MTH 102: Linear Algebra

Department of Mathematics and Statistics Indian Institute of Technology - Kanpur

Problem Set 6

Problems marked (T) are for discussions in Tutorial sessions.

1. Find the eigenvalues and corresponding

 eigenvectors
 of matrices
  −1 2 2
1 1
(a) (b)  2 2 2 
4 1
−3 −6 −6

Solution:

(a) (1 − λ)2 − 4 = 0 ⇒ (λ − 3)(λ + 1) = 0 ⇒ λ1 = 3, λ2 = −1. Also, v1 = [ 1/2 1 ]T , v2 =

[ −1/2 1 ]T .
(b) λ1 = 0, λ2 = −2, λ3 = −3 and v1 = [ 0 − 1 1 ]T , v2 = [ −2 1 0 ]T , v3 = [ −1 0 1 ]T .

2. Construct
 a basisof R3 consistingof eigenvectorsof the following matrices
0 0 2 1 1 −1
(a)  0 2 0  (b)  −1 1 1 .
2 0 3 −1 1 1

Solution:

(a) Eigenvalues are λ1 = −1, λ2 = 2, λ3 = 4 and eigenvectors are v1 = [ 1 0 − 1/2 ]T , v2 =

[ 0 1 0 ]T , v3 = [ 1 0 2 ]T . Since eigenvectors corresponding to different eigenvalues are
linearly independent, {v1 , v2 , v3 } is a basis of R3 .
(b) Similar to (a).

3. (T) This question deals with the following symmetric matrix A:

 
1 0 1
A =  0 1 -1  .
1 -1 0

One eigenvalue is λ = 1 with the line of eigenvectors x = (c, c, 0).

(a) That line is the null space of what matrix constructed from A?
Solution: The eigenvectors of λ = 1 makes the null space of A − I.
(b) Find the other two eigenvalues of A and two corresponding eigenvectors.
Solution: A has trace 2 and determinant −2. So the two eigenvalues after λ1 = 1 will add
to 1 and multiply to −2. Those are λ2 = 2 and λ3 = −1. Corresponding eigenvectors are :
   
1 1
v2 =  -1  , v3 =  -1  .
1 -2
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(c) The diagonalization A = SΛS −1 has a specially nice form because A = At . Write all entries
in the three matrices in the nice symmetric diagonalization of A.
Solution: Every symmetric matrix has the nice form A = QΛQt with an orthogonal matrix
Q. The columns of Q are orthonormal eigenvectors.
 √ √ √   
1/√2 1/ √3 1/ √6 1 0 0
Q =  1/ 2 −1/√ 3 −1/√6  Λ =  0 2 0  .
0 1/ 3 −2/ 6 0 0 -1

4. Let A be an n×n invertible matrix. Show that eigenvalues of A−1 are reciprocal of the eigenvalues
of A, moreover, A and A−1 have the same eigenvectors.
1
Solution: Ax = λx ⇒ x = λA−1 x ⇒ A−1 x = λx (Note that λ 6= 0 as A is invertible implies
that det(A) 6= 0).

5. Let A be an n×n matrix and α be a scalar. Find the eigenvalues of A−αI in terms of eigenvalues
of A. Further show that A and A − αI have the same eigenvectors.

Solution: If λ is an eigenvalue of A − αI with eigenvector v, then

Av = (A − αI)v + αv = (λ + α)v.

Thus, A and A−αI have same eigenvectors and eigenvalues of A−αI is µ−α if µ is an eigenvalue
of A.

6. (T) Let A be an n × n matrix. Show that At and A have the same eigenvalues. Do they have
the same eigenvectors?

Solution: Follows directly from det(A − λI) = det((A − λI)t ) = det(At − λI). Eigenvectors are
not same. Here is a counter example :
 
0 0
A= .
1 0

7. Let A be an n × n matrix. Show that:

(a) If A is idempotent (A2 = A) then eigenvalues of A are either 0 or 1.

Solution: Let Av = λv. Then λv = Av = A2 v = λ2 v ⇒ λ(λ − 1)v = 0. Result follows.
(b) If A is nilpotent (Am = 0 for some m ≥ 1) then all eigenvalues of A are 0.
Solution: Let Av = λv. Then Am v = λm v. Now, Am = 0 ⇒ λm = 0 ⇒ λ = 0.
(c) If A∗ = A then, the eigenvalues are all real.
Solution: Let (λ, x) be an eigenpair. Then

λx∗ x = x∗ (λx) = x∗ (Ax) = (x∗ Ax)∗ = x∗ A∗ x = x∗ Ax = λx∗ x = λx∗ x.

Hence, the required result follows.

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(d) If A∗ = −A then, the eigenvalues are either zero or purely imaginary.

Solution: Proceed as in the above problem.
(e) Let A be a unitary matrix (AA∗ = I = A∗ A). Then, the eigenvalues of A have absolute
value 1. It follows that if A is real orthogonal then the eigenvalues of A have absolute
value 1. Give an example to show that the conclusion may be false if we allow complex
orthogonal.
Solution: Let (λ, x) be an eigenpair of A. Then
kxk2 = x∗ x = x∗ (A∗ A)x = (x∗ A∗ )(Ax) = (Ax)∗ (Ax) = (λx)∗ (λx) = x∗ λλx = |λ|2 kxk2 .
√ 
2 2 √i
So |λ| = 1. For counter example, take A = .
−i 2

8. (T) Suppose that A155×5 = 0. Show that there exists a unitary matrix U such that U AU is
∗

upper triangular with diagonal entries 0.

Solution: There exists U unitary such that U ∗ AU = T , upper triangular with diag(T ) =
{λ1 , . . . , λ5 }. Hence T 15 has diagonal entries λ15 15 ∗ 15
1 , . . . , λ5 . As 0 = U A U = T
15 we see that
15
λi = 0. So, λi = 0 for all i.
9. (T) Suppose that A29
17×17 = 0. Show that A
17 = 0.

Solution: There exists U unitary such that U ∗ AU = T , upper triangular with diag(T ) =
{λ1 , . . . , λ17 }. As A29 = 0, it follows that λi = 0. So, A = U T U ∗ , A2 = U T 2 U ∗ , A3 = U T 3 U ∗
and so on. Also, verify that as T is upper triangular with zeroes on the diagonal, we must have
T 17 = 0. So, the result follows.
Alternate: As each eigenvalue of A is 0, the characteristic polynomial, namely pA (x) = x17 .
So, by Cayley Hamilton theorem, A17 = 0.
 
1 1
10. The matrix A = is NOT diagonalizable.
0 1
 
1 1
11. The matrix A = is diagonalizable.
0 2

12. Show that Hermitian, Skew-Hermitian and unitary matrices are normal.
13. Suppose that A = A∗ . Show that rankA = number of nonzero eigenvalues of A. Is this true for
each square matrix? Is this true for each square symmetric complex matrix?

Solution: By spectral theorem, there exists U , unitary such that U ∗ AU = D, diagonal. Since
U is invertible, we see that rankA = rankU ∗ AU = rankD = number of nonzero entries of D =
eigenvalues of A.
 
0 1
It is not true for general square matrices, consider A = . Here rankA = 1, whereas both
0 0
eigenvalues are 0.
 
1 i
It is not true for a general complex symmetric matrix, consider A = . Here rankA = 1,
i −1
whereas both eigenvalues are 0 (as det A = 0, trA = 0).
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 
2 -1 0
14. Show that A =  -1 2 0  is diagonalizable. Find a matrix S such that S −1 AS is a diagonal
2 2 3
matrix.

Solution: det(A − λI) = (1 − λ)(3 − λ)2 . Therefore, eigen-values are 1 and 3. The eigen spaces
(null space of A−λI), are given by E1 = {x : Ax = x} = {(x1 , x2 , x3 ) : x2 = x1 , x3 = −2x1 , x1 ∈
R} = LS({(1, 1, −2)}) and E3 = {(x1 , −x1 , x3 ) : x1 , x3 ∈ R} = LS({(1, −1, 0), (0, 0, 1)}).
Clearly, {(1, 1, −2), (1, −1, 0), (0, 0, 1)} are linearly independent and hence A is diagonalizable.
 
1 1 0
S =  1 -1 0  .
-2 0 1
 
7 -5 15
15. Let A =  6 -4 15 . Find a matrix S such that S −1 AS is a diagonal matrix and hence
0 0 1
6
calculate A .

Solution: det(A−λI) = (λ−1)2 (λ−2). Therefore, eigen-values are 1 and 2. E1 = {(x1 , x2 , x3 ) :

6x1 − 5x2 + 15x3 = 0} = LS({(1, 0, −6/15), (0, 1, 1/3)}). E2 = {(x1 , x1 , 0) : x1 ∈ R} =
LS({(1, 1, 0)}). For  
1 0 1
S =  0 1 1 ,
6 1
- 15 3 0
we have  
1 0 0
S −1 AS =  0 1 0  .
0 0 2
Therefore  
1 0 0
A6 = S −1  0 1 0  S.
0 0 26

16. Consider the 3 × 3 matrix  

a b c
A =  1 d e .
0 1 f
Determine the entries a, b, c, d, e, f so that:

• the top left 1 × 1 block is a matrix with eigenvalue 2;

• the top left 2 × 2 block is a matrix with eigenvalue 3 and -3;
• the top left 3 × 3 block is a matrix with eigenvalue 0, 1 and -2.
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Solution: Let Ai denote the top left i × i block of A. The matrix A1 is the matrix [a]. Since a
is the only eigenvalue of this matrix, we conclude that a = 2.
 
2 b
We now move onto determining the entries of the matrix A2 : A2 = .
1 d
Since the sum of the eigenvalues of A2 is 0 by hypothesis, and it is also equal to the trace of A2 ,
we obtain that 2 + d = 0 or d = −2. Moreover the product of the eigenvalues of A2 is −9 by
hypothesis, and it is qual to the determinant of A2 . Thus we have

−9 = 2d − b = −4 − b
 
2 5
and we deduce that b = 5 and therefore A2 = .
1 -2
Finally, consider A = A3 . Again, the sum of the eigenvalues of A is −1 and it is also equal to
the trace of A. We deduce that f = −1. We still need to determine the entries c and e of A and
we have  
2 5 c
A =  1 -2 e  .
0 1 -1
The characteristic polynomial of this matrix is

−λ3 − λ2 + (e + 9)λ + c − 2e + 9.

We know that the roots of this polynomial must be 0, 1 and −2. Setting λ = 0 and λ = 1, we
obtain

c − 2e + 9 = 0
−1 − 1 + (e + 9) + c − 2e + 9 = 0

which is equivalent to

c − 2e = −9
c − e = −16.

Thus c = −7 and e = 9 and we conclude

 
2 5 -7
A =  1 -2 9  .
0 1 -1

17. NOT for mid-sem or end-sem

(a) Find the eigenvalues and eigenvectors (depending on c) of

 
0.3 c
A= .
0.7 1 − c

For which value of c is the matrix A not diagonallizable (so A = SΛS −1 is impossible)?
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Solution: Eigen values are λ = 1 and λ = 0.3 − c. The eigenvector for λ = 1 is in the null
space of  
-0.7 c
A−I =
0.7 -c
so  
c
x1 = .
0.7
Similarly, the eigenvector for λ = 0.3 − c is in the null space of
 
c c
A − (0.3 − c)I =
0.7 0.7
so  
1
x2 = .
-1
A is not diagonalizable when its eigen values are equal : 1 = 0.3 − c or c = −0.7.
(b) What is the largest range of values of c (real number) so that An approaches a limiting
matrix A∞ as n → ∞?
Solution:  
n n −1 1 0
A = SΛ S =S S −1 .
0 (0.3 − c)n
This approaches a limit if |0.3 − c| < 1. We could write that out as −0.7 < c < 1.3.
(c) What is the limit of An (still depending on c)? You could work from A = SΛS −1 to find
An .
Solution: The eigen vectors are in S. As n → ∞, the smaller eigen value λn2 goes to zero,
leaving
   
∞ c 1 1 0 1 1
A = /(c + 0.7)
0.7 −1 0 0 0.7 −c
 
c c
= /(c + 0.7).
0.7 0.7