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The question states:

Show that λ is an eigenvalue of A, and find out a basis for the eigenspace $E_{λ}$

$$ A=\begin{bmatrix}1 & 0 & 2 \\ -1 & 1 & 1 \\ 2 & 0 & 1\end{bmatrix} , \lambda = 1 $$

Can someone show me how to find the basis for the eigenspace? So far I have, Ax = λx => (A-I)x = 0,

$$ A=\begin{bmatrix}1 & 0 & 2 \\ -1 & 1 & 1 \\ 2 & 0 & 1\end{bmatrix} - \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} =\begin{bmatrix}0 & 0 & 2 \\ -1 & 0 & 1 \\ 2 & 0 & 0\end{bmatrix}$$

Also, how can I show that 1 is in fact an eigenvalue of A?

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    $\begingroup$ 1 is an eigenvalue of $A$ because $A-I$ is not invertible. By definition of an eigenvalue and eigenvector, it needs to satisfy $Ax = \lambda x$, where $x$ is non-trivial, there can only be a non-trivial $x$ if $A-\lambda I$ is not invertible. $\endgroup$ – JessicaK Nov 14 '14 at 5:48
  • $\begingroup$ Thank you! That helps for one part of the question. $\endgroup$ – CloudN9ne Nov 14 '14 at 5:50
  • $\begingroup$ I made a small technical mistake in my language and edited my comment. $\endgroup$ – JessicaK Nov 14 '14 at 5:52
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To find the eigenvectors of $A$ corresponding to the eigenvalue $\lambda = 1$ solve:
$Ax = \lambda x \Rightarrow (A - \lambda I)x = 0$.

$\begin{bmatrix} 0 & 0 & 2\\ -1 & 0 & 1\\ 2 & 0 & 0 \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = 0 \Rightarrow\\ 2x_3 = 0\\ x_3 - x_1 = 0\\ 2x_1 = 0$

Or $x_1 = x_3 = 0$. Thus, $x_2$ can be any value, so the eigenvectors (for $\lambda = 1$) are all multiples of $\begin{bmatrix}0\\1\\0\end{bmatrix}$, which means this vector forms a basis for the eigenspace for $\lambda = 1$.

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