Properties of eigen values

Properties of eigen values

1. A square matrix A and its transpose AT have the same eigen values.

Proof 

Eigen values of A are the roots of its characteristic equation

This shows that the characteristic polynomial of A and AT are the same. Hence the characteristic equations of A and AT is (1).

⸫ A and AT have the same eigen values.

2. Sum of the eigen values of a square matrix A is equal to the sum of the elements on its main diagonal.

Proof 

Let A be a square matrix of order n.

Note 

Sum of the diagonal elements of A is called the trace of A.

⸫ Sum of the eigen values = trace of A

3. Product of the eigen values of a square matrix A is equal to │A│.

Proof 

Let A be a square matrix of order n.

Then its characteristic equation is  | A-λI | = 0

If λ1, λ2,...., λn are the n roots of (1), then from theory of equations, 

Note 

If at least one eigen value is 0, then |A|= 0

⸫ A is a singular matrix. If all the eigen values are non-zero, then |A| ≠ 0

⸫ A is a non-singular if all the eigen values are non-zero.

4. If λ1, λ2,...., λn are non-zero eigen values of square matrix of order n, then are eigen values of A-1.

Proof 

Let λ be any non-zero eigen value of A, then there exists a non-zero column matrix X such that AX = λX. Since all the eigen values are non-zero, A is non-singular.

So 1/λ is an eigen value of A-1.

This is true for all the eigen values of A.

⸫ 1/λ1, 1/λ2, …, 1/λn are the eigen values of A-1.

Note that the eigen vector for A corresponding to 1/λ is also X.

5. If λ1, λ2,...., λn are the eigen values of A, then

(i) cλ1, cλ2,...., cλn are the eigen values of cA, where c ≠ 0

(ii) λ1m, λ2m,...., λnm are the eigen values of Am, where m is a positive integer.

Proof 

Let λ be any eigen value of A, then there exists a non-zero column matrix X such that 

AX = λ X

(i) Multiply by c ≠ 0 then c(AX) = c(λX)

⇒ (cA) X = (cλ) X

⸫ cλ is an eigen value of cA. This is true for all eigen values of A.

⸫ c λ1, c λ2, ..., c λn are the eigen values of cA.

(ii) Now A2X = A(AX) = A(λ X) [using (1)]

= λ (AX)

= λ (λ X)

⸫ A2X = λ2 X ⇒ λ2 is an eigen value of A2. 

Similarly A3X = A(A2X)

= A(λ2X)

= λ2 (AX)

= λ2 (λX)

A3X = λ3X ⇒ λ3 is an eigen value of A3.

Proceeding in this way, we have AmX = λmX for any positive integer m. This is true for all eigen values.

⸫ λ1m, λ2m,...., λnm are eigen values of Am.

6. If λ1, λ2,...., λn are the eigen values of A, then

(i) λ1 – k, λ2 – k,..., λn – k are the eigen values of A - KI.

(ii) α0 λ12 + α1λ1 + α2, α0 λ22 + α1λ2 + α2,..., ‚ α0 λn2 + α1λn + α2 are the eigen values of α0A2 + α1A + α2I.

Proof 

Let λ be any eigen value of A

then AX = λ X            (1)

where X ≠ 0 is a column matrix.

⸫ AX - KX = λ X - KX

⇒   (A – KI)X = (λ – K)X

⸫  λ - K is an eigen value of A – KI. This is true for all eigen values of A.

⸫ λ1 — K, λ2 — K,..., λn -K are the eigen values of A – KI.

(ii) We have AX = λ X and A2X = λ2X.

⸫ α0 (A2X) = α0 (λ2X)

and α1 (AX) = α1(λX)

⸫ α0 (A2X) + α1 (AX) = α0 (λ2X) + α1(λX)

Adding α2X on both sides, we get

α0 (A2X) + α1 (AX) + α2X = α0 (λ2X) + α1 ( λ X) + α2X

⇒  (α0 A2 + α1A + α2I)X = (α0λ2 + α1λ + α2)X

This means α0λ2 + α1λ + α2 is an eigen value of a α0 A2 + α1A + α2I. 

This is true for all eigen values of A.

⸫ α0 λ12 + α1λ1 + α2, α0 λ22 + α1λ2 + α2,..., ‚ α0 λn2 + α1λn + α2 are the eigen values of α0 A2 + α1A + α2I.

Note

1. The eigen values of the unit matrix are 1, 1, 1 and the corresponding eigen vectors are which are independent.

2. The eigen values of a triangular matrix are the main diagonal elements λ1, λ2,λ3

3. If λ is an eigen value of A then AX = λX. We have seen

A2X = λ2X, ..., AmX = λmX.

Thus the eigen values of A, A2, …, Am are λ, λ2, ..., λm which are all different. But they all have the same eigen vector X. 

Similarly, λ and α0 λ2 + α1λ + α2 are eigen values of A and α0 A2 + α1A + α2I. But they have the same eigen vector X.

WORKED EXAMPLES

Example 14

Find the sum and product of the eigen values of the matrix

Solution

Sum of the eigen values = Sum of the elements on the main diagonal

= 1 + 0 + (-3) = −2

Product of the eigen values = |A| =

= 1(0 + 3) − 2 (−3 + 6) − 2(−1 – 0)

= 3 – 6 + 2 = −1

Example 15

If 2 and 3 are eigen values of A = find the eigen values of A-1 and A3.

Solution

Given 2 and 3 are two eigen values of A.

Let λ be the 3rd eigen value.

We know, sum of the eigen values = sum of the diagonal elements.

⇒ 2 + 3 + λ = 3 + (−3) + 7

⇒ λ = 2

So, eigen values of A are 2, 2, 3

⸫ Eigen values of A-1 are 1/2, 1/2, 1/3 and eigen values of A3 are

23, 23, 33 ⇒ 8, 8, 27.

Example 16

If is an eigen vector of the matrix find the corresponding eigen value.

Solution

Example 17

If A = find the eigen values of A2 - 2A + I .

Solution

Expanding by C1, (3 - λ) (2 - λ) (5 - λ) = 0

⇒ λ = 3, 2, 5 are the eigen values of A.

⸫ the eigen values of A2 - 2A + I are

32 – 2 ⸱ 3 + 1, 22 - 2 ⸱ 2 + 1, 52 - 2.5 + 1 

i.e., the eigen values of A2 - 2A + I are 4, 1, 16.

Example 18

The product of two eigen values of the matrix A = is 16. Find the third eigen value.

Solution

Let λ1, λ2, λ3 be the eigen values of A.

Given λ1 ⸱ λ2 = 16

We know that λ1⸱ λ2⸱ λ3 = |A|

= 6(9 − 1) + 2(−6 + 2) + 2(2 − 6)

⇒ 16 λ3 = 48 – 8 – 8 - 32 ⇒ λ3 = 2

Example 19

Find the eigen values of the matrix Hence find the matrix whose eigen values are 1/6 and −1.

Solution

Let A =

The characteristic equation of A is |A — λI| = 0

Example 20

If α, β are the eigen values of form the matrix whose eigen values are α3, β3.

Solution

Let A = Since α, β are the eigen values of A, by property 5(ii), α3, β3 are eigen values of A3.

EXERCISE 1.1

Find eigen values and eigen vectors of the following matrices.

ANSWERS TO EXERCISE 1.1

1. λ = 1, 3, -4; eigen vectors 

2. λ = 0, −1, 2; eigen vectors 

3. λ= -3, -3, 5; eigen vectors 

4. λ = 3, 6, 9; eigen vectors 

5. λ = 2, 2, 8; eigen vectors 

6. λ = −2, 3, 6; eigen vectors 

7. λ = 1, 1, 4; eigen vectors