Eigen Values and Eigen Vectors

EIGEN VALUES AND EIGEN VECTORS

Definition 1.2.1 Let A be a square matrix of order n. A number λ is called an eigen value of A if there exists a non-zero column matrix X such that AX = λX. Then X is called an eigen vector of A corresponding to λ.

This will represent a system of linear homogeneous equations in x1, x2, ..., xn. Since X ≠ 0 at least one of the xi ≠ 0. Hence the homogeneous system has nontrivial solutions.

⸫ the determinant of coefficients | A - λ I | = 0. This equation is called the characteristic equation of A. The determinant | A - λ I |, on expansion, will be a nth degree polynomial in λ and is known as the characteristic polynomial of A. The roots of the characteristic equation are the eigen values of A.

Definition 1.2.2 Characteristic equation and characteristic polynomial If λ is a characteristic root of a square matrix A, then | A - λ I | = 0 is called the characteristic equation of A.

The polynomial | A - λ I | in λ is called the characteristic polynomial of A.

Note

(1) The word 'eigen' is German, which means 'characteristic' or 'proper'. So, an eigen value is also known as characteristic root or proper value. Sometimes it is also known as latent root.

(2) If then the characteristic equation of A is

where S1 = a11 + a22 = sum of the diagonal elements of A.

S2 = a11 . a22  - a21 . a12 = | A |

(3) If then the characteristic equation of A is

Expanding this determinant we will get

where

S1 = sum of the diagonal elements of A

S2 = sum of the minors of elements of the main diagonal 

S3 = |A|

We will use this formula in problems.

Definition 1.2.3 The set of all distinct eigen values of the square matrix A is called the spectrum of A. The largest of the absolute values of the eigen values of A is called the spectral radius of A. The set of all eigen vectors corresponding to an eigen value λ of A, together with zero vector, forms a vector space which is called the eigenspace of A corresponding to λ.

Properties of eigen vectors

Theorem 1.2.1

(1) Eigen vector corresponding to an eigen value is not unique.

(2) Eigen vectors corresponding to different eigen values are linearly independent.

Proof (1) Let λ be an eigen value of a square matrix A of order n.

Let X be an eigen vector corresponding to λ.

Then  AX = λ X

Multiply by a constant C

⸫ C(AX) = C(λX)

⇒  A(CX) = λ (CX)

Since

C ≠ 0, X ≠ 0 we have CX  ≠ 0

⸫ CX is an eigen vector corresponding to λ for any C ≠ 0. Hence eigen vector is not unique for the eigen value λ.

(2) Let λ1, λ2, be two different eigen values of A.

Let X1, X2 be corresponding eigen vectors.

⸫ X1 and X2 are linearly independent.

Note

(1) If all the n eigen values λ1, λ2,…, λn of A are different, then the corresponding eigen vectors X1, X2, ..., Xn are linearly independent. 

(2) A given eigen vector of A corresponds to only one eigen value of A. 

(3) Eigen vectors corresponding to equal eigen values may be linearly independent or dependent.

WORKED EXAMPLES

Example 3

Find the eigen values and eigen vectors of the matrix

Solution

Let A = The characteristic equation of A is

which are the eigen values of A.

To find eigen vectors:

Let be an eigen vector of A corresponding to λ.

Case (i) If λ = 1, then equations (I) become

3x1 + x2 = 0

3x1 + x2 = 0

⸫     x2 = -3x1

Choosing x1 = 1, we get x2 = -3

⸫ eigen vector is   

Case (ii) If λ = 5, then equations (I) become

- x1 + x2 = 0 and 3x1 - 3x2 = 0

⸫    x1 = x2 

Choosing x1 = 1, we get x2 = 1

⸫ eigen vector is 

Thus eigen values of A are 1, 5 and the corresponding eigen vectors are

Note 

In case (i) we have only one equation 3x1 + x2 = 0 to solve for x1 and x2. So, we have infinite number of solutions x1 = k, x2 = -3k, for any k ≠ 0.

We have chosen the simplest solution. In fact is an eigen vector for λ = 1 for any k ≠ 0. So, for λ = 1 there are many eigen vectors. This verifies property 1.

Example 4

Show that the real matrix has two eigen vectors where b ≠ 0.

Solution

which are the eigen values of A.

To find eigen vectors

Let be an eigen vector of A corresponding to λ.

Example 5

Find the eigen values and eigen vectors of the matrix

Solution

where

S1 = sum of main diagonal elements of A

= 3 + (-2) + 3 = 4

S2 = sum of minors of diagonal elements of A

= -6 + 4 + 9 – 4 + (-6) + 4 = -2 + 5 + (-2) =1 −2 + 5 + (−2) = 1

S3 = |A| = 3(−6 + 4) + 4 (3 - 4) + 4 (-1+ 2) = - 6 – 4 + 4 = -6

⸫ The characteristic equation is λ3 - 4 λ2 + λ + 6 = 0

We choose integer factors of constant term 6 for trial solution. We find

λ = -1 is a root. To find the other roots we perform synthetic division Other roots are given by

To find eigen vectors:

Let be an eigen vector corresponding to the eigen value λ.

By rule of cross multiplication

Choosing x1 = 1, x2 = 1, x3 = 0, we get an eigen vector

Case (ii) If λ = 2, then equations (I) become

x1 - 4x2 + 4x3 = 0

x1 - 4x2 + 4x3 = 0

x1 - x2 + x3 = 0

⸫ the different equations are x1 - 4x2 + 4x3 = 0

x1 - x2 + x3 = 0

By the rule of cross multiplication, we get

Choosing x1 = 0, x2 = 1, x3 = 1, we get an eigen vector

Case (iii) If λ = 3, then equations (I) become

x1 - 4x2 + 4x3 = 0 ⇒ 0x1 - x2 + x3 = 0

x1 - 5x2 + 4x3 = 0 

x1 - x2 + 0x3 = 0 

The equations are different, but only two of them are independent. So, we can choose any two of them to solve. From the first two equations, we get

Note

(1) We are using the following integer root theorem for trial solution. "For the equation of the form

xn + an-1 xn-1 + an-2 xn-2 + ··· + a1x + a0 = 0 with integer coefficients ai, any rational root is an integer and is a factor of the constant term a0". So, it is enough we try factors of the constant term for integer solutions.

(2) In the above problem the eigen values -1, 2, 3 are different. So by property (2) the eigen vectors are linearly independent. We shall verify this:

Consider with the eigen vectors as rows.

Then |B| =1.0-1(-1)+0=1≠0

⸫ X1, X2, X3 are linearly independent.

Example 6

Find the eigen values and eigen vectors of

Solution

where S1 = sum of the diagonal elements A

= 2 + 3 + 2 = 7

S2 = sum of minors of the diagonal elements of determinant A

= 6 - 2 + 4 - 1 + 6 - 2 = 4 + 3 + 4 = 11

S3 = |A| = 2(6 - 2) − 2(2 - 1) +1(2 - 3) = 8 - 2 - 1 = 5

⸫ The characteristic equation is λ3 - 7 λ2 + 11 λ - 5 = 0 

Choose the integer factors of constant term -5 for trial. 

The integer factors of -5 are -5, 1, or -1,5.

We find λ = 1 is a root.

Other roots are given by

⸫ the eigen values are λ = 1, 1, 5 (Two equal eigen values)

To find eigen vectors:

Let be an eigen vector of A corresponding to the eigen value λ.

Case (i) If λ = 5, then the equations (I) become

- 3x1 + 2 x2 + x3  = 0              (1)

x1 - 2 x2 + x3  = 0                    (2)

x1 + 2 x2 + 3x3  = 0                 (3)

These 3 equations are different, but only 2 of them are independent.

So, we can choose any two of them to solve for x1, x2, x3. 

From (2) and (3), by rule of cross multiplication we get

Case (ii) If λ = 1, then the equations (I) become

x1 + 2x2 + x3 = 0

x1 + 2x2 + x3 = 0

x1 + 2x2 + x3 = 0

We have only one equation x1 + 2x2 + x3 = 0 to solve for x1, x2, x3.

Assign arbitrary values for two variables and solve for the third.

Choose x3 = 0, then x1 + 2x2 = 0 ⇒ x1 = −2x2

Choose x2 = 1,

⸫ x1 = -2, we get an eigen vector

We shall find one more solution from x1 + 2x2 + x3 = 0

Choose x2 = 0 then x1 + x3 = 0 ⇒ x3 = -x1

Choose x1 = 1     ⸫ x3 = − 1 

⸫ another eigen vector corresponding to λ = 1 is

Thus eigen values of A are 5, 1, 1 and the corresponding eigen vectors are

Note 

Though the eigen values are not different, we could find independent eigen vectors.

For, consider with the vectors as rows

Then |B| = 1(-1 - 0) — 1(2 − 0) + 1(0 − 1)

= -1 - 2  - 1 = -4 ≠ 0

⸫ X1, X2, X3 are linearly independent.

Example 7

Find the eigen values and eigen vectors of the matrix

Solution

where S1 = sum of the diagonal elements of A

= 3 + (-3) + 7 = 7

S2 = sum of minors of elements of the diagonal of A

= (−21 + 20) + (21−15) + (−9 + 20)

= -1 + 6 + 11 = 16

S3 = |A| = 3(-21 + 20) - 10(-14 + 12) + 5(-10 + 9)

= -3 + 20 - 5 - 12

⸫ the characteristic equation is λ3 - 7 λ2 + 16λ – 12 = 0 

Choose the integer factors of constant term -12 for trial.

We find λ = 2 is a root.

Other roots are given by

⸫ The eigen values are λ = 2, 2, 3. (Two equal eigen values)

To find eigen vectors:

Let be an eigen vector corresponding to the eigen value λ.

Case (i) If λ = 3, then the equations (I) become

0x1, + 10x2 + 5x3 = 0 ⇒ 0x1 + 2x2 + x3 = 0              (1)

-2x1 - 6x2 - 4x3 = 0 ⇒ x1 + 3x2 + 2x3 = 0                  (2)

3x1 + 5 x2 + 4x3 = 0                                                   (3)

These 3 equations are different, but only 2 of them are independent. From (1) and (2), by rule of cross multiplication we get

Choosing x1 = 1, x2 = 1, x3 = −2, we get an eigen vector

Case (ii) If λ = 2, then equations (I) become

x1 + 10x2 + 5x3  = 0              (1)

-2x1 - 5x2 -4x3  = 0                (2)

3x1 + 5x2 + 5x3  = 0               (3)

From (1) and (2), by rule of cross multiplication we get

In this case we cannot choose another vector independent of X2. So corresponding to the equal eigen values 2, 2 the eigen vectors are equal

Thus the eigen values of A are 3, 2, 2 and the corresponding eigen vectors are

Note 

Here X1 and X2 are linearly independent vectors as they belong to different eigen values, but X1, X2, X3 are linearly dependent vectors. See note (3) below properties. (Page 1.6)

Example 8

Find the eigen values and eigen vectors of the matrix

Solution

where S1 = sum of the diagonal elements of A

= 6 + (-13) + 4 = -3 

S2 = sum of minors of elements of the diagonal of A

= (−52 + 60) + (24 − 35) + (-78 + 84) = 8 – 11 + 6 = 3

S3 = |A| = 6(−52 + 60) + 6(56 − 70) + 5(−84 + 91)

= 48 + 6(-14) + 5(7) = 48 - 84 + 35 - 1

⸫ the characteristic equation is λ3 + 3 λ2 + 3 λ + 1 = 0

⇒  ( λ + 1)3 = 0 

⇒ λ = -1, -1, -1

(3 equal eigen values)

To find eigen vectors:

Let X = be an eigen vector corresponding to the eigen value X.

If λ = -1, then the equations (1) become

7x1 −6x2 + 5x3 = 0

14 x1 −12 x2 + 10 x3 = 0 ⇒ 7 x1 −6 x2 + 5 x3 = 0

7x1 − 6x2  + 5x3 = 0

We have only one equation 7x1 - 6x2 + 5x3 = 0

Assign arbitrary values to two variables and find the third. 

We shall find 3 vectors.

Putting x1 = 0, we get −6x2 + 5x3 = 0

Note 

If with the eigen vectors as rows,

then |B| = 0 - 5(0 - 42) + 6(-35 - 0) = 210 – 210 = 0

⸫ The vectors X1, X2, X3 are linearly dependent. However, any two of them are linearly independent. Geometrically, it means that all the vectors are coplanar, but any two of them are non-collinear.

In this example we have seen −1 is the only eigen value of 3 × 3 matrix and two linearly independent eigen vectors.

Example 9

Find the eigen values and eigen vectors of

Solution

where S1 = −5 + 9 + (−7) = −3

= (−63 + 54) + (35 −18) + (−45 + 40) = −9 + 17 – 5 = 3

S3 = |A| = −5(−63 + 54) + 5(−56 + 36) — 9(−24 + 18)

= 45 – 100 + 54 = -1

⸫ the characteristic equation is λ3 + 3 λ2 + 3λ + 1 = 0

⇒  (λ + 1)3 = 0

⇒  λ = -1, -1, -1

If X = is an eigen vector of the eigen value λ,

Note 

In this example, 3 × 3 matrix has one eigen value −1 and one linearly independent eigen vector.

Example 10

Find the eigen values and eigen vectors of

Solution

⸫ the characteristic equation is λ3 — 5 λ2 + 7 λ − 3 = 0 

Since the sum of the coefficients is zero, λ = 1 is a root.

The other roots are given by

To find eigen vectors:

Let be an eigen vector corresponding to an eigen value λ of A.

Case (i) If λ = 1, then equations (I) become

x1 + x2 + x3 = 0

x1 + x2 + x3 = 0

0x3 = 0

Thus we have x1 + x2 + x3 = 0

Put x1 = 1, x2 = 0, then x3 = −1

Since λ = 1 is a repeated eigen value, we have to find another eigen vector. Put x1 = 0, x2 = 1 ⸫ x3 = -1

So, another eigen vector is

Case (ii) If λ = 3 then equations (I) become

-x1 + x2 + x3 = 0

⇒   x1 - x2 - x3 = 0

and -2x3 = 0 ⇒ x3 = 0

x1 - x2 = 0 ⇒ x1 = x2

Take x2 = 1   ⸫ x1 = 1, x2 = 1, x3 = 0

⸫ an eigen vector is

Thus the eigen values are λ = 1, 1, 3 and the eigen vectors are

Example 11

Find the eigen values and eigen vectors of the matrix

Solution

Where S1 = sum of diagonal elements of A

= 11+ (-2) + (− 6) = 3

= (12 - 20) + (−66 + 70) + (−22 + 28) = -8 + 4 + 6 = 2

-88 +32 +56 = 0

⸫ The characteristic equation is λ3 − 3λ2 + 2λ - 0 = 0

⇒ λ (λ2 − 3λ + 2) = 0 ⇒ λ (λ − 1)( λ − 2) = 0 ⇒ λ = 0, 1, 2.

To find eigen vectors:

If be an eigen vector of A corresponding to λ.

Take x1 = 1, x2 = 1, x3 = 1.

⸫ an eigen vector is

Case (ii) If λ = 1, then (I) ⇒ 10x1 − 4x2 − 7x3 = 0

7x1 − 3x2 − 5x3 = 0

10x1 − 4x2 − 7x3 = 0

From the first two equations, by rule of cross multiplication we get

Case (iii) If λ = 2, then (I) ⇒ 9x1 − 4x2 − 7x3 = 0

7x1 − 4x2 − 5x3 = 0

10x1 − 4x2 − 8x3 = 0

From the first two equation, by rule of cross multiplication we get

⸫ The eigen values are 0, 1, 2 and the eigen vectors are

Example 12

Find the eigen values and eigen vectors of the matrix

Solution

Let

The characteristic equation of A is

Where s1 = sum of the diagonal elements = 6

s2 = sum of the minors of diagonal elements of A

⸫ the characteristic equation is λ3 - 6 λ2 + 11λ − 6 = 0. By trail 1 is a root. The other roots are given by

λ3 - 5λ2 + 6 = 0 

⇒  (λ − 1)( λ − 3) = 0 ⇒ λ = 2, 3.

⸫ the eigen values of A are λ = 1, 2, 3.

To find eigen vectors:

Let X = be an eigen vector corresponding to eigen value λ

Case (i) If λ = 1, then the equations (1) become x1 + x3 = 0, x2 = 0 and x1 + x3 = 0 ⇒ x3 = -x1

Choose x1 = 1, then x3 = −1 ⸫ x1 = 1, x2 = 0, x3 = −1

Case (ii) If λ = 2, then the equations (I) become x3 = 0, 0.x2 = 0, x1 = 0.

So, x2 can take any value.

Choose x2 = 1

⸫ x1 = 0, x2 = 1, x3 = 0

⸫ an eigen vector is X2

Case (iii) If λ = 3, then the equations (I) become

-x1 + x3 = 0, -x2 = 0, x1 - x3 = 0

x3 = 1, x2 = 0

Choose x3 = 1, then x = 1

⸫  an eigen vector is

⸫ the eigen values of A are 1, 2, 3 and the corresponding eigen vectors are

Example 13

Find the eigen values and the eigen vectors of the matrix

Solution:

Case (i) If λ = -2, then (I) ⇒

3x1 + x2 + 3x3 = 0

x1 + 7x2 + x3 = 0

3x1 + x2 + 3x3 = 0

From the first two equations, by rule of cross multiplication, we get

Case (ii) If λ = 3 then (I)  ⇒ 2x1 + x2 + 3x3 = 0

                                               x1 + 2x2 + x3 = 0

                                                3x1 + x2 - 2x3 = 0

From first two equations, by rule of cross multiplication, we get

Case (iii) If λ = 6 then (1) ⇒ 5x1 + x2 + 3x3 = 0

                                               x1 - x2 + x3 = 0

                                                3x1 + x2 - 5x3 = 0

From the first two equations, by rule of cross multiplication, we get