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
\[XY-YX=I.\]
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Suppose that such matrices exist and consider the trace of the matrix $XY-YX$.

Recall that the trace of a square matrix $A$ is the sum of the diagonal entries of $A$.

Proof.

We use the following basic properties of the trace of matrices.
Let $A, B$ be $n\times n$ matrices and let $c$ be a scalar.

$\tr(A+B)=\tr(A)+\tr(B)$.

$\tr(cA)=c\tr(A)$.

$\tr(AB)=\tr(BA)$.

Seeking a contradiction, we assume that there are matrices $X$ and $Y$ such that $XY-YX=I$.

Then we take the trace of both sides and obtain
\begin{align*}
n&=\tr(I)\\
&=\tr(XY-YX)\\
&=\tr(XY)-\tr(YX) \qquad \text{ (by property (1), (2) of the trace)}\\
&=\tr(XY)-\tr(YX) \qquad \text{ (by property (3) of the trace)}\\
&=0.
\end{align*}

Since $n$ is a positive integer, this is a contradiction.
Therefore, such matrices $X, Y$ do not exist.

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Did we assume tr(XY−YX)=n in the proof?

Note that the trace of the $n\times n$ identity matrix is $n$: $\tr(I)=n$.

By assumption, we have $XY-YX=I$. Combining these we have $n=tr(XY-YX)$.