Chapter 12.4, Problem 1CP

Use MATLAb’S svd command to find the best rank-one approximation of the following matrices:

$\begin{array}{l}\text{a}\text{. }\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]\\ \text{b}\text{. }\left[\begin{array}{cc}1& 4\\ 2& 3\end{array}\right]\\ \text{c}\text{. }\left[\begin{array}{ccc}1& 2& 4\\ 1& 3& 3\\ 0& 0& 1\end{array}\right]\\ \text{d}\text{. }\left[\begin{array}{ccc}1& 5& 3\\ 2& -3& 2\\ -3& 1& 1\end{array}\right]\end{array}$

To determine

Use MATLAB’s svd command to find the best rank-one approximation of the given matrx.

Answer to Problem 1CP

The best rank-one approximation of the matrix $A=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$ is $\left[\begin{array}{cc}1.1707& 1.8944\\ 1.8944& 3.0656\end{array}\right]$

Explanation of Solution

Given Information: $\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$

Calculation:

Use commands ‘A=[1 2;2 3]’ to define the matrix A and ‘[U S V]=svd(A)’ to find the singular value decomposition (SVD) of the matrix A in the MATLAB.

  

  

Therefore,

$u_1=\left[\begin{array}{c}-0.5257\\ -0.8507\end{array}\right]$

$s_1=4.2361$

$v_1=\left[\begin{array}{c}-0.5257\\ -0.8507\end{array}\right]$

The first rank-one matrix is $s_1{u}_1{v}_1^T$ .

  

  

Thus, the best rank-one approximation of the matrix $A=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$ is $\left[\begin{array}{cc}1.1707& 1.8944\\ 1.8944& 3.0656\end{array}\right]$ .

To determine

Use MATLAB’s svd command to find the best rank-one approximation of the given matrx.

Answer to Problem 1CP

The best rank-one approximation of the matrix $A=\left[\begin{array}{cc}1& 4\\ 2& 3\end{array}\right]$ is $\left[\begin{array}{cc}1.1707& 1.8944\\ 1.8944& 3.0656\end{array}\right]$

Explanation of Solution

Given Information: $\left[\begin{array}{cc}1& 4\\ 2& 3\end{array}\right]$

Calculation:

Use commands ‘A=[1 4;2 3]’ to define the matrix A and ‘[U S V]=svd(A)’ to find the singular value decomposition (SVD) of the matrix A in the MATLAB.

  

  

Therefore,

$u_1=\left[\begin{array}{c}-0.7555\\ -0.6552\end{array}\right]$

$s_1=5.3983$

$v_1=\left[\begin{array}{c}-0.3827\\ -0.9239\end{array}\right]$

The first rank-one matrix is $s_1{u}_1{v}_1^T$ .

  

  

Thus, the best rank-one approximation of the matrix $A=\left[\begin{array}{cc}1& 4\\ 2& 3\end{array}\right]$ is $\left[\begin{array}{cc}1.5608& 3.7680\\ 1.3536& 3.2678\end{array}\right]$ .

To determine

Use MATLAB’s svd command to find the best rank-one approximation of the given matrx.

Answer to Problem 1CP

The best rank-one approximation of the matrix $A=\left[\begin{array}{ccc}1& 2& 4\\ 1& 3& 3\\ 0& 0& 1\end{array}\right]$ is $\left[\begin{array}{ccc}1.0105& 2.5122& 3.6435\\ 0.9551& 2.3745& 3.4438\\ 0.1786& 0.4440& 0.6440\end{array}\right]$

Explanation of Solution

Given Information: $\left[\begin{array}{ccc}1& 2& 4\\ 1& 3& 3\\ 0& 0& 1\end{array}\right]$

Calculation:

Use commands ‘A=[1 2 4;1 3 3;0 0 1]’ to define the matrix A and ‘[U S V]=svd(A)’ to find the singular value decomposition (SVD) of the matrix A in the MATLAB.

  

  

Therefore,

$u_1=\left[\begin{array}{c}-0.7208\\ -0.6813\\ -0.1274\end{array}\right]$

$s_1=6.2980$

$v_1=\left[\begin{array}{c}-0.2226\\ -0.5534\\ -0.8026\end{array}\right]$

The first rank-one matrix is $s_1{u}_1{v}_1^T$ .

  

  

Thus, the best rank-one approximation of the matrix $A=\left[\begin{array}{ccc}1& 2& 4\\ 1& 3& 3\\ 0& 0& 1\end{array}\right]$ is $\left[\begin{array}{ccc}1.0105& 2.5122& 3.6435\\ 0.9551& 2.3745& 3.4438\\ 0.1786& 0.4440& 0.6440\end{array}\right]$ .

To determine

Use MATLAB’s svd command to find the best rank-one approximation of the given matrx.

Answer to Problem 1CP

The best rank-one approximation of the matrix $A=\left[\begin{array}{ccc}1& 5& 3\\ 2& -3& 2\\ -3& 1& 1\end{array}\right]$ is $\left[\begin{array}{ccc}-0.5141& 5.2343& 1.9953\\ 0.2070& -2.1073& -0.8033\\ 0.1425& -1.4507& -0.5530\end{array}\right]$

Explanation of Solution

Given Information: $\left[\begin{array}{ccc}1& 5& 3\\ 2& -3& 2\\ -3& 1& 1\end{array}\right]$

Calculation:

Use commands ‘A=[1 5 3;2 -3 2;-3 1 1]’ to define the matrix A and ‘[U S V]=svd(A)’ to find the singular value decomposition (SVD) of the matrix A in the MATLAB.

  

  

Therefore,

$u_1=\left[\begin{array}{c}0.8984\\ -0.3617\\ -0.2490\end{array}\right]$

$s_1=6.2614$

$v_1=\left[\begin{array}{c}-0.0914\\ 0.9305\\ 0.3547\end{array}\right]$

The first rank-one matrix is $s_1{u}_1{v}_1^T$ .

  

  

Thus, the best rank-one approximation of the matrix $A=\left[\begin{array}{ccc}1& 5& 3\\ 2& -3& 2\\ -3& 1& 1\end{array}\right]$ is $\left[\begin{array}{ccc}-0.5141& 5.2343& 1.9953\\ 0.2070& -2.1073& -0.8033\\ 0.1425& -1.4507& -0.5530\end{array}\right]$ .