Why isn't this Matrix Diagonalization

Question

I have this matrix below:

Eigenvalues is 0 and the eigenvector is (0,1) what i don't understand is why this Matrix is not Diagonzible and i have found some answer here on Math Stackexchange but i don't understand what it means below:

"A matrix is diagonalizable if and only if for each eigenvalue the dimension of the eigenspace is equal to the multiplicity of the eigenvalue."

What does it means multiplicity of the eigenvalue?? Should i apply the Rank-Theorem??

Answer 1

The multiplicity of an eigenvalue $\lambda$ is the largest natural number $m$ such that the polynomial $(t-\lambda)^m$ divides the characteristic polynomial $\det(A-tI)$. The rank theorem, whatever that is, doesn't seem relevant in computing said multiplicity. In this case, I would just factor the characteristic polynomial, if I were you.

Remark: Many authors call this multiplicity "algebraic multiplicity", as opposed to the "geometric multiplicity" which would be the dimension of the eigenspace of $\lambda$.

Answer 2

You seem to know that the eigenvalues of the matrix -call it $A$ - are $0,0$.

So if $A$ were diagonalisable we would have $S^{-1}AS=\begin{pmatrix}0 & 0\\0 & 0\end{pmatrix}$.

Hence we would have $A=S\begin{pmatrix}0 & 0\\0 & 0\end{pmatrix}S^{-1}=\begin{pmatrix}0 & 0\\0 & 0\end{pmatrix}$.

But $A$ is not $O$.

Answer 3

I like to think about these kinds of problems geometrically. Think about what effect your matrix has when it acts on $\mathbb{R}^2$. If you consider the standard basis $e_1=(1,0)$ and $e_2=(0,1)$, your matrix maps $e_1\mapsto e_2$ and it maps $e_2\mapsto 0$, so the whole plane gets crushed to the line $\mathbb{R}e_2$. Moreover, if we apply the matrix twice, the whole plane is actually mapped to $0$.

In the other hand, suppose your matrix was diagonalisable, in other words, if you choose a different basis $\{e_1',e_2'\}$ for $\mathbb{R}^2$, then the matrix can be written as $\textrm{Diag}(\lambda_1,\lambda_2)$. But from this description, the effect of the matrix on $\mathbb{R}^2$ is to scale the $e_1'$ direction by $\lambda_1$, and scale the $e_2'$ direction by $\lambda_2$. Thus applying the matrix twice would scale each of these directions twice. The only way for this to result in the whole plane being mapped to $0$ is if $\lambda_1=\lambda_2=0$. But the only matrix which is conjugate to the $0$ matrix is the $0$ matrix, a contradiction.