Chapter 6, Problem 1E

The top matrix on the menu is the diagonal matrix
$A=\left(\begin{array}{cc}\frac{5}{4}& 0\\ 0& \frac{3}{4}\end{array}\right)$
Initially, when you select this matrix, the vectors x and Ax should both be aligned along the positive x-axis. What information about an eigenvalue-eigenvector pair is apparent from the initial figure positions? Explain. Rotate x counterclockwise until x and Ax are parallel, that is, until they both lie along the same line through the origin. What can you conclude about the second eigenvalue-eigenvector pair? Repeat this experiment with thesecond matrix. How can you determine the eigen-values and eigenvectors of a $2\times 2$ diagonal matrix by inspection without doing any computations? Does this also work for $3\times 3$ diagonal matrices? Explain.

To determine

To find:

By rotate x counterclockwise until x and Ax are parallel, the eigenvalues and eigenvectors of a $2\times 2$ diagonal matrix is determined.

Answer to Problem 1E

In first figure, the eigenvalue is $\frac{5}{4}$ and the eigenvector is $\left(\begin{array}{c}1\\ 0\end{array}\right)$

In second figure, the eigenvalue is $\frac{3}{4}$ and the eigenvector is $\left(\begin{array}{c}0\\ 1\end{array}\right)$

In third figure the eigenvalue is $4$ and the eigenvector is $\left(\begin{array}{c}1\\ 0\end{array}\right)$

In third figure, the eigenvalue is $6$ and the eigenvector is $\left(\begin{array}{c}0\\ 1\end{array}\right)$

The eigenvalues are $\pm \left|Ax\right|$ and the eigenvectors are $\left\{\left(\begin{array}{c}1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\end{array}\right)\right\}$ for a $2\times 2$ diagonal matrix.

The eigenvectors are $\left\{\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right\}$ for a $3\times 3$ diagonal matrix

Explanation of Solution

Given:

The diagonal matrix

  $A=\left(\begin{array}{cc}\frac{5}{4}& 0\\ 0& \frac{3}{4}\end{array}\right)$

the vectors x and Ax should both be aligned along the positive x-axis.

Calculation:

Now using the eigshow command in MATLAB, we draw the vectors “x” and “Ax” as follows:

  $\begin{array}{l}>>A=[5/4,0;0,3/4]\\ >>eigshow(A)\end{array}$

  

From the figure, we see that the vectors x and Ax both are aligned along the positive x-axis.

As both the vectors are parallel, one vector is multiple of the other.

Therefore,

  $Ax=\lambda x$ [ $\lambda$ is the eigenvalue and $\lambda >0$ ]

  $\begin{array}{l}\left|Ax\right|=\left|\lambda x\right|\\ \left|Ax\right|=\left|\lambda \right|\left|x\right|\\ \left|\lambda \right|=\frac{\left|Ax\right|}{\left|x\right|}\end{array}$

From the figure above we obtain as:

  $\left|Ax\right|=1.25$ and $\left|x\right|=1$

Therefore,

  $\begin{array}{l}\left|\lambda \right|=\frac{\left|Ax\right|}{\left|x\right|}\\ =\frac{1.25}{1}\\ =1.25\\ =\frac{5}{4}\end{array}$

Thus for this position the eigenvalue is $\frac{5}{4}$ and the eigenvector is $\left(\begin{array}{c}1\\ 0\end{array}\right)$

Now, rotate x counterclockwise until x and Ax are parallel as follows:

  

Again, as both the vectors are parallel, one vector is multiple of the other.

Hence,

  $Ax=\lambda x$ [ $\lambda$ is the eigenvalue and $\lambda >0$ ]

  $\begin{array}{l}\left|Ax\right|=\left|\lambda x\right|\\ \left|Ax\right|=\left|\lambda \right|\left|x\right|\\ \left|\lambda \right|=\frac{\left|Ax\right|}{\left|x\right|}\end{array}$

From the second figure above it is obtain as:

  $\left|Ax\right|=0.75$ and $\left|x\right|=1$

Therefore,

  $\begin{array}{l}\left|\lambda \right|=\frac{\left|Ax\right|}{\left|x\right|}\\ =\frac{0.75}{1}\\ =0.75\\ =\frac{3}{4}\end{array}$

Thus for this position the eigenvalue is $\frac{3}{4}$ and the eigenvector is $\left(\begin{array}{c}0\\ 1\end{array}\right)$

Let the matrix:

  $A=\left(\begin{array}{cc}4& 0\\ 0& 6\end{array}\right)$

Now using the eigshow command in MATLAB, we draw the vectors x and Ax as follows:

  $\begin{array}{l}>>A=[4,0;0,6]\\ >>eigshow(A)\end{array}$

  

Since both the vectors are parallel, we can say that one vector is multiple of the other.

So,

  $Ax=\lambda x$ [ $\lambda$ is the eigenvalue and $\lambda >0$ ]

  $\begin{array}{l}\left|Ax\right|=\left|\lambda x\right|\\ \left|Ax\right|=\left|\lambda \right|\left|x\right|\\ \left|\lambda \right|=\frac{\left|Ax\right|}{\left|x\right|}\end{array}$

From the figure above we obtain as:

  $\left|Ax\right|=4$ and $\left|x\right|=1$

Therefore,

  $\begin{array}{l}\left|\lambda \right|=\frac{\left|Ax\right|}{\left|x\right|}\\ =\frac{4}{1}\\ =4\end{array}$

Thus for this position the eigenvalue is $4$ and the eigenvector is $\left(\begin{array}{c}1\\ 0\end{array}\right)$

Now, we rotate x counterclockwise until x and Ax are parallel as follows:

  

Again, as both the vectors are parallel, one vector is multiple of the other.

Hence,

  $Ax=\lambda x$ [ $\lambda$ is the eigenvalue and $\lambda >0$ ]

  $\begin{array}{l}\left|Ax\right|=\left|\lambda x\right|\\ \left|Ax\right|=\left|\lambda \right|\left|x\right|\\ \left|\lambda \right|=\frac{\left|Ax\right|}{\left|x\right|}\end{array}$

From the second figure above it is obtain as:

  $\left|Ax\right|=6$ and $\left|x\right|=1$

Hence,

  $\begin{array}{l}\left|\lambda \right|=\frac{\left|Ax\right|}{\left|x\right|}\\ =\frac{6}{1}\\ =6\end{array}$

Thus for this position the eigenvalue is $6$ and the eigenvector is $\left(\begin{array}{c}0\\ 1\end{array}\right)$

The eigenvalues are $\pm \left|Ax\right|$ and the eigenvectors are $\left\{\left(\begin{array}{c}1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\end{array}\right)\right\}$ for a $2\times 2$ diagonal matrix without doing any computation.

This experiment also works for a $3\times 3$ diagonal matrix. The eigenvalues are the diagonal entries of the matrix and the eigenvectors are $\left\{\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right\}$ for a $3\times 3$ diagonal matrix without doing any computation.

Conclusion:

In first figure, the eigenvalue is $\frac{5}{4}$ and the eigenvector is $\left(\begin{array}{c}1\\ 0\end{array}\right)$

In second figure, the eigenvalue is $\frac{3}{4}$ and the eigenvector is $\left(\begin{array}{c}0\\ 1\end{array}\right)$

In third figure the eigenvalue is $4$ and the eigenvector is $\left(\begin{array}{c}1\\ 0\end{array}\right)$

In third figure, the eigenvalue is $6$ and the eigenvector is $\left(\begin{array}{c}0\\ 1\end{array}\right)$

The eigenvalues are $\pm \left|Ax\right|$ and the eigenvectors are $\left\{\left(\begin{array}{c}1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\end{array}\right)\right\}$ for a $2\times 2$ diagonal matrix.

the eigenvectors are $\left\{\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right\}$ for a $3\times 3$ diagonal matrix