# An Example of a Matrix that Cannot Be a Commutator

## Problem 565

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
$ABA^{-1}=-B. \tag{*}$ Taking the trace, we have
$-\tr(B)=\tr(-B)=\tr(ABA^{-1})=tr(BAA^{-1})=\tr(B),$ hence the trace $\tr(B)=0$.
Thus, the characteristic polynomial of $B$ is
$x^2-\tr(B)x+\det(B)=x^2+1.$ Hence the eigenvalues of $B$ are $\pm i$.

Note that the matrix $\begin{bmatrix} 0 & -1\\ 1& 0 \end{bmatrix}$ has also eigenvalues $\pm i$.
Thus this matrix is similar to the matrix $B$ as both matrices are similar to the diagonal matrix $\begin{bmatrix} i & 0\\ 0& -i \end{bmatrix}$.
Let $P$ be a nonsingular matrix such that
$B’:=P^{-1}BP=\begin{bmatrix} 0 & -1\\ 1& 0 \end{bmatrix}.$ Let $A’=P^{-1}AP$.

The relation (*) is equivalent to $AB=-BA$.
Using this we have
\begin{align*}
A’B’&=(P^{-1}AP)(P^{-1}BP)=P^{-1}(AB)P\\
&=P^{-1}(-BA)P=-(P^{-1}BP)(P^{-1}AP)=-B’A’.
\end{align*}

Let $A’=\begin{bmatrix} a & b\\ c& d \end{bmatrix}$.
Then $A’B’=-B’A’$ gives
\begin{align*}
\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}
\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}
=-\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}\\[6pt] \Leftrightarrow
\begin{bmatrix}
b & -a\\
d& -c
\end{bmatrix}=\begin{bmatrix}
c & d\\
-a& -b
\end{bmatrix}.
\end{align*}
Hence we obtain $d=-a$ and $c=b$.

Then
\begin{align*}
1=\det(A)=\det(PA’P^{-1})=\det(A’)=\begin{vmatrix}
a & b\\
b& -a
\end{vmatrix}=-a^2-b^2,
\end{align*}
which is impossible.
Therefore, the matrix $-I$ cannot be written as a commutator $[A, B]$ for any $2\times 2$ matrices $A, B$ with determinant $1$.

Let $A$ and $B$ be $n\times n$ skew-symmetric matrices. Namely $A^{\trans}=-A$ and $B^{\trans}=-B$. (a) Prove that $A+B$ is skew-symmetric. (b)...