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## A symmetric matrix

Linear Algebra
Riemann
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### A symmetric matrix

Let $A$ be square matrix over a field $\mathbb{F}$. If
$$A^2 = A A^\top$$
holds , then prove that $A$ is symmetric.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Tolaso J Kos
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### Re: A symmetric matrix

Let $A$ be an $n \times n$ square matrix over a field $\mathbb{F}$ such that

$$A^2 =AA^{\top}$$

Taking transposed matrices back at $(1)$ we get that

\begin{align*}
\left ( A^2 \right )^\top = \left ( A A^\top \right )^\top &\Rightarrow \left ( A^\top \right )^2 = \left ( A^\top \right )^\top A^\top
\\ &\Rightarrow \left ( A^2 \right )^2 = A A^\top
\end{align*}

and thus

$$A^2 = \left( A^\top \right)^2$$

On the other hand it holds that

$$\left ( A A^\top - A^\top A \right )^2 = \mathbb{O}_{n \times n}$$

since

\begin{align*}
\left ( A A^\top - A^\top A \right )^2 &= \left ( A A^\top - A^\top A \right ) \left ( A A^\top - A^\top A \right ) \\
&=A A^\top A A^\top - A A^\top A^\top A - \\
&\quad \quad -A^\top A A A^\top + A^\top A A^\top A \\
&\overset{(2)}{=} \cancel{A A A A - A A AA} - \\ &\quad \quad -A^\top AA A^\top +A^\top A A^\top A\\
&=-A^\top AA A^\top +A^\top A A^\top A \\
&\overset{(2)}{=} \cancel{A^\top A^\top A^\top A^\top - A^\top A^\top A^\top A^\top} \\ &=\mathbb{O}_{n \times n}
\end{align*}

Of course it holds that if a matrix $M$ is symmetric or antisymmetric and $M^2=\mathbb{O}_{n \times n}$ then $M=\mathbb{O}_{n \times n}$. The proof is left as an exercise to the reader.

We can safely conclude using the above observations that for our matrix $A$ it holds that

$$A A^\top = A^\top A$$

But then for the matrix $A-A^\top$ it holds that

\begin{align*}
\left ( A - A^\top \right )^2 &= \left ( A - A^\top \right ) \left ( A - A^\top \right ) \\
&=A A - A A^\top - A^\top A + A^\top A^\top \\
&\mathop {=} \limits_{(3)}^{(2)} A A - A A^\top -A A^\top + A A \\
&\overset{(2)}{=} A A - AA - AA + AA\\
&= \mathbb{O}_{n \times n}
\end{align*}

and since the matrix $A- A^\top$ is antisymmetric we conclude that $A - A^\top = \mathbb{O}_{n \times n}$ and thus $A = A^\top$. Hence the result.
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