# Conditions on matrix entries for the matrix to be diagonalizable

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!

• You got the answer, when you get the null space $A-I$, put this directly, compute the dimension in terms of $a,b,c,d$. In fact $A-I$ is rank $2$ as $A-2I$ for any $a,b,c,d$. – Toni Mhax Nov 12 '19 at 4:01

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, $$a_{12}=1$$.)