Diagonalize a matrix 14 –14Diagonalize the following matrix and show each step: 0 26.L0 0 3

The given matrix is, A=144-14026003 So, calculating the eigen values of the given matrix A is,…

Answered: Г1 4 —14 Diagonalize the following…

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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Diagonalize a matrix

14 –14
Diagonalize the following matrix and show each step: 0 2
6.
L0 0 3

Expert Solution

Step 1

The given matrix is,

$A = [\begin{matrix} 14 & 4 & - 14 \\ 0 & 2 & 6 \\ 0 & 0 & 3 \end{matrix}]$

So, calculating the eigen values of the given matrix $A$ is,

$\begin{array}{rcl} |A - λ I| & = & 0 \\ |\begin{matrix} (1 - λ) & 4 & - 14 \\ 0 & (2 - λ) & 6 \\ 0 & 0 & (3 - λ) \end{matrix}| & = & 0 \\ (1 - λ) [(2 - λ) (3 - λ) - 0] - 4 (0 - 0) - 14 (0 - 0) & = & 0 \\ (λ - 1) (λ - 2) (λ - 3) & = & 0 \\ λ & = & 1, 2 or 3 \end{array}$

So, the eigen values are 1, 2 and 3.

Step 2

Now taking for eigen value, $λ = 1$

$\begin{array}{rcl} A - λ I & = & [\begin{matrix} 1 & 4 & - 14 \\ 0 & 2 & 6 \\ 0 & 0 & 3 \end{matrix}] - 1 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \\ = & [\begin{matrix} 0 & 4 & - 14 \\ 0 & 1 & 6 \\ 0 & 0 & 2 \end{matrix}] \end{array}$

Now reducing the obtained matrix.

$\begin{array}{rcl} A - λ I & = & [\begin{matrix} 0 & 4 & - 14 \\ 0 & 1 & 6 \\ 0 & 0 & 2 \end{matrix}] \\ = & [\begin{matrix} 0 & 1 & \frac{- 7}{2} \\ 0 & 0 & \frac{19}{2} \\ 0 & 0 & 2 \end{matrix}] \begin{matrix} R_{1} \to \frac{R_{1}}{4} \\ R_{2} \to R_{2} - R_{1} \end{matrix} \\ = & [\begin{matrix} 0 & 1 & \frac{- 7}{2} \\ 0 & 0 & 1 \\ 0 & 0 & 2 \end{matrix}] R_{2} \to R_{2} \times \frac{2}{19} \\ = & [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 2 \end{matrix}] R_{1} \to R_{1} + \frac{7}{2} R_{2} \\ = & [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}] R_{3} \to R_{3} - 2 R_{2} \\ (A - λ I) [\begin{matrix} X_{1} \\ X_{2} \\ X_{3} \end{matrix}] & = & [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} X_{1} \\ X_{2} \\ X_{3} \end{matrix}] \\ = & [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}] \end{array}$

Step 3

Therefore we can write,

$\begin{array}{rcl} X_{2} & = & 0, X_{3} = 0 \\ So, that \\ V & = & [\begin{matrix} X_{1} \\ 0 \\ 0 \end{matrix}] \\ Let, us consider X_{1} & = & 1 \\ V & = & [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}] \end{array}$

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