# 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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### 1 Response

1. Kuldeep Sarma says:

This problem can also be done in this way. Let C=AB and D=BA. Then both C,D are 2 by 2 matrices and according to given condition C^2=0. Hence C is a nilpotent matrix. So all eigenvalues of C=AB are 0 and we know for square matrices,AB and BA have same set of eigenvalues, So all eigenvalues of BA are zero. Suppose p(t) be the characteristic polynomial of D=BA. Then we have p(t)=t^2. And by cayley hamilton theorem, we have p(D)=0 or D^2=0 or (BA)^2=0.

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