Linear Algebra, Numerical and Complex Analysis (MA11004 )

Department of Mathematics
Indian Institute of Technology Kharagpur
Solution of Tutorial sheet 4, Spring 2024

1. Prove the following statements:

|A|
(a) If λ is an eigenvalue of a non-singular matrix A, then is an eigenvalue of adj A,
λ
where |A| denotes the determinant of the matrix A.
(b) If A and B are two invertible matrices, then AB and BA have same characteristic
roots.
(c) If λ is an eigenvalue of algebraic multiplicity r of A, then 0 is an eigenvalue of
algebraic multiplicity r of the matrix A − λIn .
Solution:
adj A
(a) A is non-singular, so A−1 exists and A−1 = .
|A|
Now,
|A − λIn | , where n is the order of A.
= A − λAA−1
adj A
= A − λA
|A|
λ
=|A| In − adj A
|A|
λn |A|
=|A| n In − adj A
|A| λ
λn |A|
= n−1 In − adj A
|A| λ

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Since A is non-singular, 6= 0.
λ
Therefore, the characteristic equation
|A − λIn | = 0
|A|
=⇒ In − adj A = 0
λ
|A|
It shows that is an eigen value of adj A when λ is an eigen value of A.
λ

(b) Now,
AB = In AB = B −1 B(AB) = B −1 (BA)B

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 Now,
|AB − λIn | = B −1 (BA)B − λIn
= B −1 (BA)B − B −1 λIn B
=|B −1 | |(BA) − λIn | |B|
= |(BA) − λIn | |B −1 B|
= |(BA) − λIn | |In |
= |(BA) − λIn |

It shows that AB and BA have same characteristic equations.

Hence, they have same characteristic root.

(c) Since, λ is an r−fold eigen value of A,

|A − xIn | = (x − r)r ψ(x),
where ψ(x) is a polynomial in x of degree (n − r) and ψ(λ) 6= 0.
Now, characteristic polynomial of A − xIn is
|A − xIn − λIn |
= |A − (x + λ)In |
=(λ + x − λ)r ψ(λ + x)
=xr ψ(λ + x)
=xr µ(x),
where µ(0) = ψ(λ) 6= 0.
This proves that 0 is a root of multiplicity of r of the characteristic polynomial of A − xIn .

2. Find all the eigenvalues and the corresponding eigenvectors of the following matrices:
   
    3 10 5 2 −1 1
−5 2 0 −1
(a) (b) (c) −2 −3 −4 (d) −1
   2 −1 .
2 −2 1 0
3 5 7 1 −1 2
Solution:
 
−5 2
(a) Here, A = .
2 −2
The characteristic polynomial of A is given by
|A − λIn | = 0
 
−5 − λ 2
=⇒ det =0
2 −2 − λ
=⇒ (−5 − λ)(−2 − λ) − 4 = 0
=⇒ λ2 + 7λ + 6 = 0
=⇒ (λ + 1)(λ + 6) = 0
=⇒ λ = −1, −6

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The eigen values of the given matrix are -1 and -6.
When λ = −1, the corresponding eigenvectors are given by
    
−5 + 1 2 x1 0
=
2 −2 + 1 x2 0
    
−4 2 x1 0
=⇒ =
2 −1 x2 0

=⇒ 2x1 − x2 = 0
1
=⇒ x1 = x2
2
Let x1 = k, then x2 = 2k.  
k
Hence, eigenvector corresponding to λ = −1 is X1 = .
2k
When λ = −6, the corresponding eigenvectors are given by
    
−5 + 6 2 x1 0
=
2 −2 + 6 x2 0
    
1 2 x1 0
=⇒ =
2 4 x2 0

=⇒ x1 + 2x2 = 0
=⇒ x1 = −2x2
Let x2 = k1 , then x1 = −2k1 .  
−2k1
Hence, eigenvector corresponding to λ = −6 is X2 = .
k1

 
0 −1
(b) Here, A = .
1 0
The characteristic polynomial of A is given by

|A − λIn | = 0
 
0 − λ −1
=⇒ det =0
1 0−λ
=⇒ λ2 + 1 = 0
=⇒ λ = ±i

A is a real matrix and the eigen values of A are not real numbers. Therefore, the real
matrix A has no eigen vector.
But if A be considered as a complex matrix, then the eigenvectors corresponding to the
eigen values i and −i can be obtained.

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x1
Let X1 = be an eigen vector corresponding to λ = i,
x2
then,
AX1 = iX1
(
−ix1 − x2 = 0
=⇒
x1 − ix2 = 0
The equivalent system is x1 − ix2 = 0 =⇒ x1 = ix2 .
Let x2 = k then x1 = ik.  
i
Hence, eigenvector corresponding to λ = i is X1 = k , where k is a non-zero complex
1
number.  
x1
Let X2 = be an eigen vector corresponding to λ = −i,
x2
then,
AX2 = −iX2
(
ix1 − x2 = 0
=⇒
x1 + ix2 = 0
The equivalent system is ix1 − x2 = 0 =⇒ ix1 = x2 .
Let x1 = k1 then x2 = ik1 .  
1
Hence, eigenvector corresponding to λ = −i is X2 = k1 , where k1 is a non-zero
i
complex number.

(c) The characteristic equation of the given matrix A is

|A − λI| = 0
3−λ 10 5
−2 −3 − λ −4 =0
3 5 7−λ
⇒(3 − λ) −21 + 3λ − 7λ + λ2 + 20 − 10[−14 + 2λ + 12] + 5[−1 + 3λ] = 0
 

⇒(3 − λ) λ2 − 4λ − 1 − 10(2λ − 2) + 5(3λ − 1) = 0

⇒λ3 − 7λ2 + 16λ − 12 = 0

⇒(λ − 3)(λ − 2)(λ − 2) = 0
⇒λ = 2, 2, 3
 >
Eigenvalues of the matrix A are 3, 2, 2. Let X1 = x1 x2 x3 be the eigenvector
corresponding to λ = 3. Then we have
    
3−3 10 5 x1 0
(A − I)X1 =  −2 −3 − 3 −4   x2 = 0 
 
3 5 7−3 x3 0
⇒ −2x1 − 6x2 − 4x3 = 0 ...........(1)
3x1 + 5x2 + 4x3 = 0 ...............(2)

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solving (1) & (2) by cross multiplication method we have,
x1 x2 x3
= =
−24 + 20 −12 + 8 −10 + 18
x1 x2 x3
⇒ = = = k (say)
−4 −4 8
then x1 = −4k, x2 = −4k, x3 = 8k
 
−1
Therefore, the eigenvector x1 = 4k  −1  , k ∈ R\{0}
2
 >
Let X2 = x1 x2 x3 be the eigenvector corresponding to λ = 2
    
3−2 10 5 x1 0
(A − I)x2 =  −2 −3 − 2 −4   x2 = 0 
 
3 5 7−2 x3 0
⇒ x1 + 10x2 + 5x3 = 0 .......(3)
− 2x1 − 5x2 − 4x3 = 0 .......(4)

solving (3) & (4) by cross multiplication method we get,

x1 x2 x3
= =
−40 + 25 −10 + 4 −5 + 20
x1 x2 x3
⇒ = = = k1 (say)
−15 −6 15
⇒ x1 = −15k1 , x2 = −6k1 , x3 = 15k1

 
−5
Therefore, the eigen vector X2 = 3k1  −2  , k1 ∈ R \ {0}.
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We get one eigenvector corresponding to repeated root λ2 = 2 = λ3 .
Eigenvectors corresponding to λ2 = 2 = λ3 are not linearly independent.

(d) The characteristic equation of the given matrix is.

2 − λ −1 1
−1 2 − λ −1 =0
1 −1 2 − λ
⇒ (2 − λ) (2 − λ)2 − 1 + 1[−2 + λ + 1] + 1[1 − 2 + λ] = 0
 

⇒ (2 − λ) 4 − 1λ + λ2 − 1 + (λ − 1) + λ − 1 = 0

⇒ 8 − 8λ + 2λ2 − 2 − 4λ + 4λ2 − λ3 + λ + 2λ − 2 = 0
⇒ λ3 − 6λ2 + 9λ − λ = 0
⇒ (λ − 1)(λ − 1)(λ − 9) = 0
⇒ λ = 1, 1, 4

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Hence, the eigen values are 1,1,4.
 >
Let X1 = x1 x2 x3 be the eigenvector corresponding to λ = 4 then we have,
    
2 − 4 −1 1 x1 0
 −1 2 − 4 −1   x2  =  0 
1 −1 2 − 4 x3 0
⇒ −2x1 − x2 + x3 = 0 ...........(1)
x4 − x2 − 2x3 = 0 ...........(2)
cross multipling (1) & (2) we get,
x1 x2 x3 x4 x2 x3
= = ⇒ = = = k(say)
2+1 1−4 2+1 1 −1 1
then x1 = k, x2 = −k, x3 = k.

1
Therefore the corresponding eigenvector X1 = k  −1  , k ∈ R\{0}.
1
 >
Let X2 = x1 x2 x3 be the eigenvector corresponging to λ = 1.
then we have,     
2 − 1 −1 1 x1 0
 −1 2 − 1 −1   x2 = 0 
 
1 −1 2−1 x3 0
    
1 −1 1 x1 0
⇒  −1 1 −1   x2  =  0 
1 −1 1 x3 0
    
1 −1 1 x1 0
⇒ 0 0 0   x2  =  0 
0 0 0 x3 0
⇒ x1 − x2 + x3 = 0
Let x1 = k1 and x2 = k2 and x3 = k2 − k1 .
 
k2
Therefore the corresponding eigenvector X2 = k  k1 .
k2 − k1
 
1
=⇒ X2 = k  1  taking k1 = 1, k2 = 1.
0
 >
Let X3 = l m n , then as X3 in orthoronal to X1 since A is symmetric.
 
  l
Now, 1 −1 1  m  = 0
n
⇒ l − m + n = 0 ......(2)
Similarly, X3 is orthogonal to X1 gives
 
  l
1 −1 0  m  = 0
n

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 ⇒ l + m = 0 ......(3)

−1
l m n
solving (2) &(3) we have −1 = 1 = 2 ∴ x3 =  1 .
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3. Find two different 2 × 2 matrices A and B such that

 both have the same eigenvalues
1
λ1 = λ2 = 2 and both have the same eigenvector corresponding to 2.
0

Solution:
   
a b 1
Let A = . since is an eigenvelutor with eigenvalue 2, we have
c d 0
    
a b 1 1
=2
c d 0 0
⇒ a = 2, c = 0
Since a + d = trace(A) = 2 + 2 = 4 [∵ trace(A) = λ1 + λ2 ]
⇒d=4−a=2
 
2 b
Thus, A = . It is very easy to check such m matrix with λ1 = λ2 = 2.
0 2
Finally, the condition that A has only one eigenvector implies b 6= 0.
Similarly, B has the same form as of A. So we can take different values of b for A and B.
   
2 1 2 3
For example, A = &B =
0 2 0 2

   
1 a b
4. Let a + b = c + d. Show that is an eigenvector of and find the eigenvalues.
1 c d

Solution:

Here a + b = c + d
      
a b 1 a+d 1
Now, we can write = = (a + b)
c d 1 c+d 1
 
1
Therefore is an eigenvector of given matrix corresponding to λ1 = (a + b). Again,
1
we know that trace = α + d = λ1 + λ2

⇒ λ2 = (a + d) − λ1
⇒ λ2 = d − b

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  
−i 3 + 2i −2 − i
5. Show that the matrix A = −3 + 2i 0 3 − 4i  is skew-Hermitian.
2−i −3 − 4i −2i

Solution:
 
i 3 − 2i −2 + i
Here Ā =  −3 − 2i 0 3 + 4i 
2+i −3 + 4i 2i
 
i −3 − 2i 2+i
⇒ (Ā)0 =  3 − 2i 0 −3 + 4i 
−2 + i 3 + 4i 2i
 
i −3 − 2i 2+i
⇒ Aθ =  3 − 2i 0 −3 + 4i 
−2 + i 3 + 4i 2i
 
−i 3 + 2i −2 − i
= −  −3 + 2i 0 3 − 4i 
2−i −3 − 4i −2i
= −A
Hence, Aθ = −A ⇒ A is skew-Hermitian matrix.

6. Prove the following statements:

 
cos θ − sin θ
(a) If 0 < θ < π, then A = has no real eigenvalues.
sin θ cos θ
1
(b) If λ is an eigenvalue of an orthogonal matrix, then is also an eigenvalue of it.
λ
Solution:

(a). The characteristic equation of A is

cos θ − λ − sin θ
=0
sin θ cos θ − λ

⇒ (cos θ − λ)2 + sin2 θ = 0

⇒ cos2 θ + sin2 θ − 3λ cos θ + λ2 = 0
⇒ λ2 − 2λ cos θ + 1 = 0
√ √
2 cos θ ± 4 cos2 θ − 4 2 cos θ ± 2i 1 − cos2 θ
⇒λ= =
2 2
⇒ λ = cos θ ± i sin θ
Hence, the given matrix A has no real eigenvalues.

(b). Let A be an orthogonal matrix, then A> A = I, i.e., A> = A−1 . So A is also non-
singular. If λ is an eigenvalue of A, then λ 6= 0, and from Ax = λx multiplying λ−1 A−1 ,

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 we get λ−1 A−1 (AX) = λ−1 A−1 λX ⇒ λ−1 X = A−1 X. Thus λ1 is an eigenvalue of A−1 .
Therefore λ1 is ar eigenvalue of A> . The value of the determinant remains unchanged
when rows become coloumns.

∴ A> − λI = |A − λI|

i.e., the eigen-values of A> and A are same. Hence, 1

λ is an eigenvalue of A.

7. If A and B are two unitary matrices, show that AB is a unitary matrix.

Solution:

Since A is a unitary matrix

A · Aθ = Aθ A = I ..............(1)
similarly B · B = B θ · B = I .............(2)
θ
 
Now (AB) · (AB)θ = (AB) B θ · Aθ
 
= A BB θ · Aθ
= AIAθ [From (2)]
θ
= AA
=I [From (1)]

 
Again, (AB)θ · (AB) = B θ · Aθ (AB)
 B
= B θ Aθ A
= B θ IB [From (2]]
θ
=B B
=I [From (1)]

Hence, AB is a unitary matrix.

 
i 2 − 3i 4 + 5i
8. Express the matrix A = 6 + i 0 4 − 5i as the sum of a Hermitian and a skew
−i 2−i 2+i
Hermitian matrix.

Solution:

Here we have  
i 2 − 3i 4 + 5i
A= 6+i 0 4 − 5i  .......(1)
−i 2−i 2+i

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  
−i 2 + 3i 4 − 5i
⇒ Ā =  6 − i 0 4 + 5i 
−i 2+i 2−i
 
−i 6−i −i
⇒ (Ā)0 =  2 + 3i 0 2+i 
4 − 5i 4 + 5i 2 − i
 
−i 6−i i
⇒ Aθ =  2 + 3i 0 2 + i  ...........(2)
4 − 5i 4 + 5i 2 − i
on adding (1) & (2) we get,
 
0 8 − 4i 4 + 6i
A + Aθ =  8 + 4i 0 6 − 4i 
4 − 6i 6 + 4i 4
 
0 4 − 2i 2 + 3i
1
Let R = (A + Aθ ) =  4 + 2i 0 3 − 2i  ..........(3)
2
2 − 3i 3 + 2i 2
on subtracting (2) from (1) we get.
 
2i −4 − 2i 4 + 4i
A − Aθ =  4 − 2i 0 2 − 6i 
−4 + 4i −2 − 6i 2i
 
i −2 − i 2 + 2i
1  
A − Aθ =  2 − i 0 1 − 3i  ............(4)
2
−2 + 2i −1 − 3i i

From (3) & (4), we have

   
0 4 − 2i 2 + 3i i −2 − i 2 + 2i
A =  4 + 2i 0 3 − 2i  +  2 − i 0 1 − 3i 
2 − 3i 3 + 2i 2 −2 + 2i −1 − 3i i

Hermitian matrix Skew-Hermilian matix

 
0 1 + 2i
9. If N = , then show that (I − N )(I + N )−1 is a unitary matrix, where I
−1 + 2i 0
is the identity matrix of order 2.

Solution:
 
0 1 + 2i
We have N = .
−1 + 2i 0
Now,

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1 0 0 1 + 2i 1 −1 + 2i
I −N = − = ...............(1)
0 1 −1 + 2i 0 1 − 2i 1

Now we have to find (I − N )−1

     
1 0 0 1 + 2i 1 1 + 2i
I +N = + =
0 1 −1 + 2i 0 −1 + 2i 1
⇒ |I + N | = 1 − (−1 − 4) = 6
 
1 −1 − 2i
⇒ adj(I + N ) =
1 − 2i 1
 
−1 adj(I + N ) 1 1 −1 − 2i
(I + N ) = = ..............(2)
|I + N | 6 1 − 2i 1

For unitary matrix Aθ A = I.

from (1) & (2) we get,
  
−1 1 1 −1 + 2i 1 −1 − 2i
(I − N )(I + N ) =
6 1 − 2i 1 1 − 2i 1
 
1 −4 −2 − 4i
= = B (say)
6 2 − 4i −4
 
> 1 −4 2 + 4i
Now (B̄) =
6 −2 + 4i −4
  
> 1 −4 2 + 4i −4 −2 − 4i
∴ (B) B =
36 −2 + 4i −4 2 − 4i −9
 
1 36 0
=
36 0 36
=I

Hence, it is a unitary matrix.

10. Let A be a n × n unitary matrix and λ be an eigen value of A. Show that |λ| = 1.

Solution:

If v is an eigenvector of A with eigenvalue λ, then

v ∗ v = v ∗ (A∗ A)v = (Av)∗ (Av) = (λv)∗ (λv) = λλv ∗ v.

Note that v ∗ v > 0 whenever v is a non-zero vector. Therefore |λ| = 1.

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