Finding Eigenvectors

This is a tedious process

Process

1. Find the characteristic polynomial of the matrix $M$
2. Determine the roots of the polynomial, these are your Eigenvalues
3. For each eigenvalue $\lambda$, compute the eigenvector $\overrightarrow{x}$ given by $(M-(\lambda )I)\overrightarrow{x}=0$

Example

Find all eigenvalues and eigenvectors of

$$M=\left[\begin{array}{cc}1& 2\\ 4& 3\end{array}\right]$$

1. $X_M(\lambda )=\det (M-\lambda I)$
2. $=(1-\lambda )(3-\lambda )-8$
3. $=(\lambda -5)(\lambda +1)$
4. The roots are $\lambda =5,\lambda =-1$
5. For $\lambda =5$: we find $(M-(5)I)\overrightarrow{x}=0$

\left[\begin{array}{cc|c} -4 & 2 & 0 \ 4 & -2 & 0 \ \end{array}\right]

\left[\begin{array}{cc|c} 1 & -\frac{1}{2} & 0 \ 0 & 0 & 0 \ \end{array}\right]

1. Then, $v = (1,2)$ is a non-zero soln 6. For $\lambda = -1$, we find $(M - (-1)I) \overrightarrow{x} = 0$

\left[\begin{array}{cc|c} 2 & 2 & 0 \ 4 & 4 & 0 \ \end{array}\right]

\left[\begin{array}{cc|c} 1 & 1 & 0 \ 0 & 0 & 0 \ \end{array}\right]

1. Then, $v = (1,-1)$ is a non-zero soln 7. Thus, we get $(1,2)$ is a $5$-eigenvector 8. Thus, we get $(1,-1)$ is a $(-1)$-eigenvector