Question

[M] In Exercises 37–40, use a matrix program to find the eigenvalues of the matrix. Then use the method of Example 4 with a row reduction routine to produce a basis for each eigenspace.

39. $\left(\begin{array}{ccccc}4& -9& -7& 8& 2\\ -7& -9& 0& 7& 14\\ 5& 10& 5& -5& -10\\ -2& 3& 7& 0& 4\\ -3& -13& -7& 10& 11\end{array}\right)$

Short answer

The eigenvalues are $\lambda =\left(-2,-2,5,5,5\right)$. The basis for eigenspace of $\lambda =-2$ is $N=\left\{\left(\begin{array}{c}-2\\ 7\\ -5\\ 5\\ 0\end{array}\right),\left(\begin{array}{c}3\\ 7\\ -5\\ 0\\ 5\end{array}\right)\right\}$. The basis for eigenspace of $\lambda =5$ is $N=\left\{\left(\begin{array}{c}2\\ -1\\ 1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}-1\\ 1\\ 0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}2\\ 0\\ 0\\ 0\\ 1\end{array}\right)\right\}$.

Step-by-step solution

Command for input the matrix in MATLAB

Write the command given below to input the matrix

$>>\mathrm{A}=\left(4-9-782;-7-90714;5105-5-10;-23704;-3-13-71011\right)$

Command for finding the eigenvalues of the matrix

$>>\mathrm{e}\mathrm{v}=\mathrm{e}\mathrm{i}\mathrm{g}\left(\mathrm{A}\right)$

The output will be$\mathrm{e}\mathrm{v}=\left(-2,-2,5,5,5\right)$.

Hence, the eigenvalues are $\mathrm{e}\mathrm{v}=\left(-2,-2,5,5,5\right)$.

Command for finding the null basis for each eigenvalue

Write the command given below to find the null basis corresponding to $\lambda =-2$:

$>>\mathrm{N}=\mathrm{A}-\mathrm{e}\mathrm{v}\left(1\right)\ast \mathrm{e}\mathrm{y}\mathrm{e}\left(5\right)$

The output is given below:

$N=\left(\begin{array}{cc}-2& 3\\ 7& 7\\ -5& -5\\ 5& 0\\ 0& 5\end{array}\right)$

Hence, the basis for eigenspace of $\lambda =-2$ is $N=\left\{\left(\begin{array}{c}-2\\ 7\\ -5\\ 5\\ 0\end{array}\right),\left(\begin{array}{c}3\\ 7\\ -5\\ 0\\ 5\end{array}\right)\right\}$.

Write the command given below to find the null basis corresponding to $\lambda =5$:

$>>\mathrm{N}=\mathrm{A}-\mathrm{e}\mathrm{v}\left(2\right)\ast \mathrm{e}\mathrm{y}\mathrm{e}\left(5\right)$

The output is given below:

$N=\left(\begin{array}{ccc}2& -1& 2\\ -1& 1& 0\\ 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$

Hence, the basis for eigenspace of $\lambda =5$ is $N=\left\{\left(\begin{array}{c}2\\ -1\\ 1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}-1\\ 1\\ 0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}2\\ 0\\ 0\\ 0\\ 1\end{array}\right)\right\}$.