Consider the matrix
$$A=\left(\begin{array}{cccc} 1 &1 &-3 &0 \\ -1 &-1 &1 &-2 \\ -1 &-1 &-1 &-2 \\ 1 &1 &2 &3 \\ \end{array}\right)$$
The characteristic polynomial of $A$ is $\chi _A(x)=x^2(x-1)^2.$ Is $A$ diagonalizable over $\mathbb{C}?$ Explain.
I'm having a hard time understanding diagonalization. I believe that the characteristic polynomial yields 2 distinct eigenvalues $(0$ and $1)$. From what I understand, if there were $4$ distinct eigenvalues, you would know for sure that it's diagonalizable. However, since there are only $2$, where do you go from here to decide if it's digonalizable or not?
Let $\lambda$ be an eigenvalue. You should check the eigenspaces, i.e the kernel of the matrix $(A-\lambda I_4)$. If a vector $v$ is in the kernel $\text{Ker}(A-\lambda I_4)$ of the matrix, then
$$(A-\lambda I_4)(v)=0 \iff A(v)-\lambda v=0 \iff A(v)=\lambda v$$
Vectors like $v$ are called eigenvectors. If you can find two linearly independent vectors $v_1,v_2 \in \text{Ker}(A-\lambda_1)$ for the first eigenvalue, another two linearly independent vectors $w_1,w_2 \in \text{Ker}(A-\lambda_2)$ for the second value you can form a basis $\mathcal{V}=(v_1,v_2,w_1,w_2)$. Since the columns of a matrix are the images of your basis' vectors, the matrix will be diagonal. It should look like this:
$$\begin{pmatrix}\lambda_1 & 0 & 0 & 0\\ 0 & \lambda_1 & 0 & 0 \\ 0 & 0 & \lambda_2 & 0 \\ 0 & 0 & 0 & \lambda_2 \end{pmatrix} = \begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$
In a non-algebraically closed field (i.e. $\mathbb{R}$) you should also check if all the roots of the characteristic polynomial lie in the field. This is an intrinsic characteristic of the vector space we are working on. Here, because we are working on $\mathbb C$, we have no problems.
If the geometric multeplicity $M_g (\lambda):= \text{dim(Ker }(A-\lambda)$ i.e. the dimension of the kernel of the matrix $A-\lambda I_4$ is strictly smaller than the algebraic multeplicity $M_a (\lambda)$ of $\lambda$ as a root of the characteristic polynomial, the matrix will not be diagonalizable. We can verify through algebra that $M_g (\lambda)\leq M_a (\lambda)$ is always true, but the matrix is diagonalizable if and only if equality holds.
What are the dimensions of the two kernels? Can you find these four vectors?