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?
Best Answer
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 $\lambda = 1$.