
1 0
3z
y z
0
0
.
and hence the eigenspace corresponding to
1
1
1
3
is a basis for
7
4 7
4
1
1
0.
4 4
ie 7
7
4
4 44 4
4 1 16 4 7
3 18 2 81
ie
108
ie
3 2 15
36
ie
3
3
12
Hence the eigenvalues are
3 and
12.
x y z is an eigenvector corresponding to
3 if and only if
7x 4y z
3x i.e.
4x 4y z
0
4x 7y z
3y
4x 4y z
0
4x 4y 4z
3z
4x 4y z
0
So x y z is an eigenvector if 4x 4y z 0.
1
Let z
,y
then x
4 .
2
span
consisting of eigenvectors of the matrix.
4
(b) The characteristic equation is det
3 is E 3
0
0
0
0
1
1
1
.
1
4
x
y
z
So
1
4
1
1
0
0
1
and hence the eigenspace corresponding to
3 is E 3
span
1
0
4
1
1
0
.
x y z is an eigenvector corresponding to
12 if and only if
7x 4y z
12x i.e.
5x 4y z
0
4x 7y z
12y
4x 5y z
0
4x 4y 4z
12z
4x 4y 8z
0
has the same solutions as the systems corresponding to the following:
1
5
4
1
4
5
2 0
1 0
1 0
r2
r3
r2
r3
5r1
4r1
i.e. eigenvectors x y z satisfy
x
Let z
.
then y
and x
1
0
0
y
1
9
9
2z
y z
0
0
Hence the eigenvectors corresponding to
1
1
1
span
1
0
4
2 0
9 0
9 0
x
y
z
12 are
1
1
1
.
1
1
0
1
1
1
is a basis for
2
3
consisting of eigenvectors of the matrix.
2
(c) The characteristic equation is det
10
5
4
3
5
0.
4 6
ie 2
4
6
20 2 10 6
25 3 40 5 4
3 4 2 5
ie
2
2
ie
1
3
2
2
ie
1
2
Hence the eigenvalues are
1 and
2.
x y z is an eigenvector corresponding to
1 if and only if
2x 2y 3z
x i.e.
x 2y 3z
0
10x 4y 5z
y
10x 5y 5z
0
5x 4y 6z
z
5x 4y 5z
0
has the same solutions as the systems corresponding to the following:
1
2
5
2 3 0
1 1 0
4 5 0
r2
r3
r2
r3
1
0
0
2r1
5r1
i.e. eigenvectors x y z satisfy x
Let z
i.e. E 12
then y
5
3
and x
1
3
2y
3y
2
3
6
3z
5z
.
3
3 0
5 0
10 0
0
0
0
0
0
0
Hence the eigenvectors corresponding to
1
3
5
3
x
y
z
1 are
i.e. E 1
span
1
5
3
span
4
15
10
1
.
x y z is an eigenvector corresponding to
2 if and only if
2x 2y 3z
2x i.e.
2y 3z
0
10x 4y 5z
2y
10x 6y 5z
0
5x 4y 6z
2z
5x 4y 4z
0
has the same solutions as the systems corresponding to the following:
5
10
0
4 4 0
6 5 0
2 3 0
r2
r2
5
0
0
2r1
4
2
2
x y z is an eigenvector if and only if 5x
Let z
then y
3
2
2
5
and x
3. Suppose A
dimE 2
2
d1 0 0
0 d2 0
0 0 d3
..
..
.
.
0 0 0
4y
2y
4z
3z
.
2 are
3
i.e. E 2
1
consisting of eigenvectors of the ma-
0
0
0
..
.
dn
d1
0
0
0
..
.
The characteristic equation is det
0
0
d2
0
..
.
0
i.e. d1
d2
dn
So the eigenvalues are d1 d2
2
5
3
2
x
y
z
3 so there is no basis for
..
0
0
.
Hence the eigenvectors corresponding to
dimE 1
trix.
4 0
3 0
3 0
0
0
0
..
.
d3
..
0
0
.
0
dn
0
dn .
4. Let be an eigenvalue of A and x an eigenvector corresponding to . i.e. Ax
x
Hence
xt Ax
xt x
where x denotes the complex conjugate of x
Now
xt Ax
xt x
xt x
so
xt Ax
xt x
A A since A is real
t
t
t
t
so
x Ax
xx
t
t
ie
xAx
xt x
xt x C
Hence
xt Ax
xt x
At
A
t
t
t
So from
xx
x x i.e.
x x 0. Thus
and so is purely imaginary.
1
5. The characteristic equation is det
1
1
5
1
2 3
4
1
2
0.
.
2 2
ie 1
1 1
7 1 3 5
0
3
2
ie
3
3
0
2
ie
1
4
3
0
ie
1
1
3
0
Hence the eigenvalues are
1,
1 and
3.
x y z is an eigenvector corresponding to
1 if and only if
x y z
x i.e.
2x y z
0
x y 2z
y
x 2z
0
5x 2y 3z
z
5x 2y 4z
0
has the same solutions as the systems corresponding to the following:
1
2
5
0
1
2
2 0
1 0
4 0
r2
r3
r2
r3
2r1
5r1
i.e. eigenvectors x y z satisfy
x
y
1
0
0
2z
3z
0
1
2
2 0
3 0
6 0
0
0
Let z
then y
3 and x 2 .
So 2 3 1 is an eigenvector corresponding to
1.
x y z is an eigenvector corresponding to
1 if and only if
x y z
x i.e.
y z
0 1
x y 2z
y
x 2y 2z
0 2
5x 2y 3z
z
5x 2y 2z
0 3
2
3 gives 4x 0 so x 0 and 1 gives y z.
So 0 1 1 is an eigenvector corresponding to
1.
x y z is an eigenvector corresponding to
3 if and only if
x y z
3x i.e.
2x y z
0
x y 2z
3y
x 4y 2z
0
5x 2y 3z
3z
5x 2y
0
has the same solutions as the systems corresponding to the following:
1
2
5
4
1
2
2 0
1 0
0 0
r2
r3
r2
r3
2r1
5r1
i.e. eigenvectors x y z satisfy
x
1
0
0
4y
9y
4
9
18
2z
5z
0
0
2
Let z
then y 59 and x
9 .
So 2 5 9 is an eigenvector corresponding to
3.
2 0
3 1
1 1
So, by the diagonalisation theorem, if P
P 1 AP
2 0
5 0
10 0
2
5
9
and D
D.
6. A has characteristic equation
1
4
3
ie
4
Hence, by the Cayley-Hamilton theorem A4
(a) Rearranging
we get
so A
1
4
2
1
2
2
4
4
2
3A 4A
3
4A3
A4 4A3 3A2 4A
A3 4A2 3A 4I
5
0
0
4I
0
4I
I
1 0 0
0 1 0
0 0 3
then
So A
1
4
1
A3
4A2
3A
4I .
A4
A5
(b) From
so
4A3 3A2 4A 4I
4A4 3A3 4A2 4A
4 4A3 3A2 4A 4I 3A3 4A2 4A
13A3 16A2 12A 16I
13A4 16A3 12A2 16A
13 4A3 3A2 4A 4I 16A3 12A2 16A
36A3 51A2 36A 52I
and so A6
7.
is an eigenvalue of A
i.e. iff
3
3
det
3
2
8
1
2
4
0
3
0
0
1
x
y
3x
2x
is an eigenvector corresponding to
4y
3y
i.e. iff x
x
y
3x
2x
x
y
i.e.
2x
2x
4y
4y
1 iff
0
0
2y. Hence eigenvectors are multiples of
is an eigenvector corresponding to
4y
3y
i.e. iff x
x
y
i.e.
4x
2x
4y
2y
0
0
Now, by the diagonalisation theorem, if we let P
PDP 1
PD20 P
P
PIP
I
120
0
.
1 iff
y. Hence eigenvectors are multiples of
D.
So
A
20
Hence A
2
1
1
1
2 1
1 1
and D
1
0
1
0
1
20
P
1
1
8. Suppose is an eigenvalue of A. Then x 0 such that Ax
2x
Now A2 x A Ax
A x
Ax
k
k
1
k
1
kx
Similarly, A x A
Ax
A x
But Ak 0 so k x 0 hence k 0 (since x 0). So
0.
Hence all the eigenvalues of A are zero.
x.
9. By the diagonalisation theorem there exists an invertible matrix P such that
6
0
1
then P 1 AP
0
1
P 1 AP
2
D where D
..
.
.
0
So A PDP
Hence detA
n
1
det PDP 1
detPdetDdetP 1
detPdetP 1 detD
det PP 1 detD
detIdetD
detD
0
1
..
det
.
0
1 2
n
n
10. Let xk population of London in k years time.
Let yk population outside London in k years time.
(x0 and y0 are the populations now).
The population sizes after k 1 years are determined by the populations after k years as follows:
i.e. xk
1
xk
yk
Axk where xk
xk
1
0 9xk
0 05yk
yk
1
0 1xk
0 95yk
0 9 0 05
0 1 0 95
and A
Hence x1 Ax0
x2 Ax1 A2 x0
x3 Ax2 A3 x0
and in general xk Ak x0
We investigate Ak x0 using the eigenvalues and vectors of A.
09
01
is an eigenvalue of A iff det
0 05
0 95
2 1 85
ie
0 85
0
ie
0 85
1
0
So the eigenvalues are
0 85 and
x
y
0 05y
0 95y
i.e. iff x
y
x
y
0 9x
0 1x
1.
is an eigenvalue corresponding to
0 9x
0 1x
0 85x
0 85y
i.e.
1
1
0. Hence
0 85 iff
0 05x 0 05y
0 1x 0 1y
x
y
i.e.
0
0
is an eigenvalue corresponding to
is an eigenvalue corresponding to
0 05y
0 95y
0
0 1x
0 1x
1 iff
0 05y
0 05y
7
0
0
0 85.
i.e. iff
2x
Let P
Hence Ak x0
y
0. Hence
1 1
1 2
. Now A
1
2
is an eigenvalue corresponding to
PDP
1
0 85 0
0 1
where D
1.
.
PDP 1 k x0
PDk P 1 x0
1
3
1
3
1 1
0 85 k 0
1
3
1 2
0
1k
0 85 k 1
2x0 y0
0 85 k 2
x0 y0
k
x0 y0 0 85 2x0 y0
2 x0 y0
0 85 k 2x0 y0
2
1
1
1
x0
y0
1
1
k 2x
Hence xk
0 y0
3 x0 y0
3 0 85
2
1
k
yk
2x0 y0
3 x0 y0
3 0 85
1
2
As k
, xk
3 x0 y0 and yk
3 x0 y0 . i.e. eventually
2
live in London and 3 outside London.
8
1
3
of the total population will