linear algebra - Find the eigenvalues and associated eigenvectors for this matrix

$$A = \begin{pmatrix} 1 & -1 & 4 \\ 3 & 2 & -1 \\ 2 & 1 & -1\end{pmatrix}$$

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?

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 $\lambda_3$!