Finding the eigenvalues and a basis for the eigenspaces of a $3\times3$ matrix. For the  matrix $A \in M_{3\times3}(\mathbb{R})$ below, I need to find the eigenvalues and a basis for the corresponding eigenspaces:
$$\begin{bmatrix}\
1 & -3 & 3 \\           
3 & -5 & 3 \\
6 & -6 & 4 \\
\end{bmatrix}$$
I have tried to use the formula $\det(I\lambda - A) = 0$ but I ended up with equation $\lambda^3 - 12\lambda - 16 = 0$, of which I can't seem to find the solutions to.
 A: Oher answers already explain how you can factorize the cubic. This is to complement those answers because sometimes it's possible to efficiently use properties of determinants to avoid having to factorize afterwards. I think it's useful to show and for you to try out. Note that:


*

*adding the second column to the other two;

*subtracting the first and last row from the middle one;


gives you:
$$\begin{vmatrix}
1-\lambda & -3 & 3 \\
3 & -5-\lambda & 3 \\
6 & -6 & 4-\lambda 
\end{vmatrix} = 
\begin{vmatrix}
-2-\lambda & -3 & 0 \\
-2-\lambda & -5-\lambda & -2-\lambda \\
0 & -6 & -2-\lambda 
\end{vmatrix}= 
\begin{vmatrix}
-2-\lambda & -3 & 0 \\
0 & 4-\lambda & 0 \\
0 & -6 & -2-\lambda 
\end{vmatrix}$$
Then simply:
$$\begin{vmatrix}
-2-\lambda & -3 & 0 \\
0 & 4-\lambda & 0 \\
0 & -6 & -2-\lambda 
\end{vmatrix} = 0 \iff \left( -2-\lambda \right)^2 (4-\lambda) = 0 \iff \lambda = -2 \;\vee\; \lambda = 4 $$
No annoying cubics here!
You can then proceed to find the eigenvectors/eigenspaces for each eigenvalue.
A: You can use the rationnal root theorem to find a root of your polynomial : $x$ is a rationnal root of $\lambda^3 - 12\lambda - 16\Rightarrow  x$ is a divisor of $16$. You check if there is a root of $\lambda^3 - 12\lambda - 16$ in $\{\pm1,\pm2,\pm4,\pm8,\pm16\}$, and you find that $4$ is indeed a root. Thus $\lambda^3 - 12\lambda - 16=(\lambda-4)(\lambda^2+4\lambda+4)$. 
Can you finish ?
The eigenvalues are $-2$ and $4$

Now will find the eigen sace associated to $4$ for example. $X=\begin{pmatrix} x \\ y\\z \end{pmatrix}$ is in this eigenspace iff :
$$AX=4X\iff \begin{pmatrix}\
1 & -3 & 3 \\           
3 & -5 & 3 \\
6 & -6 & 4 \\
\end{pmatrix}\begin{pmatrix} x \\ y\\z \end{pmatrix}=4\begin{pmatrix} x \\ y\\z \end{pmatrix}\\\iff\begin{cases}x-3y+3z=4x\\3x-5y+3z=4y\\6x-6y+4z=4z\end{cases}\\\iff\begin{cases}-3x-3y+3z=0\\3x-9y+3z=0\\6x-6y=0\end{cases}\\\iff\begin{cases}x=y\\z=2x\end{cases}$$
So the eigenspace associated to $4$ is $\text{Vect}\left(\begin{pmatrix} 1 \\ 1\\2 \end{pmatrix}\right)$.
You can do the same for the eigenspace associated to $-2$.
A: For textbook problems, usually at least one root of a cubic polynomial will be rational.  The Rational Root Theorem says that a rational root $p/q$, with $p$ and $q$ integers must have $p$ as a factor of the constant term (in this case $-16$) and $q$ as a factor of the leading coefficient (in this case $1$).  So your choices for roots are $\pm 1, \pm 2, \pm 4, \pm 8, \pm 16$.  One would start plugging these in until one gets $0$.  Once you know a root $r$ you can divide by $x-r$ and get a quadratic, which is easy to solve.
