If 2 by 2 Matrices Satisfy $A=AB-BA$, then $A^2$ is Zero Matrix

Linear algebra problems and solutions

Problem 337

Let $A, B$ be complex $2\times 2$ matrices satisfying the relation

Prove that $A^2=O$, where $O$ is the $2\times 2$ zero matrix.

LoadingAdd to solve later


  1. Find the trace of $A$.
  2. Use the Cayley-Hamilton theorem


We first calculate the trace of the matrix $A$ as follows. We have

Thus $\tr(A)=0$ and it follows from the Cayley-Hamilton theorem (see below) for the $2\times 2$ matrix $A$ that
where $I$ is the $2\times 2$ identity matrix.

Thus, we obtain
\[A^2=-\det(A)I. \tag{*}\]

Next, we compute $A^2$ in two ways.
We have
Adding these two, we have
& \stackrel{(*)}{=} (-\det(A)I)B-B(-\det(A)I)\\

As a result, we obtain $A^2=O$. This completes the proof.

The Cayley-Hamilton theorem for a $2\times 2$ matrix

Let us add the proof of the fact we used in the proof about the Cayley-Hamilton theorem.
Let $A=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$ be a $2\times 2$ matrix.

Then its characteristic polynomial is
a-x & b\\
c& d-x
since $\tr(A)=a+d$ and $\det(A)=ad-bc$.

The Cayley-Hamilton theorem says that the matrix $A$ satisfies its characteristic equation $p(x)=0$.
Namely we have
\[A^2-\tr(A)A+\det(A)I=O.\] This is the equality we used in the proof.


As a variation of this problem, consider the following problem.

Let $A, B$ be $2\times 2$ matrices satisfying $A=AB-BA$.
Then prove that $\det(A)=0$.

LoadingAdd to solve later

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Linear Algebra
Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra
Normal Nilpotent Matrix is Zero Matrix

A complex square ($n\times n$) matrix $A$ is called normal if \[A^* A=A A^*,\] where $A^*$ denotes the conjugate transpose...