Answered: 1 1 1 3 1 0 0 1 -1 1 1 | -1 -O -/2 1/3…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

A diagonalisation of a matrix A is given in the form P-1AP=D.List the eigenvalues of A and the bases for the corresponding eigenspaces

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Expert Solution

Step 1

A diagonalization of a matrix A is given in the form $P^{- 1} A P = D .$

Here $P^{- 1} = [\begin{matrix} \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\ \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} \\ \frac{1}{3} & \frac{1}{3} & - \frac{2}{3} \end{matrix}], A = [\begin{matrix} 1 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}], P = [\begin{matrix} 3 & 1 & 0 \\ 1 & - 1 & 1 \\ 2 & 0 & - 1 \end{matrix}], D = [\begin{matrix} 2 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & - 1 \end{matrix}]$

To find the eigenvalues of $A$ and the bases for corresponding eigenspaces.

Step 2

Here $A = [\begin{matrix} 1 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$

To find eigenvalues let us consider $|A - λ I| = 0 .$

$\Rightarrow |\begin{matrix} 1 - λ & 1 & 1 \\ 0 & - λ & 1 \\ 1 & 1 & - λ \end{matrix}| = 0$

Here add column $3$ multiplied by $λ$ to column $2 : C_{2} = C_{2} + λ C_{3}$ .

$\Rightarrow |\begin{matrix} 1 - λ & λ + 1 & 1 \\ 0 & 0 & 1 \\ 1 & 1 - λ^{2} & - λ \end{matrix}| = 0$

Now expand along row $2,$

$\Rightarrow (0) {(- 1)}^{2 + 1} |\begin{matrix} λ + 1 & 1 \\ 1 - λ^{2} & - λ \end{matrix}| + (0) {(- 1)}^{2 + 2} |\begin{matrix} 1 - λ & 1 \\ 1 & - λ \end{matrix}| + (1) {(- 1)}^{2 + 3} |\begin{matrix} 1 - λ & λ + 1 \\ 1 & 1 - λ^{2} \end{matrix}| = 0$

$\Rightarrow - [(1 - λ) (1 - λ^{2}) - (λ + 1)] = 0$

$\Rightarrow - [1 - λ^{2} - λ + λ^{3} - λ - 1] = 0$

$\Rightarrow - [- λ^{2} - 2 λ + λ^{3}] = 0$

$\Rightarrow - λ^{3} + λ^{2} + 2 λ = 0$

$\Rightarrow λ (- λ^{2} + λ + 2) = 0$

$\Rightarrow λ (- λ^{2} + 2 λ - λ + 2) = 0$

$\Rightarrow λ (- λ (λ - 2) - 1 (λ - 2)) = 0$

$\Rightarrow λ (λ - 2) (- λ - 1) = 0$

$\Rightarrow λ = 0, λ = - 1, λ = 2$

Thus eigenvalues of a matrix $A$ are $- 1, 0, 2 .$

Now to find bases for corresponding eigenspaces.

For $λ = 2$

$\Rightarrow A - 2 I = [\begin{matrix} - 1 & 1 & 1 \\ 0 & - 2 & 1 \\ 1 & 1 & - 2 \end{matrix}]$

Perform $R_{1} = - R_{1}$

$\Rightarrow A - 2 I = [\begin{matrix} 1 & - 1 & - 1 \\ 0 & - 2 & 1 \\ 1 & 1 & - 2 \end{matrix}]$

Now $R_{3} = R_{3} - R_{1}$

$\Rightarrow A - 2 I = [\begin{matrix} 1 & - 1 & - 1 \\ 0 & - 2 & 1 \\ 0 & 2 & - 1 \end{matrix}]$

Here $R_{2} = - \frac{R_{2}}{2}$

$\Rightarrow A - 2 I = [\begin{matrix} 1 & - 1 & - 1 \\ 0 & 1 & - \frac{1}{2} \\ 0 & 2 & - 1 \end{matrix}]$

Now $R_{1} = R_{1} + R_{2}$

$\Rightarrow A - 2 I = [\begin{matrix} 1 & 0 & - \frac{3}{2} \\ 0 & 1 & - \frac{1}{2} \\ 0 & 2 & - 1 \end{matrix}]$

Here $R_{3} = R_{3} - 2 R_{2}$

$\Rightarrow A - 2 I = [\begin{matrix} 1 & 0 & - \frac{3}{2} \\ 0 & 1 & - \frac{1}{2} \\ 0 & 0 & 0 \end{matrix}]$

This is recued row echelon form of matrix.

To find the null space, solve the matrix equation $[\begin{matrix} 1 & 0 & - \frac{3}{2} \\ 0 & 1 & - \frac{1}{2} \\ 0 & 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$ .

If we take $x_{3} = t \in ℝ$ then $x_{1} = \frac{3 t}{2}, x_{2} = \frac{t}{2}$

$\Rightarrow x = [\begin{matrix} \frac{3 t}{2} \\ \frac{t}{2} \\ t \end{matrix}] = [\begin{matrix} \frac{3}{2} \\ \frac{1}{2} \\ 1 \end{matrix}] t .$

Thus basis for the eigenspace corresponding to eigenvalue $λ = 2$ is $\{[\begin{matrix} \frac{3}{2} \\ \frac{1}{2} \\ 1 \end{matrix}]\} .$

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