What is an Eigenbasis and how do I calculate it with the information below.
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:
$$E_{\lambda = 2} = \operatorname{span}\left(\begin{bmatrix} -1\\ 1\\ 0\\ \end{bmatrix} \begin{bmatrix} -1\\ 0\\ 1\\ \end{bmatrix} \right) $$
$$E_{\lambda = 8} = \operatorname{span}\left(\begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix}\right)$$
$\endgroup$2 Answers
$\begingroup$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 $A$.
$\endgroup$ 5 $\begingroup$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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