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$A= \begin{pmatrix} 0 & -1 & 0\\ 4 & 4 & 0\\ 2 & 1 & 2 \end{pmatrix}$ 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 $2$?

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Please can you tell me if this is done correctly?

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1 Answer

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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.

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