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The matrix is symmetric

Let A be a square matrix and for that it holds that

A^2 =AA^{\top}

Prove that A is symmetric.

Solution

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

(1)   \begin{equation*} A^2 =AA^{\top} \end{equation*}

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

(2)   \begin{equation*} A^2 = \left( A^\top \right)^2 \end{equation*}

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

(3)   \begin{equation*} A A^\top = A^\top A \end{equation*}

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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