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I have the matrix

$$A = \begin{bmatrix} 4 & 2 & 2\\ 2 & 4 & 2\\ 2 & 2 & 4 \end{bmatrix}$$

I've calculated the Eigenvalues and Eigenvectors as follows with help in a previous question:

$\lambda = 2$ or $8$

$$E_{\lambda = 2} = \left\{ \begin{bmatrix} v_1\\ v_2\\ v_3\\ \end{bmatrix} = a\begin{bmatrix} -1\\ 1\\ 0\\ \end{bmatrix} + b \begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix} \mid a, b \in \mathbb R \right\}$$

$$E_{\lambda = 8} = \left\{ \begin{bmatrix} v_1\\ v_2\\ v_3\\ \end{bmatrix} = a \begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix} \mid \mathbb R\right\}$$

Is the Eigenbasis simply:

$$E_{\lambda = 2} = \operatorname{span}\left(\begin{bmatrix} -1\\ 1\\ 0\\ \end{bmatrix} \begin{bmatrix} -1\\ 0\\ 1\\ \end{bmatrix} \right) $$

$$E_{\lambda = 8} = \operatorname{span}\left(\begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix}\right)$$

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An eigenbasis is a basis in which every vector is an eigenvector.

In your case, $$ \left\{ \pmatrix{-1\\1\\0}, \pmatrix{-1\\0\\1}, \pmatrix{1\\1\\1} \right\} $$ is an eigenbasis for your matrix $A$.

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  • $\begingroup$ The question I'm trying to answer says: Find a basis for each eigenspace of A. Is it the same answer? $\endgroup$ – Gabi Aug 25 '15 at 18:13
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    $\begingroup$ The first two form a basis of one eigenspace, and the second two form a basis of the other. So this isn't quite the same answer, but it is certainly related. $\endgroup$ – Omnomnomnom Aug 25 '15 at 18:17
  • $\begingroup$ does it matter which pairs you pick or can it be any two of the three? Do the first two provide a basis in R^2 and the second and third provide a basis in R^3 $\endgroup$ – Gabi Aug 25 '15 at 18:20
  • $\begingroup$ Yes, it does matter which pair (or more generally, which groups) you pick. The first two form a basis of $E_{\lambda = 2}$ because they are a maximal linearly independent set of eigenvectors associated with $\lambda = 2$. In other words, you need to group the eigenvectors according to their eigenvalues. $\endgroup$ – Omnomnomnom Aug 25 '15 at 18:22
  • $\begingroup$ All of these vectors are elements of $\Bbb R^3$. $E_{\lambda = 2}$ is a 2-dimensional subspace of $\Bbb R^3$, while $E_{\lambda = 8}$ is a 1-dimensional subspace of $\Bbb R^3$. If we take all of them together, the eigenvectors we've chosen form a basis of $\Bbb R^3$. $\endgroup$ – Omnomnomnom Aug 25 '15 at 18:24
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To help add some important concepts to eigenvalues and eigenvectors I will drag in another matrix.

$${\bf A} = \left[\begin{array}{cc}2&2\\0&2\end{array}\right]$$

if we solve

$$|{\bf A}-\lambda {\bf I}| = 0 \Leftrightarrow (2-\lambda)^2 = 0$$, we see that 2 is an eigenvalue that occurs twice. But if we solve $${\bf A}-2 {\bf I} = 0 \Leftrightarrow \left[\begin{array}{cc|c}0&2&0\\0&0&0\end{array}\right]$$ So only first component is actually $\neq 0$. This brings up the concepts of geometric dimensionality and algebraic dimensionality. $[0,1]^t$ is a Generalized eigenvector belonging to the same generalized eigenspace as $[1,0]^t$ which is the "true eigenvector".

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