True or False: If $A, B$ are 2 by 2 Matrices such that $(AB)^2=O$, then $(BA)^2=O$
Problem 537
Let $A$ and $B$ be $2\times 2$ matrices such that $(AB)^2=O$, where $O$ is the $2\times 2$ zero matrix.
Determine whether $(BA)^2$ must be $O$ as well. If so, prove it. If not, give a counter example.
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Proof.
It is true that the matrix $(BA)^2$ must be the zero matrix as we shall prove now.
For notational convenience, let $C:=AB$.
As $C$ is a $2\times 2$ matrix, it satisfies the relation
\[C^2-\tr(C)C+\det(C)I=O \tag{*}\]
by the Cayley-Hamilton theorem.
Here $I$ is the $2\times 2$ identity matrix.
We compute the determinant of $C$ as follows.
We have
\begin{align*}
\det(C)^2=\det(C^2)=\det((AB)^2)=\det(O)=0.
\end{align*}
Hence $\det(C)=0$.
Since $C^2=O$ and $\det(C)=0$, the Cayley-Hamilton relation (*) becomes
\[\tr(C)C=O.\]
It follows that we have either $C=O$ or $\tr(C)=0$.
In either case, we have $\tr(C)=0$.
Note that
\begin{align*}
\det(BA)&=\det(AB)=\det(C)=0\\
\tr(BA)&=\tr(AB)=\tr(C)=0.
\end{align*}
Applying the Cayley-Hamilton theorem to the matrix $BA$, we obtain
\begin{align*}
O=(BA)^2-\tr(BA)\cdot BA+\det(BA)I=(BA)^2.
\end{align*}
Thus, we obtain $(BA)^2=O$ as claimed.
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