I am trying to understand for which $a\in\mathbb{R}$ the matrix $\begin{bmatrix} a & 2 & -1\\ 1 & a & -1\\ 0 & 0 & 2\end{bmatrix}$ is diagonalizable and, in that case, which are its eigenvectors.
My solution:
The characteristic polynomial of the matrix is $p(\lambda)=(2-\lambda)[(a-\lambda)^2-2]$ which is zero for $\fbox{$\lambda =2$}$ and, if $a\neq 2\pm \sqrt{2}$, we have other two distinct eigenvalues $\fbox{$ \lambda_{+}=a+\sqrt{2},\lambda_{-}=a-\sqrt{2}$}$ thus three distinct eigenvectors $\fbox{$v_{2}=(2,-1,2), v_{+}=(\sqrt{2},1,0), v_{-}=(\sqrt{2},-1,0)$}$ for $a=2$ and $\fbox{$v_{2}=(-\frac{3}{a-2},\frac{1+a}{(a-2)^2},1), v_{+}=(\sqrt{2},1,0), v_{-}=(\sqrt{2},-1,0)$}$ for $a\neq 2$.
Thus, the matrix is diagonalizable for $a\neq 2\pm\sqrt{2}$.
Does this make sense? Is there a better way to solve this problem? Thanks
After having got that the eigenvalues are $2$, $a+\sqrt2$, and $a-\sqrt2$, then you know that, unless $a=2\pm\sqrt2$, then you have $3$ distinct eigenvalues. So, the matrix is diagonalizable; there is no need to compute the eigenvectors.
And if $a=2\pm\sqrt2$, then it is not diagonalizable. Simple check than, in each case, the eigenspace corresponding to the eigenvalue $2$ is $1$-dimensional. But $2$ is then a double root of the characteristic polynomial. So, if your matrix was indeed diagonalizable, the eigenspace corresponding to the eigenvalue $2$ would be $2$-dimensional.