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Suppose we have a matrix $A \in M_2(\mathbb{C})$ such that it's characteristic polynomial is $p_{A}(t) = t^2$. Prove that $A$ is either similar to the zero matrix or similar to $\begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}$.

So far what I have done. Let $$\begin{bmatrix} a & b\\ c & d \\ \end{bmatrix}$$ Then since $Tr(A)=0$ and $det(A)=0$ $$a+d=0$$ $$ad-bc=0$$ So $A$ actually has the from $\begin{bmatrix} a & b \\ c & -a \\ \end{bmatrix}$. Second the only matrix that is similar to the zero matrix would be the case when $A =0_2$ since if $AS = S0_2 = 0_2$ then A=0_2 as $S$ is invertible. So now it comes down to solving the system $$AS = S \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}$$ $$\begin{bmatrix} a & b \\ c & -a\\ \end{bmatrix} \begin{bmatrix} x & y \\ z & w \\ \end{bmatrix} = \begin{bmatrix} x & y \\ z & w \\ \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix} $$ Which boils down to solving this set of equations:

$$ax +bz=0$$ $$cx -aw=0$$ $$ay+bw=x$$ $$cy-aw=z$$ $$xw -yz \neq 0$$

I have not been able to figure out a solution to this system. Any help is appreciated thanks!

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Since $\det A=0$, there is a non-null vector $v=(v_1,v_2)$ such that $A.v=0$. Let $w=(w_1,w_2)$ be some vector of $\mathbb{C}^2$ such that $v$ and $w$ are linearly indepndent. Let$$P=\begin{bmatrix}v_1&w_1\\v_2&w_2\end{bmatrix}.$$Then, since $A.v=0$, the left column of $P^{-1}AP$ only has $0$'s. In other words$$P^{-1}AP=\begin{bmatrix}0&\alpha\\0&\beta\end{bmatrix},$$for some numbers $\alpha$ and $\beta$. But, since $A$ and $P^{-1}AP$ have the same trace, which is $0$, $\beta=0$. So,$$P^{-1}AP=\begin{bmatrix}0&\alpha\\0&0\end{bmatrix}.$$It is not hard to prove now that you can change $P$ a bit so that $\alpha=0$ or $\alpha=1$.

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  • $\begingroup$ Thanks a lot I knew there had to be a better way to do it, other than solving that horrible system! $\endgroup$ – Justin Stevenson Feb 3 at 21:11

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