TITLE: Dimension of the corresponding eigenspace?
QUESTION [8 upvotes]: I'm studying for my linear exam and would appreciate any help for this practise question:
You are given that λ = 1 is an eigenvalue of A. What is the dimension of the corresponding eigenspace?
A = $\begin{bmatrix} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 1 \end{bmatrix}$
Then with my knowing that λ = 1, I got:
$\begin{bmatrix} 0 & 0 & 0 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix}$
Which I assume right off the bat means my dimension is 0. Is that correct? If not how should I do it? 
If we had a different matrix, how would I go ahead to properly find the dimension?
In layman terms I think that it would be whichever value is linearly independent?
Thanks for the help.

REPLY [7 votes]: The dimension is two. Note that the vectors $
u=\left[
  \begin{array}{c}
    0 \\
    1 \\
    0 \\
0 \\
  \end{array}
\right]
$ and 
$v=
\left[
  \begin{array}{c}
    0 \\
    0 \\
    1 \\
0 \\
  \end{array}
\right]
$ are in the null space of $A-I_4=\begin{bmatrix} 0 & 0 & 0 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix}$, i.e. 
$$Au=u\mbox{ and } Av=v.$$
So $u$ and $v$ are eigenvectors corresponding to the eigenvalue $1$. In fact, the form a basis for the null space of $A-I_4$. Therefore, the eigenspace for $1$ is spanned by $u$ and $v$, and its dimension is two.

REPLY [7 votes]: In general, the eigenspace of an eigenvalue $\lambda$ is the set of all vectors $v$ such that $Av=\lambda v$.  This also means $Av-\lambda v=0$, or $(A-\lambda I)v=0$.  Hence, you can just calculate the kernel of $A-\lambda I$ to find the eigenspace of $\lambda$.