If $A^{\trans}A=A$, then $A$ is a Symmetric Idempotent Matrix

Problem 447

Let $A$ be a square matrix such that
\[A^{\trans}A=A,\]
where $A^{\trans}$ is the transpose matrix of $A$.
Prove that $A$ is idempotent, that is, $A^2=A$. Also, prove that $A$ is a symmetric matrix.

Recall the basic properties of transpose matrices.
For matrices $A, B$, we have

$(AB)^{\trans}=B^{\trans}A^{\trans}$.

$A^{\trans \trans}=A$.

Proof.

We first prove that $A$ is a symmetric matrix.
We have
\begin{align*}
A^{\trans}&=(A^{\trans}A)^{\trans}\\
&=A^{\trans}A^{\trans\trans} && \text{by property 1}\\
&=A^{\trans}A && \text{by property 2}\\
&=A.
\end{align*}
Hence we obtained $A^{\trans}=A$, and thus $A$ is a symmetric matrix.

Now we prove that $A$ is idempotent.
We compute
\begin{align*}
A^2&=AA\\
&=A^{\trans}A && \text{since $A$ is symmetric}\\
&=A && \text{by assumption}.
\end{align*}
Therefore, the matrix $A$ satisfies $A^2=A$, and hence it is idempotent.

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