$$A = \begin{pmatrix} 1 & -1 & 4 \\ 3 & 2 & -1 \\ 2 & 1 & -1\end{pmatrix}$$
I solved for the determinant of matrix $A$ and got the following:
$\det\begin{bmatrix}
1-\lambda & -1 & 4\\
3 & 2-\lambda & -1 \\
2 & 1 & -1-\lambda
\end{bmatrix} = -\lambda^{3}+2\lambda^{2}+5\lambda-6$
How do I find the eigenvalues from this?
Once I have the eigenvalues, do I get the eigenvectors from the bases?
Best Answer
To find eigenvalues, note that you want to find the solutions to the characteristic equation given by $\det (A - \lambda I) = 0$
As you noted, we have the equation $-\lambda^3 + 2\lambda^2 + 5 \lambda - 6 = 0$ Note that this is just $-(\lambda - 3)(\lambda + 2)(\lambda -1) = 0$ so your eigenvalues are $\lambda_1 = 3, \lambda_2 = -2,$ and $\lambda_3 = 1$.
Now, to get the eigenvectors, you want to look for $v$ such that $Av = \lambda v$ or $(A- \lambda I) v = 0$. For $\lambda_1$, $A- \lambda I = \begin{pmatrix} -2 & -1 & 4 \\ 3 & -1 & -1 \\ 2 & 1 & -4 \end{pmatrix}$ so row reducing gives $\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{pmatrix}$ In other words, for your eigenvector $v_1 = (a_1,a_2,a_3)$, \begin{align*} a_1 - a_3 &= 0 \\ a_2 - 2a_3 &= 0 \end{align*} Let $a_3 = t$, then $a_1 = t$ and $a_2 = 2t$ so a basis for the eigenspace corresponding to $\lambda_1$ is given by $\{\begin{pmatrix} 1 \\ 2 \\ 1 \\ \end{pmatrix} \}$.
So, for $\lambda_1 =3$, we get a corresponding eigenvector $(1,2,1)^T$. Try finding the eigenvectors for $\lambda_2$ and $\lambda_3$!