I have a matrix $$B= \begin{bmatrix} 1&0&0 \\ 0&0&1 \\ 0&1&0 \end{bmatrix}$$
with two eigenvalues $\lambda=-1$ with eigenvector u=$(0,-1,1)$ and the double eigenvalue $\lambda=1$ to which corresponds the eigenspace $S_2$ given by $(x,y,z)\in\mathbb{R^3}:y-z=0$
My question is:
How can I find a basis of the eigenspace $S_2$?
In general, is a basis of a degenerate eigenspace given by its eigenvectors that are equal in number to the degeneracy and are linearly independent?
Could you please show me how linear independence is required?
I would do this:
One eigenstate can be $(2,1,1)$ and the other eigenstate $(0,1,1)$, now what about the independence?
and generally speaking, how do you find the basis of an eigenspace?
$y-z=0$ means $y=z$ and $x$ is free since we have no condition on it, therefore the vectors inside $S_2$ have shape $(x,y,y)=x(1,0,0)+y(0,1,1)$. Since $(1,0,0)$ and $(0,1,1)$ are independent then these two vectors are your basis.