I am not sure about my answer. I have found similar questions to mine but they still did not help me.
I have this matrix, and I need to find the values of $a$ parameter where $D$ will be Diagonalizable:
$$ \begin{pmatrix}0& a&-a \\a&0&a\\ -a & a &0\end{pmatrix}=D.$$
I have found that $p(x)= (x-a)^2(x+2a),$
so my answer is as follows.
When $a=0$,
my Eigenvalues $0$ and the diagonal matrix = $diag(0,0,0),$
so the matrix is diagonalizable.
When $a$ any real number and not $0$, it is also diagonalizable, and its diagonal matrix = $diag(a,a,-2a).$
So it means $D$ is diagonalizable for any value of $a$.
am i wrong here? i just not sure if I answered the question at all, if I covered all the cases.
You are right. Indeed, for all $a\in\mathbb{R}\space,$ $D$ is diagonalizable. This is because $D$ is symmetric for all values of $a\in\mathbb{R}\space,$ and because real symmetric matrices are diagonalizable by a spectral theorem.