Eigenvectors and Kernel of Matrix I'm trying to take find the eigenvectors of the matrix
$$
\begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix}.
$$
I've found the eigenvalues of $1$ and $0$. I'm very confused how to find the kernel of $E0$ and $E1$. How do I go about doing this?
 A: I am assuming you want to find the eigenspace of the eigenvalues when you say $E_0$ and $E_1$.  The kernel comes into play through the following formula: $E_\lambda = \ker(A-\lambda I_n)$, where $A$ is your $n \times n$ matrix.
This means $E_0 = \ker(A - 0I_3) = \ker 
\begin{pmatrix}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{pmatrix}.$
The kernel of a matrix is the vectors $\vec{x}$ which satisfy $A\vec{x} = \vec{0}$.  You simply do reduced row echelon form of the augmented matrix, which yields
$\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}.$
We know $x_1 = 0$ and $x_2 = 0$, whereas $x_3$ is a free variable.  The solutions to the equation are of the form 
$\begin{pmatrix}
0 \\ 0 \\ x_3
\end{pmatrix} = x_3
\begin{pmatrix}
0 \\ 0 \\ 1
\end{pmatrix}$. 
As a result, $E_0 = span
\begin{pmatrix}
0 \\ 0 \\ 1
\end{pmatrix}$.
Can you do $E_1$ now?
A: Both $A$ and $A-I$ are in row echelon form and so solving the kernel for your $E_0$ and $E_1$ can be done automatically by looking at the two matrices without any row reductions.
