# Dimension of the corresponding eigenspace?

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.

• Can you recall the definition of an eigenspace? Then just apply that to your situation (and solve the equation). – Marc van Leeuwen Apr 16 '13 at 5:25
• No, the dimension of the eigenspace is the dimension of the null space of the matrix $A - \lambda I$ (the second matrix you mentioned). Note that you have two free variables, $x_2$ and $x_3$, and so the dimension is two. – Suugaku Apr 16 '13 at 5:32

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.
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 $\lambda$.