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.

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.

More from my site

• If 2 by 2 Matrices Satisfy $A=AB-BA$, then $A^2$ is Zero Matrix Let $A, B$ be complex $2\times 2$ matrices satisfying the relation $A=AB-BA.$ Prove that $A^2=O$, where $O$ is the $2\times 2$ zero matrix.   Hint. Find the trace of $A$. Use the Cayley-Hamilton theorem Proof. We first calculate the […]
• If Two Matrices are Similar, then their Determinants are the Same Prove that if $A$ and $B$ are similar matrices, then their determinants are the same.   Proof. Suppose that $A$ and $B$ are similar. Then there exists a nonsingular matrix $S$ such that $S^{-1}AS=B$ by definition. Then we […]
• Determine Whether Given Matrices are Similar (a) Is the matrix $A=\begin{bmatrix} 1 & 2\\ 0& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 1& 2 \end{bmatrix}$?   (b) Is the matrix $A=\begin{bmatrix} 0 & 1\\ 5& 3 \end{bmatrix}$ similar to the matrix […]
• Trace, Determinant, and Eigenvalue (Harvard University Exam Problem) (a) A $2 \times 2$ matrix $A$ satisfies $\tr(A^2)=5$ and $\tr(A)=3$. Find $\det(A)$. (b) A $2 \times 2$ matrix has two parallel columns and $\tr(A)=5$. Find $\tr(A^2)$. (c) A $2\times 2$ matrix $A$ has $\det(A)=5$ and positive integer eigenvalues. What is the trace of […]
• The Formula for the Inverse Matrix of $I+A$ for a $2\times 2$ Singular Matrix $A$ Let $A$ be a singular $2\times 2$ matrix such that $\tr(A)\neq -1$ and let $I$ be the $2\times 2$ identity matrix. Then prove that the inverse matrix of the matrix $I+A$ is given by the following formula: $(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.$ Using the formula, calculate […]
• An Example of a Matrix that Cannot Be a Commutator Let $I$ be the $2\times 2$ identity matrix. Then prove that $-I$ cannot be a commutator $[A, B]:=ABA^{-1}B^{-1}$ for any $2\times 2$ matrices $A$ and $B$ with determinant $1$.   Proof. Assume that $[A, B]=-I$. Then $ABA^{-1}B^{-1}=-I$ implies $ABA^{-1}=-B. […] • Matrix XY-YX Never Be the Identity Matrix Let I be the n\times n identity matrix, where n is a positive integer. Prove that there are no n\times n matrices X and Y such that \[XY-YX=I.$   Hint. Suppose that such matrices exist and consider the trace of the matrix $XY-YX$. Recall that the trace of […]
• Does the Trace Commute with Matrix Multiplication? Is $\tr (A B) = \tr (A) \tr (B)$? Let $A$ and $B$ be $n \times n$ matrices. Is it always true that $\tr (A B) = \tr (A) \tr (B)$? If it is true, prove it. If not, give a counterexample.   Solution. There are many counterexamples. For one, take \[A = \begin{bmatrix} 1 & 0 \\ 0 & 0 […]

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.

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Diagonalize the Complex Symmetric 3 by 3 Matrix with $\sin x$ and $\cos x$

Consider the complex matrix \[A=\begin{bmatrix} \sqrt{2}\cos x & i \sin x & 0 \\ i \sin x &0 &-i \sin...

Close