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 $1$.

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 $1$.

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 […]

How to Diagonalize a Matrix. Step by Step Explanation.
In this post, we explain how to diagonalize a matrix if it is diagonalizable.
As an example, we solve the following problem.
Diagonalize the matrix
\[A=\begin{bmatrix}
4 & -3 & -3 \\
3 &-2 &-3 \\
-1 & 1 & 2
\end{bmatrix}\]
by finding a nonsingular […]

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 […]

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 […]

Non-Abelian Simple Group is Equal to its Commutator Subgroup
Let $G$ be a non-abelian simple group. Let $D(G)=[G,G]$ be the commutator subgroup of $G$. Show that $G=D(G)$.
Definitions/Hint.
We first recall relevant definitions.
A group is called simple if its normal subgroups are either the trivial subgroup or the group […]

Group Generated by Commutators of Two Normal Subgroups is a Normal Subgroup
Let $G$ be a group and $H$ and $K$ be subgroups of $G$.
For $h \in H$, and $k \in K$, we define the commutator $[h, k]:=hkh^{-1}k^{-1}$.
Let $[H,K]$ be a subgroup of $G$ generated by all such commutators.
Show that if $H$ and $K$ are normal subgroups of $G$, then the subgroup […]

True or False: If $A, B$ are 2 by 2 Matrices such that $(AB)^2=O$, then $(BA)^2=O$
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 […]