The basis for an eigenspace

1. $$A$$= $$\begin{bmatrix} -4 & 8 & -12 \\ 6 & -6 & 12 \\ 6 & -8 & 14 \end{bmatrix}$$

with $$\lambda = 2.$$ What is the basis for the eigenspace of $$A$$?

Solution: $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$$ $$= x_2\begin{bmatrix} \frac{4}{3} \\ 1 \\ 0 \end{bmatrix}$$ $$+x_3 \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$$

The eigenspace is a 2 dimensional subspace of $$\mathbb{R}^{3}$$.

A basis is $$\begin{bmatrix} 4 \\ 3 \\ 0 \end{bmatrix}$$, $$\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$$.

Hello, I just wanted to check if my solution was correct please. Someone told me it was not but I'm not sure how it is wrong. Thank you.

• Yup. The system is degenerate so the eigenspace is indeed a 2 dimensional subspace of $\mathbb{R}^3$. The two eigenvectors that you find sufficiently compose the basis of that subspace. – Paichu May 8 at 2:54

Your eigenvectors$$\begin{bmatrix} 4 \\ 3 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$$ are correct.
Your eigenspace associated to $$\lambda =2$$ is the two dimensional space generated by these eigenvectors as you have mentioned.