Finding the eigenvalues and eigenvectors of $4\times 4$ matrix Find all eigenvalues and eigenvectors(and generalized eigenvectors) of the following matrix.
$$\mathbf{A} = \begin{pmatrix} -1&0&0&0\\ 5&-2&0&0\\ 0&3&1&0\\ 2&0&1&1 \end{pmatrix}$$
Okay so when I did the work I got my eigenvalues to be $\lambda= \{1,-1,-2\}$. I'm struggling with the eigenvectors. When I tried doing it for $\lambda=-1$ I got something along the lines of 
$$0\vec{v}_1+0\vec{v}_{2}+0\vec{v}_3+0\vec{v}_4 = 0$$
$$5\vec{v}_1+-1\vec{v}_{2}+0\vec{v}_3+0\vec{v}_4 = 0$$
$$0\vec{v}_1+3\vec{v}_{2}+2\vec{v}_3+0\vec{v}_4 = 0$$
$$2\vec{v}_1+0\vec{v}_{2}+1\vec{v}_3+2\vec{v}_4 = 0$$
But this way I couldn't get a solid number for my vectors. Some direction on this would be great. I'm really stuck on the vectors part.
 A: For the eigenvalue, $\lambda_1 = -1$, we get the RREF of $[A - \lambda_1 I]v_1 = [A + I]v_1 = 0$ as:
$$\left(
\begin{array}{cccc}
 1 & 0 & 0 & -\frac{4}{11} \\
 0 & 1 & 0 & -\frac{20}{11} \\
 0 & 0 & 1 & \frac{30}{11} \\
 0 & 0 & 0 & 0 \\
\end{array}
\right)v_1 = 0$$
This gives us an eigenvector of $v_1 = (4, 20, -30, 11)$.
Update 1
The steps to do that RREF are:


*

*Do row swaps to move the all zeros row to be the bottom row.

*$R_3 = R_3 - \dfrac{2}{5}R_1$ (this is read, replace row $3$ by row three minus $2/5$ row 1. 

*$R_3 = 5 R_3$ (multiply $R_3$ by $5$)

*$R_3 = R_3 - \dfrac{2}{3}R_2$

*$R_3 = 3 R_3$

*$R_3 = R_3/11$

*$R_2 = R_2 - 2 R_3$

*$R_2 = R_2/3$

*$R_1 = R_1 + R_2$

*$R_1 = R_1/5$


Update 2
For the eigenvalue, $\lambda_2 = -2$, we have a RREF for $[A+2I]v_2 = 0$ as:
$$\left(
\begin{array}{cccc}
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & -3 \\
 0 & 0 & 1 & 3 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right)v_2 = 0$$
In this case, we choose $v_2 = (0, 3, -3, 1)$.
Update 3
For the double eigenvalue, we get one independent eigenvector and have to find a second generalized one. That process is (it could change to chaining in the worst cases):


*

*Find $[A - \lambda_3 I]v_3 = [A - I]v_3 = 0$ for $v-3$.

*Solve $[A-I]v_4 = v_3$ for $v_4$.


You should get:
$$v_3 = (0, 0, 0, 1), v_4 = (0, 0, 1, 0)$$
A: You can read off the eigenvector directly from the matrix you have.
The only thing you need is that the matrix is in some triangular form.
For example for $\lambda=-1$ we have:
$$(A-\lambda I)\vec v = (A + I)\vec v = \vec 0$$
$$\begin{bmatrix} 0 &  &  &  \\
5 & -1 &  &  \\
0 & 3 & 2 & \\
2 & 0 & 1 & 2
\end{bmatrix}
\begin{pmatrix}
w \\ x \\ y \\ z
\end{pmatrix} =
\begin{pmatrix}
0 \\ 0 \\ 0 \\ 0
\end{pmatrix}$$
We can choose our own scaling factor, which means that we have a free choice for some non-zero element.
Let's try $w=1$, meaning the first line checks out.
Then, from the second line we can conclude that $5w - x = 0$, so $x=5$.
From the 3rd line we get that $3x+2y=0$, so $y=-7.5$.
Finally, from the 4th line we get that $2w + y + 2z = 0$, so $z = 2.75$.
In other words:
$$\vec v = \begin{pmatrix}
1 \\ 5 \\ -7.5 \\ 2.75
\end{pmatrix}$$
To get nice round numbers, we can scale it up by a factor of 4 and get:
$$\vec v = \begin{pmatrix}
4 \\ 20 \\ -30 \\ 11
\end{pmatrix}$$
A: Look up computing reduced row echelon form for finding the kernel of a matrix. This will give you the eigenvector for each eigenvalue because when you plug in the eigenvalue you know that the eigenvector is in the kernel of the resulting matrix.
