Prove that $\{ A \in \mathbb{R}^{n \times n} \mid A \text{ is symmetric}\}^{\bot} = \{ A \in \mathbb{R}^{n \times n} \mid A \ \text{is skew-symmetric}\}$ with $\langle A, B \rangle = \operatorname{Tr}(A^\top B)$.
(Note: I am aware of the other 3 questions posted about this problem on this forum, this one too.)
Proof.
Let $\{A\in\mathbf{R}^{n\times n}\mid A\text{ symmetric}\}=W$ and $\{A\in\mathbf{R}^{n\times n}\mid A\text{ antisymmetric}\}=V$.
We first prove the inclusion $\supseteq$. Let $A\in V$, then we have to show that $A\in W^\perp$, or equivalently $A\perp B$ for all $B\in\mathbf{R}^{n\times n}$ symmetric. We have
$$\langle A,B\rangle=\operatorname{Tr}(A^\top B)=\operatorname{Tr}(-AB)=-\operatorname{Tr}(AB^\top)=-\operatorname{Tr}(B^\top A)=-\langle B,A\rangle=-\langle A,B\rangle$$
so $\langle A,B\rangle=0$, therefore $A\in W^\top$.
Now we prove the inclusion $\subseteq$. Let $A\in W^\perp$, then we have to show that $A\in V$.
Consider the inner product $\langle A,A+A^\top \rangle$. Since $(A+A^\top)^\top=A+A^\top$, we have that $A+A^\top$ is symmetric for all $A\in \mathbf{R}^{n\times n}$ and because $A$ is orthogonal to all symmetric matrices, this inner product is equal to $0$. But we also have
$$0=\langle A,A+A^\top\rangle=\langle A,A\rangle+\langle A,A^\top\rangle \iff \langle A,A\rangle=\langle A,-A^\top\rangle$$
Here comes the problem: I want to deduce from this step that $A=-A^\top$, but I realise that this is not allowed.
How should I prove the second inclusion? I know that otherwise I can use something that uses the unique decomposition of $A\in\mathbf{R}^{n\times n}$ as sum of a symmetric and an antisymmetric matrix $A=\frac{A+A^\top}{2}+\frac{A-A^\top}{2}$, but I don't see how this would help.
Best Answer
You wish to show that for $A\in W^\bot$ you have $A+A^T = 0$. We have $$\begin{aligned} \langle A + A^T, A + A^T\rangle &=\langle A, A+A^T\rangle +\langle A^T, A+A^T\rangle\\ &=0+\langle A^T, A+A^T\rangle \end{aligned}$$ since $A + A^T$ is symmetric and $A\in W^\bot$. If you expand the second term you obtain $$\langle A^T, A+A^T\rangle = \text{Tr}\left(A(A+A^T)\right) = \text{Tr}\left((A+A^T) A\right) =\langle A+A^T, A\rangle = 0 $$ as well, where we used that $\text{Tr}(BC) = \text{Tr}(CB)$ for all matrices $B$ and $C$.
This shows that $\langle A + A^T, A + A^T\rangle=0$ and since $\langle\cdot,\cdot\rangle$ is an inner product this implies that $A + A^T=0$, which gives $A^T = -A$ as desired.