I am seeking verification of my solution to the below problem, or tips on how my solution can be improved. Here is the problem:
For which complex numbers $a,b,c,d$ is the matrix \begin{bmatrix} 1 & 0 & a & b \\ 0 & 1 & c & d \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix} diagonalizable over $\mathbb{C}$ ?
Here is my approach to the problem:
Denote the above matrix by $A$. Then $A$ is diagonalizable if and only if for every eigenvalue $\lambda$ of $A$, the algebraic multiplicity of $\lambda$ is equal to the geometric multiplicity of $\lambda$.
One can quickly calculate the characteristic polynomial of $A$, and find that it is given by $p_A(x) = (1-x)^2(2-x)^2$. Thus, we have two eigenvalues $\lambda = 1$ and $\lambda = 2$, both with algebraic multiplicity $2$. Thus, $A$ is diagonalizable if and only if the geometric multiplicity of $\lambda = 1$ is $2$ and the geometric multiplicity of $\lambda = 2$ is $2$.
$A-1I = \begin{bmatrix} 0 & 0 & a & b \\ 0 & 0 & c & d \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$. We need the dimension of the null space of $A - 1I$ to be equal to $2$ in order for $A$ to be diagonalizable. Using the rank-nullity theorem, this is equivalent to asking that the rank of $A - 1I$ is equal to $2$, i.e., that there are two nonzero rows in the row echelon form of $A-1I$. I claim this occurs if:
$1)$ $a,b,c,d$ are all equal to $0$,
$2)$ $a,d \neq 0$ and $b,c = 0$
$3$ $a,d = 0$ and $b,c \neq 0$
$A-2I = \begin{bmatrix} -1 & 0 & a & b \\ 0 & -1 & c & d \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$. Similar to the above, we need two nonzero rows in the row echelon form of $A-2I$ in order for $A$ to be diagonalizable. But I claim this happens for any $a,b,c,d$ here, so we don't gain any new conditions on $a,b,c,d$ here.
Thus, my final answer is that $A$ is diagonalizable over $\mathbb{C}$ only if one of the following occurs:
$1)$ $a,b,c,d$ are all equal to $0$,
$2)$ $a,d \neq 0$ and $b,c = 0$
$3$ $a,d = 0$ and $b,c \neq 0$
Is my solution correct? If not, where did I make an error in my logic? Is there any ways my solution can be improved?
Thanks!
I think your conditions for $\lambda=1$ should be the same as for $\lambda=2$. Because no matter what $a,b,c,d$ are, you can do row operations to remove them. Does that make sense?
Try it in this Sage cell, which also shows the Jordan canonical form over a certain subfield of the complex numbers. (And compare to this one, which now is not diagonalizable at all, with just one entry different, $a_{12}=1$.)