[Math] Calculate the dimension of the eigenspace

Question

$A= \begin{pmatrix} 0 & -1 & 0\\ 4 & 4 & 0\\ 2 & 1 & 2
\end{pmatrix}$ is the matrix.

The tripple eigenvalue is $\lambda=2$

The eigenspace is $E_{A}(2)= \left\{ \begin{pmatrix} x\\
-2x\\ z \end{pmatrix} \mid x,z \in \mathbb{R}\right\}$

What's the dimension of the eigenspace?

I think in order to answer that we first need the basis of the eigenspace:

$$\begin{pmatrix}
x\\
-2x\\
z
\end{pmatrix}= x\begin{pmatrix}
1\\
-2\\
0
\end{pmatrix}+z\begin{pmatrix}
0\\
0\\
1
\end{pmatrix}$$

So basis $B=
\begin{pmatrix}
1\\
-2\\
0
\end{pmatrix},\begin{pmatrix}
0\\
0\\
1
\end{pmatrix}$

We have $2$ vectors here thus the dimension of the eigenspace is $2$?

---

Please can you tell me if this is done correctly?

Answer 1

You don't need to find particular eigenvectors if all you want is the dimension of the eigenspace.

The eigenspace is the null space of $A-\lambda I$, so just find the rank of that matrix (say, by Gaussian elimination, but possibly only into non-reduced row echelon form) and subtract it from $3$ per the rank-nullity theorem.