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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!

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    $\begingroup$ 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 $a,b,c,d$. $\endgroup$ – Toni Mhax Nov 12 '19 at 4:01
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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, $a_{12}=1$.)

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As,eigenvalue of this matrix are 1 and 2.if matrix A is diagonalizable then minimial polynomial of A is (A-I)(A-2I) so we put A in this equation then I found that a,b,c,d be any number

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