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Let $ A = \begin{bmatrix} 4&0&0\\ 0&4&0\\ 0&0&0 \end{bmatrix}$

I understand the eigenvalues to be: $\lambda_1 = 4$, $\lambda_2 = 4$, $\lambda_3 = 0$.

I am unable to compute the corresponding eigenvectors.

For $\lambda_1 = 4:$

$(A-\lambda_1I)v = 0$

$\Rightarrow \begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&0&-4 \end{bmatrix}v=0$

after rref we get:

$\Rightarrow \begin{bmatrix} 0&0&1\\ 0&0&0\\ 0&0&0 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} $

with 2 free variables I am not sure how to compute a distictive $v_1$

I only know from wolfram that $v_1 = (1,0,0)$, $v_2 = (0,1,0)$, $v_3 = (0,0,1)$

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    Have you considered the possibility that the eigenspace of 4 is two-dimensional? – 2017-02-20

1 Answers 1

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The matrix $A$ is already diagonal so the columns are the eigenvectors.

For the $v_3 = (0,0,1)$ it is clear because that $\lambda_3 = 0$.