SOLUTION 10
Let A be an n × n matrix, then the distinct eigenvalues of A are
linearly independent.
Problem 1.
a. Let {v1 , . . . , vk } be a set of eigenvectors corresponding to distinct eigenvalues
λ1 , . . . , λk of A.
Suppose that k = 1, then we have one eigenvector v1 . Since eigenvectors are
nonzero, the equation c1 v1 = 0 has only the trivial solution c1 = 0. Thus, the set
{v1 } is linearly independent.
Because of part a, we can assume that the vectors {v1 , . . . , vp } are linearly
independent, where 1 ≤ p < k.
• If vp+1 is linearly dependent on the vectors {v1 , . . . , vp }, then vp+1 is a
linear combination of the vectors {v1 , . . . , vp }. Therefore,
b.
(0.1)
c1 v1 + · · · + cp vp = vp+1 .
• We multiply both sides of (0.1) by A, since Ax is a linear transformation
we have
(0.2)
c1 Av1 v1 + · · · + cp Avp = Avp+1 .
• Since v1 , . . . , vp are eigenvectors corresponding to the eigenvalues λ1 , . . . , λp
of A, (0.2) can be written as
(0.3)
c1 λ1 v1 + · · · + cp λp vp = λp+1 vp+1 .
• Subbing in (0.1) into (0.3) and subtract to the other side gives
(0.4)
c1 (λ1 − λp+1 )v1 + · · · + cp (λp − λp+1 )vp+1 = 0.
Since the vectors {v1 , . . . , vp } are linearly independent (0.4) must have only the
trivial solution. Moreover, the eigenvalues are distinct, thus λ1 −λp+1 6= 0, . . . , λp −
λp+1 6= 0. It follows that c1 = · · · = cp = 0. However, (0.1) then implies that
vp+1 = 0, which contradicts vp+1 being an eigenvector. Therefore, our original
assumption that vp+1 is linearly dependent is false, and it follows that the vectors
{v1 , . . . , vp , vp+1 } are linearly independent.
Remark: This concludes the proof by induction. We established our base
case in part a. In part b-c we have shown that if the result holds for the
vectors v1 , . . . , vp then it must also hold for the vectors v1 , . . . , vp , vp+1 .
Therefore, all eigenvectors v1 , . . . , vk corresponding to distinct eigenvalues
must be linearly independent.
c.
1
SOLUTION 10
2
Let A be an n × n matrix with n distinct eigenvalues. Then the
eigenvectors of A form a basis for Rn .
Problem 2.
From problem 1 we know that the eigenvectors v1 , . . . , vn corresponding
to the n distinct eigenvalues will be linearly independent. Therefore, we have n
linearly independent vectors in Rn , which therefore form a basis for Rn .
Proof.
I am assuming that A is a real matrix and all the eigenvalues are
real. This is not always the case, but can be guaranteed when the matrix
A is symmetric. If the eigenvalues are not real, then the corresponding
eigenvectors will be complex and ultimately form a basis for Cn .
Remark:
Problem 3.
a.
Let A =
4
2
5
1
.
The eigenvalues of A are the roots of the characteristic polynomial
det (A − λI) = (4 − λ)(1 − λ) − 10
= λ2 − 5λ − 6 = (λ − 6)(λ + 1).
Therefore, the eigenvalues are λ1 = −1 and λ2 = 6.
b.
• A − λ1 I =
1
is v1 =
−1
• A − λ2 I =
5 5
2 2
1
0
∼
1
0
, an eigenvector of A corresponding to λ1
.
−2 5
2 −5
5
to λ2 is v2 =
.
2
∼
−2
0
5
0
, an eigenvector of A corresponding
Since the eigenvalues are distinct, the eigenvectors of A are linearly independent.
Therefore, we have two linearly independent vectors in R2 , it follows that they must
form a basis for R2 .
c.
Problem 4.
1
−1
a.
Let S =
b.
Note that S
−1
5
2
=
.
1
7
2
1
−5
1
. Moreover,
1
S AS =
7
1 −7
=
0
7
−1
2
1
−5
1
0
42
=
−1
1
−1
0
30
12
0
6
.
SOLUTION 10
c.
Let x =
1
1
3
, we want to nd scalars c1 and c2 such that
x = c 1 v1 + c 2 v2 .
To this end, we compute
Therefore, [x]β =
1 2
c1
=
c2
7 1
3
−7
2
7
−5
1
1
1
=
− 73
2
7
.
. Moreover, we nd the image of x under T by noting
T (x) = T (c1 v1 + c2 v2 ) = c1 T (v1 ) + c2 T (v2 )
= c1 (λ1 v1 ) + c2 (λ2 v2 )
3
2
1
5
=−
−
+
6
.
−1
2
7
7
Problem 5.
a.
We will show that S is invertible by nding the inverse of S :

0
 −2
1
1
1 −1
2 0 | 0
0
0 1
0
1
0
 
1
0
0 ∼ 0
0
1
1
0
2
1
0 | 2
1
− 12
0
1
0
− 41
1
4
1
4
1
2
1
2
1
2

.
We see that S has 3 pivots and is therefore invertible, and

S −1 = 
b.
1
2
1
2
− 12
− 14
1
4
1
4
1
2
1
2
1
2

.
Compute B = S −1 AS

=
1
2
1
2
− 12
− 14
1
4
1
4
1
2
1
2
1
2

0
2 2 4
  1 3 2   −2
1
4 1 4


3
2 17
4
4
31
5 

= 0 4
.
4
0 47 − 34


1 −1
2 0 
0 1
The eigenvalues of B must be the same as the eigenvalues of A, since similarity
transformations preserve eigenvalues.
One eigenvalue of A is λ1 = 2, the other two eigenvalues of A are stored in the
smaller submatrix
31
5
c.
 =
4
7
4
4
− 43
.
We can nd the eigenvalues of  by computing the roots of the characteristic
polynomial
31
−3
35
−λ
−λ −
det  − λI =
4
4
16
= λ − 7λ − 8 = (λ − 8) (λ + 1) .
Therefore, the other two eigenvalues of A are λ2 = 8 and λ3 = −1.
2