Question

M] In Exercises 19 and 20, find (a) the largest eigenvalue and (b) the eigenvalue closest to zero. In each case, set $${\mathbf{x}}_{\mathbf{0}}\mathbf{=}\left(\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\right)$$ and carry out approximations until the approximating sequence seems accurate to four decimal places. Include the approximate eigenvector.

19.$$A\mathbf{=}\left[\begin{array}{cccc}\mathbf{10}& \mathbf{7}& \mathbf{8}& \mathbf{7}\\ \mathbf{7}& \mathbf{5}& \mathbf{6}& \mathbf{5}\\ \mathbf{8}& \mathbf{6}& \mathbf{10}& \mathbf{9}\\ \mathbf{7}& \mathbf{5}& \mathbf{9}& \mathbf{10}\end{array}\right]$$

Short answer

1. The largest eigenvalue is$$30.288685$$.

An estimate for the eigenvalue to four decimal places is $$0.01015$$ and the corresponding eigenvector is $$\left[\begin{array}{c}-0.603972\\ 1\\ -0.251135\\ 0.148953\end{array}\right]$$.

Step-by-step solution

Write the function to compute the power matrices

(a)

$$\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\left[x,\mathrm{l}\mathrm{a}\mathrm{m}\mathrm{b}\mathrm{d}\mathrm{a}\right]=\mathrm{p}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{t}(A,x_0,\mathrm{n}\mathrm{i}\mathrm{t})$$

$$x=x_0$$;

For$$n=1:\mathrm{n}\mathrm{i}\mathrm{t}$$

$$\mathrm{x}\mathrm{n}\mathrm{e}\mathrm{w}=A\ast \mathrm{x}$$

$$\mathrm{l}\mathrm{a}\mathrm{m}\mathrm{b}\mathrm{d}\mathrm{a}=\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}(\mathrm{x}\mathrm{n}\mathrm{e}\mathrm{w},\mathrm{i}\mathrm{n}\mathrm{f})/\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}(\mathrm{x},\mathrm{i}\mathrm{n}\mathrm{f});$$

$$\mathrm{f}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{f}{(}^{\prime }\mathrm{n}=\mathrm{\%}4\mathrm{d}\mathrm{l}\mathrm{a}\mathrm{m}\mathrm{b}\mathrm{d}\mathrm{a}=\mathrm{\%}\mathrm{g}\mathrm{x}=\mathrm{\%}\mathrm{g}\mathrm{\%}\mathrm{g}\mathrm{\%}\mathrm{g}\mathrm{\setminus }{\mathrm{n}}^{\prime },\mathrm{n},\mathrm{l}\mathrm{a}\mathrm{m}\mathrm{b}\mathrm{d}\mathrm{a},{\mathrm{x}}^{\prime });$$

$$\mathrm{x}=\mathrm{x}\mathrm{n}\mathrm{e}\mathrm{w};\mathrm{e}\mathrm{n}\mathrm{d}\mathrm{x}=\mathrm{x}/\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}(\mathrm{x});\mathrm{\%}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{s}\mathrm{e}\mathrm{x}\mathrm{f}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{f}{(}^{\prime }\mathrm{n}=\mathrm{\%}4\mathrm{d}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{s}\mathrm{e}\mathrm{d}\mathrm{x}=\mathrm{\%}\mathrm{g}\mathrm{\%}\mathrm{g}\mathrm{\%}\mathrm{g}\mathrm{\setminus }{\mathrm{n}}^{\prime },\mathrm{n},{\mathrm{x}}^{\prime });$$

Find the middle eigenvalue

(b)

Enter the Matrix$$B$$in MATLAB:

$$>>A=\left[10787;7565;86109;75910\right]$$

Enter the$$x_0$$in MATLAB:

$$>>x_0={\left[1000\right]}^{\prime }$$

Now find the eigenvector.

$$>>\mathrm{p}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{t}(A,x_0,7)$$

Construct the data in the table shown below:

$$k$$	$$0$$	$$1$$	$$2$$	$$3$$	$$4$$
$${\mathbf{x}}_k$$	$$\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]$$	$$\left[\begin{array}{c}1\\ .7\\ .8\\ .7\end{array}\right]$$	$$\left[\begin{array}{c}.988679\\ .709434\\ 1\\ .932075\end{array}\right]$$	$$\left[\begin{array}{c}.961467\\ .691491\\ 1\\ .942201\end{array}\right]$$	$$\left[\begin{array}{c}.958115\\ .689261\\ 1\\ .943578\end{array}\right]$$
$$A{\mathbf{x}}_k$$	$$\left[\begin{array}{c}10\\ 7\\ 8\\ 7\end{array}\right]$$	$$\left[\begin{array}{c}26.2\\ 18.8\\ 26.5\\ 24.7\end{array}\right]$$	$$\left[\begin{array}{c}29.0060\\ 20.8675\\ 30.2892\\ 28.5863\end{array}\right]$$	$$\left[\begin{array}{c}29.0060\\ 20.8675\\ 30.2892\\ 28.7887\end{array}\right]$$	$$\left[\begin{array}{c}29.0505\\ 20.8987\\ 30.3205\\ 28.6097\end{array}\right]$$
$${\mu }_k$$	$$10$$	$$26.5$$	$$30.5547$$	$$30.3205$$	$$30.2927$$

$$k$$	$$5$$	$$6$$	$$7$$
$${\mathbf{x}}_k$$	$$\left[\begin{array}{c}.957691\\ .688978\\ 1\\ .9433755\end{array}\right]$$	$$\left[\begin{array}{c}.957637\\ .688942\\ 1\\ .943778\end{array}\right]$$	$$\left[\begin{array}{c}.957630\\ .688938\\ 1\\ .943781\end{array}\right]$$
$$A{\mathbf{x}}_k$$	$$\left[\begin{array}{c}29.0060\\ 20.8675\\ 30.2892\\ 28.5863\end{array}\right]$$	$$\left[\begin{array}{c}29.0054\\ 20.8671\\ 30.2887\\ 28.5859\end{array}\right]$$	$$\left[\begin{array}{c}29.0053\\ 20.8670\\ 30.2887\\ 28.5859\end{array}\right]$$
$${\mu }_k$$	$$30.2892$$	$$30.2887$$	$$30.2887$$

The value of $${\mu }_6=30.2887={\mu }_1$$ , the largest eigenvalue is $$30.288685$$.

Write the function of MATLAB

$$\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\left[\mathrm{v},\mathrm{l}\mathrm{a}\mathrm{m}\mathrm{b}\mathrm{d}\mathrm{a}\right]=\mathrm{I}\mathrm{P}\mathrm{M}\left(\mathrm{B},\mathrm{t}\mathrm{o}\mathrm{l}\right)$$

tic;

$$\mathrm{A}=\mathrm{i}\mathrm{n}\mathrm{v}\left(\mathrm{B}\right);$$

$$\mathrm{n}=\mathrm{s}\mathrm{i}\mathrm{z}\mathrm{e}\left(\mathrm{A},1\right);$$

$$\mathrm{v}=\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{d}\left(\mathrm{n},1\right);$$

$$\mathrm{v}=\mathrm{v}/\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\left(\mathrm{v}\right);$$

$$\mathrm{r}\mathrm{e}\mathrm{s}=1;$$

$$\mathrm{w}\mathrm{h}\mathrm{i}\mathrm{l}\mathrm{e}\left(\mathrm{r}\mathrm{s}\mathrm{e}>\mathrm{t}\mathrm{o}\mathrm{l}\right)$$

$$\mathrm{W}=\mathrm{A}\ast \mathrm{v};$$

$$\mathrm{l}\mathrm{a}\mathrm{m}\mathrm{b}\mathrm{d}\mathrm{a}=\mathrm{m}\mathrm{a}\mathrm{x}\left(\mathrm{a}\mathrm{b}\mathrm{s}\left(\mathrm{W}\right)\right);$$

$$\mathrm{V}=\mathrm{W}/\mathrm{l}\mathrm{a}\mathrm{m}\mathrm{d}\mathrm{a};$$

$$\mathrm{r}\mathrm{e}\mathrm{s}=\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\left(\mathrm{A}\ast \mathrm{v}-\mathrm{l}\mathrm{a}\mathrm{m}\mathrm{b}\mathrm{d}\mathrm{a}\ast \mathrm{v}\right);$$

toc

end

Find the eigenvector

Enter the Matrix$$B$$in MATLAB:

$$>>A=\left[10787;7565;86109;75910\right]$$

Now find the Eigenvalue.

$$>>\mathrm{I}\mathrm{P}\mathrm{M}\left(\mathrm{B},\mathrm{t}\mathrm{o}\mathrm{l}\right)$$

Construct the data in the table shown below:

$$k$$	$$0$$	$$1$$	$$2$$	$$3$$	$$4$$
$${\mathbf{x}}_k$$	$$\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]$$	$$\left[\begin{array}{c}-.609756\\ .7\\ -.243902\\ .146341\end{array}\right]$$	$$\left[\begin{array}{c}-.604007\\ 1\\ -.251051\\ .148899\end{array}\right]$$	$$\left[\begin{array}{c}-.603973\\ 1\\ -.251134\\ .148953\end{array}\right]$$	$$\left[\begin{array}{c}-.603972\\ 1\\ -.251135\\ .148953\end{array}\right]$$
$$A{\mathbf{x}}_k$$	$$\left[\begin{array}{c}25\\ -41\\ 10\\ -6\end{array}\right]$$	$$\left[\begin{array}{c}-59.5610\\ 98.6098\\ -24.7561\\ 14.6829\end{array}\right]$$	$$\left[\begin{array}{c}-59.5041\\ 98.5211\\ -24.7420\\ 14.6750\end{array}\right]$$	$$\left[\begin{array}{c}-59.5044\\ 98.5217\\ -24.7423\\ 14.6751\end{array}\right]$$	$$\left[\begin{array}{c}-59.5044\\ 98.5217\\ -24.7423\\ 14.6751\end{array}\right]$$
$${\mu }_k$$	$$-41$$	$$98.6098$$	$$98.5211$$	$$98.5217$$	$$98.5217$$
$$v_k$$	$$-.0243902$$	$$-.0101410$$	$$-.0101501$$	$$-.0101500$$	$$-.0101500$$

Thus, an estimate for the eigenvalue to five decimal places is$$0.01015$$and the corresponding eigenvector is$$\left[\begin{array}{c}-0.603972\\ 1\\ -0.251135\\ 0.148953\end{array}\right]$$.