How to check if a matrix is diagonizable?

Question

So i have this $3\times 3$ matrix

$$ \begin{pmatrix} -2 & 0 & 0 \\ 3 & 1 & -6 \\ 0 & 0 & -2 \\ \end{pmatrix} $$

I want to check if the matrix is diagonizable. First thing i do is find the roots : namely $\lambda_1 = −2$ of multiplicity $2,$ and $\lambda_2 = 1$ of multiplicity $1.$

In order to ensure that the matrix is diagonalizable we need to find a basis of $\mathbb{R}^3$ whose elements are eigenvectors for $A.$ In particular, since we have the eigenvalue $\lambda_1=-2$ having multiplicity $2$ we need to check that the dimension of the eigenspace $E(\lambda_1)$ is equal to $2.$ We compute the eigenspaces of $A.$

$\DeclareMathOperator\Nul{Nul}$On the answer sheet is says that we need to find $\Nul(A+2Id)$... but why $2Id?$ Isn't is supposed to be $-2,$ so $\Nul(-2Id-A)$??

I do not understand this $\Nul(A+2Id).$ I always thought that we should compute $\Nul(-2Id-A),$ so what do i need to do ?

$\Nul(-2Id-A)$ gives me a different result from $\Nul(A+2Id).$

Does that mean that since $-2$ has a multiplicity of $2,$ we need to work with $2$ and not $-2$?

Somebody please help with a clear and simple explanation ! Thanks!

Answer 1

You can compute $\operatorname{Nul}(A+2\operatorname{Id})$ or you can compute $\operatorname{Nul}(-A-2\operatorname{Id})$. It makes no difference, since they are the same space.

By the way, your matrix is diagonalizable.

Answer 2

How did you find your roots?

Normally we find them from $\det(A-\lambda I)=0$. Filling in $\lambda=-2$ then gives us $\det(A+2I)=0$. The corresponding eigenvector(s) are in $\text{Nul}(A+2I)$.

Btw, $\text{Nul}(A+2I) = \text{Nul}(-2I-A)$.