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Is the matrix $A=\begin{bmatrix} 1&1&0&0\\ -1&-1&0&0\\ -2&-2&2&1\\ 1&1&-1&0\\ \end{bmatrix}$ similar over the field of complex numbers to a diagonal matrix?

My attempt: By exercise 10, section 6.2, $A\in M_4(\Bbb{C})$ is diagonalizable$\iff$$A$ is similar over $\Bbb{C}$ to a diagonal matrix. $A\in M_4(\Bbb{C})$ is diagonalizable if $T_A:\Bbb{C}^4\to \Bbb{C}^4$ defined by $T_A(x)=A\cdot x$, $\forall x\in \Bbb{C}^4$ is diagonalizable. Let $B=\{e_1,…,e_4\}$ be canonical basis of $\Bbb{C}^4$. Characteristic polynomial function of $T_A$ is $f:\Bbb{C}\to \Bbb{C}$ such that $f(x)=\det (xI_4-[T_A]_B)=\det (xI_4-A)$, $\forall x\in \Bbb{C}$. It’s easy to check, $\det (xI_4-A)=x^2(x-1)^2$. So $0$, $1$ are only eigenvalue of $T_A$. Let $W_0$ and $W_1$ be null space of $T_A-0\cdot I_4$ and $T_A-1\cdot I_4$, respectively. By equivalent definition, $T_A$ is diagonalizable$\iff$$f=x^2(x-1)^2$ and $\dim (W_0)=2$, $\dim(W_1)=2$. But $\dim (W_0)=\dim (N_{T_A})=1\neq 2$. So $T_A$ is not diagonalizable. Thus $A$ is not diagonalizable. Hence $A$ is not similar over $\Bbb{C}$ to a diagonal matrix. Is my proof correct?

user264745
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    Your proof is perfectly fine. To make it lighter, you could directly reason on $A$ rather than on $T_A$, since the characteristic polynomial and its properties regarding diagonalizability are well-defined over matrices. I would also add an argument on why $\dim \mathrm{Ker}(A) = 1$, although that is easy to verify. – Zag Dec 24 '22 at 11:41
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    @Zag Yeah you’re right. I feel more comfortable working with map than matrices. – user264745 Dec 24 '22 at 14:19

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The proof in OP is correct. The matrix has a block form $$A=\begin{pmatrix} A_1 & 0\\ B & A_2 \end{pmatrix},\quad A_1=\begin{pmatrix} 1 & 1\\ -1 & -1\end{pmatrix},\quad A_2=\begin{pmatrix} 2 & 1\\ -1 & 0\end{pmatrix}$$ The characteristic polynomial of $A$ is equal to the characteristic polynomial of $$\tilde{A}=\begin{pmatrix} A_1 & 0\\ 0 & A_2 \end{pmatrix}$$ Diagonalizability of $A$ would imply that property for $A_1$ and $A_2.$ However none of the matrices $A_1$ and $A_2$ is diagonalizable. The characteristic polynomial of $A_1$ is equal $x^2$ and $A_1\neq 0.$ Similarly the characteristic polynomial of $A_2$ is equal $(x-1)^2$ and $A_2\neq I.$

Ryszard Szwarc
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