Proof verification: $X^{r}=\left(\begin{array}{ll}0 & 1 \\0 & 0\end{array}\right)$ has no solution in $M_{2}(\mathbb {C})$ for $ r \geq 2$

Question

Let $ r\geq 2$ be a natural number. Show this the equation

$X^{r}=\left(\begin{array}{ll}0 & 1 \\0 & 0\end{array}\right)$

has no solution in $M_{2}(\mathbb {C})$.

Proof:

I tried proving with induction. So the equation isn't true for $n = 2$ and let's consider the induction step with $r \rightsquigarrow r+1$

$X^{r+1}=\left(\begin{array}{ll}0 & 1 \\0 & 0\end{array}\right) \\ \Leftrightarrow X^{r}X=\left(\begin{array}{ll}0 & 1 \\0 & 0\end{array}\right) \\ \Leftrightarrow X^{r}X=X^{r}$

Which is not true for $ r \geq 2$.

Also, I'm not sure if I'm on the right track, it just seems I'm missing out on something.

I think there might be another way of proving this with eigenvalues but I couldn't get on with that. I'd be very happy seeing a solution with this!

Answer 1

Your attempt is flawed. For instance:

• $X^{r}X=\left(\begin{array}{ll}0 & 1 \\0 & 0\end{array}\right) \not\Leftrightarrow X^{r}X=X^{r}$
• "$X^{r}X=X^{r}$ is not true for $ r \geq 2$" has no reason to hold.

Here is a proof (by contradiction):

If $X^r=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ then $X^{2r}=\begin{pmatrix}0&1\\0&0\end{pmatrix}^2=0_2$ (the $2\times2$ zero matrix) hence $X$ is a nilpotent $2\times2$ matrix, i.e. $X^2=0_2,$ and therefore (since $r\ge2$) $X^r=0_2\ne\begin{pmatrix}0&1\\0&0\end{pmatrix},$ a contradiction.