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  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.

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  • $\begingroup$ 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. $\endgroup$ – Paichu May 8 at 2:54
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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.

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  • $\begingroup$ Thank you for the feedback! $\endgroup$ – bengraham12 May 8 at 3:42

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