Orthogonal Transformation of a Symmetric Matrix to Diagonal Form: Example Solved Problems

ORTHOGONAL TRANSFORMATION OF A SYMMETRIC MATRIX TO DIAGONAL FORM

WORKED EXAMPLES

 

Example 1

Find an orthogonal matrix P such that P^TAP is diagonal matrix where A = .

Solution:

Given that A = 

The characteristic polynomial of A is |A− λI| = 0

⇒ (1 −λ) [(1 − λ) (5 − λ) − 4] − 0[0 (5 – λ) + 2]  −1[0 (2)+1(1− λ)] = 0

(1 − λ) [5 − λ − 5λ + λ² − 4] − 1 [1 − λ] = 0

 (1 − λ) [λ² − 6λ + 1] −1 [1‒ λ] = 0

 λ² − 6λ + 1 − λ³ + 6λ² − λ −1 + λ =0

 −λ³ + 7λ² − 6λ = 0

⇒ λ³ − 7λ² + 6λ = 0

==>

λ=0 (or) λ²−7λ + 6 = 0

 λ=0 (or) (λ−1) (λ − 6) = 0

  The eigenvalues are λ = 0, 1, 6.

To find the eigenvectors

Let us consider [ A− λI ]X=0 where 

 (1−λ)x₁+0x₂−x₃=0

0x₁ + (1‒λ)x₂ + 2x₃ = 0

 − x₁+2x₂+(5−λ) x₃ = 0

                      ... (1)

Case (i)

When λ=0, the system of equations (1) becomes

 x₁+0x₂−x₃=0

 0x₁ + x₂+2x₃=0

 −x₁ + 2x₂+5x₃ = 0

Here all the three equations are different. Hence we can take any two of these three equations.

  Let us consider first two equations.

 x₁+0x₂−x₃=0

 0x₁ + x₂ + 2x₃ = 0

The corresponding eigenvector is 

Case (ii)

When λ= 1, the system of equations (1) becomes

 0x₁+0x₂−x₃=0

 0x₁+0x₂+2x₃=0

 − x₁ + 2x₂+4x₃ = 0

Let us consider the last two equations.

 0x₁+0x₂+2x₃ = 0

  ‒x₁ + 2x₂ + 4x₃ = 0

The corresponding eigenvector is 

Case (iii)

When λ=6, the system of equations (1) becomes

 −5x₁+0x₂−x₃=0

 0x₁‒5x₂+2x₃=0

−x₁+2x₂−x₃=0

Now let us consider the first two equations

 −5x₁+0x₂−x₃ = 0

 0x₁‒5x₂+2x₃=0

The corresponding eigenvector is

  The three eigenvectors are respectively

 These three eigenvectors are orthogonal.

Now || X₁ || = √ [(1)² + (− 2)² + (1)² ] = √[1 + 4+1] = √6

|| X₂ || = √[(2)² + (1)² + (0)² ] = √[4 + 1 + 0] = √5

|| X₃ || = √ [ (− 1)² + (2)² + (5)² ] = √[1 +4 + 25] = √30

 || X₁ ||² = 6, ||X₂||² = 5 and ||X₃||² = 30

We have to compute the values of P^TAP = D

where D is the diagonalized matrix.

 

Example 2

Orthogonally diagonalize the given symmetric matrix

Solution:

Given that A = 

The characteristic equation is |A− λI| = 0

 |A− λI| = 

 (8− λ) [(5− λ) (5− λ)−16] + 2 [−2 (5− λ)−8]+2 [−8−2 (5− λ)]=0

 (8−λ) [25 + λ² − 10λ − 16] + 2 [−10+2λ−8]+2 [−8−10+2λ] = 0

 (8−λ) [λ²−10λ +9]+2 [2λ−18]+2 [2λ−18]=0

 8λ²−80λ+72− λ³ +10λ²−9λ+4λ−36+4λ−36=0

 − λ³ + 18λ²−81λ = 0

  λ³−18λ²+81λ=0

 λ[λ² − 18λ +81]=0

  λ=0 (or) λ²−18λ +81=0

  λ=0 (or) (λ‒9) (λ−9)=0 .

The eigenvalues are λ=0, 9, 9

To find the eigenvectors:

Let us consider [A− λI]X=0

 (8−λ)x₁−2x₂ + 2x₃ = 0

 ‒2x₁+(5−λ)x₂+4x₂ = 0

 2x₁ + 4x₂ + (5−λ)x₃ = 0

                           ... (1)

Case (i)

When λ=0, the system of equation (1) becomes

 8x₁‒2x₂+2x₃=0

−2x₁ +5x₂+4x₃ = 0

 2x₁ + 4x₂ + 5x₃ = 0

Let us consider the first two equations

8x₁ − 2x₂+2x₃ = 0

−2x₁ +5x₂ + 4x₃ = 0

  The corresponding eigenvector is 

Case (ii)

 When λ=9, the system of equation (1) becomes

 −x₁−2x₂+2x₃=0

 −2x₁−4x₂ + 4x₃ = 0

 2x₁ + 4x₂ − 4x₃ = 0

Here all the three equations are similar

 x₁+2x₂−2x₃=0

Let us put x₁ = 0 and x₂ = 1.

If we put x₁ = 0 and x₂=1 then 0+2 −2x₃ = 0

⇒     2x₃=2     ⇒ x₃ = 1

Hence

 x₁ =0, x₂ = 1 and x₃ = 1

 The corresponding eigenvector is 

Case (iii)

 When λ=9, the system of equation (1) has again a similar equation x₁+2x₂‒2x₃=0

 Let  which is orthogonal to X₁ and X₂.

  .−l−2m+2n=0 and 0l+m+n=0

Solving these two equations;

These three eigenvectors are orthogonal.

|| X₁ || = √[(− 1)² + (− 2)² +(2)² ] = √[1 +4 +4] = √9=3

|| X₂ || = √[ (0)² + (1)² + (1)² ] = √[0 + 1 + 1] = √2

||X₂|| = √[ (−4)² + (1)² + (− 1)² ] = √[ 16 + 1 + 1] = √18

  || X₁ ||² = 9,  || X₂||² = 2 and || X₃||² = 18

 

Example 3

For matrix A =  find an orthogonal matrix P such that P^TAP is diagonal.

Solution:

Let A = 

The characteristic polynomial is |A−λI|=0

On expanding the determinant we get λ³ – 18λ² + 45λ = 0

 λ[λ² − 18λ +45]=0

 λ=0 (or) λ²−18λ +45 = 0

  λ=0 (or) (λ −3) (λ −15) = 0

 The eigenvalues are λ=0, 3, 15

To find the eigenvectors

The eigenvectors are given by [A−λI] X=0

 (8−λ)x₁−6x₂+2x₃=0

 −6x₁+(7−λ)x₂‒4x₃=0

 2x₁‒4x₂ + (3− λ)x₃ = 0

                  ………(1)

Case (i)

When λ=0, the system of equations (1) becomes

 8x₁−6x₂+2x₃=0

 −6x₁ + 7x₂ − 4x₃ = 0

 2x₁4x₂+3x₃ = 0

By solving any two of these equations, we get

x₁/5 = x₂/10 = x₃/10

 The corresponding eigenvector is X₁ = 

Case (ii)

 When λ=3, the system of equations (1) becomes

 5x₁−6x₂+2x₃ = 0

 −6x₁+4x₂− 4x₃ = 0

 2x₁−4x₂ +0x₃ = 0

Solving any two of these equations we get

x₁/2 = x₂/1 = x₃/‒2

 The eigenvector corresponding to λ=3 is X₂: 

Case (iii)

When λ= 15, the system of equations (1) becomes

−7x₁−6x₂+2x₃=0

 −6x₁−8x₂−4x₃=0

 2x₁−4x₂−12x₃=0

By solving any two of these equations, we get

x₁/2 = x₂/‒2 = x₃/1

 The eigenvector corresponding to λ= 15 is X₃ = 

 The eigenvectors are orthogonal.

|| X₁ || = √[1² + 2² + 2²] = √[1 +4 +4] = √9 = 3

|| X₂ || = √[ 2² + 1² + (−2)² ] = √[4+1+4] = √9=3

|| X₃ || = √[2² + (−2)² + 1²] = √[4 +4 + 1] = √9 = 3

 || X₁ ||² = 9, || X₂ ||² = 9 and || X₃ ||² = 9

 

Example 4

For the matrix A =  find an orthogonal matrix P  such that P^TAP is diagonal.

Solution:

Let A = 

The characteristic polynomial is |A−λI} = 0

 |A ‒ λI| = 

On expanding the above determinant we have

 λ³ – 9λ² + 24λ − 16 = 0

By solving this equation,

The eigenvalues are λ = 1, 4, 4

To find the eigenvectors:

The eigenvectors are given by [A−λI]X=0

 (3−λ)x₁ + x₂ + x₃ = 0

 x₁+ (3−λ)x₂ − x₃ = 0

  x₁−x₂ + (3−λ)x₃ = 0

                    ... (1)

Case (i)

When λ= 1, the system of equations (1) becomes

2x₁+x₂+x₃=0

x₁+2x₂−x₃=0

x₁−x₂+2x₃=0x

Solving the first two equations, we get

x₁/‒3 = x₂/3 = x₃/3

 The eigenvector corresponding λ=1 is X₁ = 

Case (ii)

When λ=4, the system of equations (1) becomes

−x₁+x₂+x₃=0

x₁−x₂−x₃=0

x₁−x₂−x₃=0

All the equations are similar x₁ − x₂ − x₃ = 0

Put x₂ = 1 and x₃ = 0, then we get x₁ = 1.

 The eigenvector corresponding to λ=4 is X₂ = 

Case (iii)

When λ=4, again we have the same similar equation x₁ − x₂ − x₃ = 0

Let X₃ = 

Since X₃ is orthogonal to X₂, we get 1+m+0n = 0

Since X₃ satisfies l−m−n=0. Solving these two equations

We get l/‒1 = m/1 = n/‒2

 The eigenvector corresponding to λ=4 is X₃ = 

Hence X₁^TX₂=0, X₂^TX₃=0 and X₃^TX₁ =0.

All the three eigenvectors are orthogonal.

|| X₁ || = √[(− 1)² + 1² + 1² ] = √3

|| X₂ || = √[ 1² + 1² + 0² ] = √2

|| X₃ || = √[ (−1)² + 1² + (−2)² ] = √6

 || X₁ ||² = 3, || X₂ ||² = 2 and || X₃ ||² = 6.

 

Example 5

For the matrix A = , find an orthogonal matrix P such that P^TAP is diagonal.

Solution:

It is given that A= 

The characteristic polynomial is |A−λI|=0

On expanding the above determinant, we get

 λ³−11λ² +36λ−36=0

⇒ (λ−2) (λ−3) (λ−6)=0

  The eigenvalues are λ=2, 3, 6.

To find the eigenvectors:

Let us consider [A−λI] X=0

 (3−λ)x₁−x₂+x₃=0

 −x₁+(5−λ)x₂−x₃=0

 x₁‒x₂+(3−λ)x₃=0

                      ...(1)

Case (i)

When λ=2, the system of equations (1) becomes

 x₁‒x₂+x₃=0

 −x₁+3x₂−x₃=0

 x₁‒x₂+x₃=0

Solving the first two equations we get

 x₁/‒2 = x₂/0 = x₃/2

 The eigenvector corresponding to λ=2 is X₁ = 

Case (ii)

When λ=3, the system of equations (1) becomes

 0x₁−x₂+x₃=0

 −x₁+2x₂−x₃=0

 x₁−x₂+0x₃=0

From the first and last equations, we have x₁ = x₂=x₃.

Let x₁ = 1, x₂=1 and x₃ = 1 (say)

Then the eigenvector when λ=3 is X₂ = 

Case (iii)

When λ=6, the system of equations (1) becomes

−3x₁x₂+x₃=0

−x₁−x₂−x₃=0

x₁−x₂−3x₃=0

Solving these equations, we get X₃ = 

The three eigenvectors are orthogonal

|| X₁ || = √ [1² + 0² + (−1)² ] = √2

|| X₂ || = √[1² + 1² + 1²] = √3

|| X₃ || = √[1² + ( − 2)² + 1²] = √6

 || X₁ ||² = 2, || X²||² = 3 and || X₃ ||² = 6

P^TAP = Diagonal matrix

 

Example 6

Find the principal axes for the quadratic form q= x²₁+x²₂−4x₁x₂.

Solution:

In order to use the concept of diagonalization, we first express q in matrix form.

The quadratic form can be converted into matrix form with the following formula.

Here the obtained matrix A is symmetric.

Now let us find the eigenvalues of the obtained matrix

We know that the characteristic polynomial is |A−λI|=0

 |A – λI| =  = (1 − λ) (1 − A) ‒ 4 = 0

 (1−λ)²−4=0

 (1+λ²−2λ)−4=0

λ²−2λ−3=0

 (λ+1) (λ−3)=0

 The eigenvalues are λ=−1 and 3

To find the eigenvectors:

The eigenvectors are obtained as [A−λI] X=0, where 

 (1 − λ) x₁ − 2x₂ = 0

 −2x₁+ (1 − λ) x₂ = 0

               …………..(1)

Case (i)

When λ=−1, the above system of equations (1) becomes

2x₁−2x₂=0

 ‒2x₁ + 2x₂ = 0

⇒ x₁=x₂

If we put x₁ = 1, then x₂ = 1.

 The eigenvector when λ=−1 is 

Case (ii)

When λ=3, the above system of equations (1) becomes

− 2x₁−2x₂ = 0

−2x₁−2x₃=0

⇒ x₁ = x₂

If we put x₁ = 1, then x₂ =−1.

 The eigenvector when λ=3 is 

These two eigenvectors are orthogonal.

|| X₁ || = √2 and || X₂ ||=√2

⇒ || X₁ ||² = 2 and || X₂ ||² = 2

 q = ‒y₁² + 3y₂²

 

Example 7

Find the principal axes for the quadratic form q = 10x²₁ + 2x²₂ + 5x²₃ + 6x₂x₃ − 10x₁x₃ − 4x₁x₂

Solution:

The given quadratic form is

 q=10x²₁+2x²₂+5x²₃ + 6x₂x₃ − 10x₁x₃ − 4x₁x₂

To transform the quadratic form to matrix form, we use the formula

The matrix A is a real symmetric matrix.

To find the eigenvalues

The characteristic polynomial is | A−λI |=0.

On expanding the determinant we get λ³ − 17λ² + 42λ = 0.

By solving this equation, we have λ = 0, 3, 14.

To find the eigenvectors

Let us consider [A− λI] X = 0 to find the eigenvectors.

(10−λ)x₁−2x₂−5x₃=0

−2x₁ + (2−λ)x₂ + 3x₃ = 0

−5x₁+3x₂+(5−λ)x₃=0

Case (i)

When λ=0, the equations (1) become

10x₁ − 2x₂−5x₃=0

−2x₁+2x₂+3x₃=0x

−5x₁+3x₂+5x₃=0

Solving the first two equations, we get x₁ = 1, x₂=−5 and x₃ = 4.

 The eigenvector for λ=0 is X₁ = 

Case (ii)

When λ=3, the equations (1) become

7x₁‒2x₂−5x₃=0

−2x₁−x₂+3x₃=0

−5x₁+3x₂+2x₃ = 0

By solving the last two equations, we get x₁ = −11, x₂= −11 and x₃ = −11

 The eigenvector for λ=3 is X₂ = 

Case (iii)

When λ= 14, the equations (1) become,

 −4x₁−2x₂−5x₃=0

  −2x₁−12x₂+3x₃=0

 −5x₁ +3x₂−9x₃=0

By solving the first and last equations, we get

 x₁ = 33, x₂ = −11 and x₃=−22,

 The eigenvector for λ= 14 is X₃ = 

The eigenvectors are orthogonal pairwise

|| X₁ || = √[1² + (− 5)² +4²] = √[1 +25 + 16] = √42

|| X₂ || = √[1² + 1² + 1²] = √3

|| X₃ || = √[3² + (− 1)² + (− 2)²] = √[9 +1 + 4] = √14

This is equivalent to X=PY (since P⁻¹=PT)

In terms of y₁, y₂, y₃, q takes the form, q=Y^T(D) Y

 q = 0y²₁ + 3y²₂ + 14y²₃