Find an SVD of the indicated matrix. 3A =4

Answered: 3 A = 4

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

Find an SVD of the indicated matrix.

Expert Solution

Step 1

Let $A = [\begin{matrix} 3 \\ 4 \end{matrix}]$

To find singular value decomposition (SVD) of $A .$

The singular value decomposition of a matrix is factorization of matrix into the product of three matrices.

First find transpose of matrix $A$ .

$\Rightarrow A^{T} = [\begin{matrix} 3 & 4 \end{matrix}]$

Step 2

Consider $W = A A^{T}$

$\Rightarrow W = [\begin{matrix} 3 \\ 4 \end{matrix}] [\begin{matrix} 3 & 4 \end{matrix}]$

$\Rightarrow W = [\begin{matrix} 9 & 12 \\ 12 & 16 \end{matrix}]$

Now find the eigenvalues and eigenvectors of $W$ .

For that consider $|W - λ I| = 0$ .

$\Rightarrow |\begin{matrix} 9 - λ & 12 \\ 12 & 16 - λ \end{matrix}| = 0$

$\Rightarrow (9 - λ) (16 - λ) - 144 = 0$

$\Rightarrow λ^{2} - 25 λ = 0$

$\Rightarrow λ (λ - 25) = 0$

$\Rightarrow λ = 0, λ = 25$

Now we will find eigenvectors corresponding to each eigenvalue.

For $λ = 25$

Consider $W - 25 I = [\begin{matrix} - 16 & 12 \\ 12 & - 9 \end{matrix}]$

Perform $R_{1} = - \frac{R_{1}}{16}$

$\Rightarrow W - 25 I = [\begin{matrix} 1 & - \frac{3}{4} \\ 12 & - 9 \end{matrix}]$

Here $R_{2} = R_{2} - 12 R_{1}$

$\Rightarrow W - 25 I = [\begin{matrix} 1 & - \frac{3}{4} \\ 0 & 0 \end{matrix}]$

To find the null space solve the matrix equation $[\begin{matrix} 1 & - \frac{3}{4} \\ 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]$ .

If we take $x_{2} = t \in ℝ$ then $x_{1} = \frac{3 t}{4}$ .

Thus eigenvector corresponding to eigenvalue $λ = 25$ is $[\begin{matrix} \frac{3}{4} \\ 1 \end{matrix}] .$

For $λ = 0$

Consider $W - 0 I = [\begin{matrix} 9 & 12 \\ 12 & 16 \end{matrix}]$

Perform $R_{1} = \frac{R_{1}}{9}$

$\Rightarrow W - 0 I = [\begin{matrix} 1 & \frac{4}{3} \\ 12 & 16 \end{matrix}]$

Here $R_{2} = R_{2} - 12 R_{1}$

$\Rightarrow W - 0 I = [\begin{matrix} 1 & \frac{4}{3} \\ 0 & 0 \end{matrix}]$

To find the null space solve the matrix equation $[\begin{matrix} 1 & \frac{4}{3} \\ 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]$ .

If we take $x_{2} = t \in ℝ$ then $x_{1} = - \frac{4 t}{3}$ .

Thus eigenvector corresponding to eigenvalue $λ = 0$ is $[\begin{matrix} - \frac{4}{3} \\ 1 \end{matrix}] .$

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