I'm trying to find an eigenbasis for matrix A = $\begin{bmatrix}1&-1&1\\-1&1&-1\\1&-1&1\end{bmatrix}$ so that I can use the result to diagonalize A.
Because the characteristic equation is $\\-λ^3+3λ^2 = 0$, the eigenvalues are 0 and 3.
λ = 0 yields the eigenvectors $\begin{bmatrix}1\\1\\0\end{bmatrix}$, $\begin{bmatrix}1\\0\\-1\end{bmatrix}$, and λ = 3 yields the eigenvector $\begin{bmatrix}1\\-1\\1\end{bmatrix}$.
Is it correct that since all three eigenvectors are linearly independent, they form an eigenbasis for A? If so, how do I use that to diagonalize A?