# When is a matrix similar to $\begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}$

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!

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 $$\alpha=1$$.

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