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I attempted to find the eigenvalues and corresponding eigenvectors for this matrix. $$ A= \begin{bmatrix} \frac13 & \frac13 & \frac13 \\ \frac13 & \frac13 & \frac13 \\ \frac13 & \frac13 & \frac13 \\ \end{bmatrix} $$

The eigenvalues are easily computed as $\lambda_1 = 0, \lambda_2 = 0,$ and $\lambda_3 = 1$.

The corresponding eigenspace with $\lambda_3$ I calculated is $\{(\alpha,\alpha,\alpha):\alpha \in \mathbb{R} \} = span(1,1,1).$

However, when I try to find the two eigenvectors corresponding to the zero eigenvalue, I get three eigenvectors when I know there should only be two:

$(A-I\lambda)x = \begin{bmatrix} \frac13 & \frac13 & \frac13 \\ \frac13 & \frac13 & \frac13 \\ \frac13 & \frac13 & \frac13 \\ \end{bmatrix} $ $ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix}= $ $ \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} $

$\rightarrow x_1+x_2+x_3 = 0 $

Which yields:

$\{(\alpha,-\alpha,0):\alpha \in \mathbb{R} \} = span(1,-1,0).$

$\{(\alpha,0,-\alpha):\alpha \in \mathbb{R} \} = span(1,0,-1).$

$\{(0,\alpha,-\alpha):\alpha \in \mathbb{R} \} = span(0,1,-1).$

Why am I computing 3 instead of 2 eigenspaces for eigenvalue of $0$?

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Note that these $3$ vectors are linearly dependent. Just subtract any two of them..
So they generate a 2d subspace.

Geometrically, $x_1+x_2+x_3=0$ is the plane through the origin orthogonal to $(1,1,1)^T$.

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  • $\begingroup$ Oh of course! thank you $\endgroup$ – JBL Jun 1 '18 at 19:06

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