Let $\displaystyle \text{Sym}_n\left(\mathbb{R}\right)$ be the set of all symmetric $\displaystyle n\times n$ matrices and $\displaystyle \text{ASym}_n\left(\mathbb{R}\right)$ the set of all anti-symmetric matrices (that is matrices $\displaystyle A$ with $\displaystyle A^\top=-A$). I claim that $\displaystyle \text{Sym}_n\left(\mathbb{R}\right)^\perp=\text{AS ym}_n\left(\mathbb{R}\right)$. Indeed, assuming that you are imposing the inner product on $\displaystyle \text{Mat}_n\left(\mathbb{R}\right)$ by identifying it with $\displaystyle \mathbb{R}^{n^2}$ you can readily prove (if you're really desperate a proof can be gleaned from

here) that $\displaystyle \left\langle A,B\right\rangle=\text{tr}\left(AB^{\top}\right)$ from where it's immediate that $\displaystyle \text{ASym}_n\left(\mathbb{R}\right)\subseteq\text {Sym}_n\left(\mathbb{R}\right)^\perp$ since if

$\displaystyle A\in\text{ASym}_n\left(\mathbb{R}\right),S\in\text {Sym}_n\left(\mathbb{R}\right)$ then

$\displaystyle \left\langle A,B\right\rangle=\text{tr}\left(AB^{\top}\right)= \text{tr} \left(-AB\right)=-\text{tr}\left(AB\right)$

but with equal validity

$\displaystyle \left\langle A,B\right\rangle=\left\langle B,A\right\rangle=\text{tr}\left(BA^\top\right)= \text{tr} \left(BA\right)$

and so $\displaystyle \left\langle A,B\right\rangle=0$. Now, to finish the argument note that every matrix $\displaystyle M\in\text{Mat}_n\left(\mathbb{R}\right)$ may be written as

$\displaystyle \displaystyle M=\underbrace{\frac{M+M^\top}{2}}_{\text{symmetric }}+\underbrace{\frac{M-M^\top}{2}}_{\text{anti-symmetric}}$

and so $\displaystyle \text{Mat}_n\left(\mathbb{R}\right)=\text{Sym}_n \left(\mathbb{R}\right)+\text{ASym}_n\left(\mathbb {R}\right)$ that said it's evident that $\displaystyle \text{Sym}_n\left(\mathbb{R}\right)\cap\text{ASym} _n\left(\mathbb{R}\right)=\{\bold{0}\}$ since if $\displaystyle M$ is in the intersection then $\displaystyle M=M^\top=-M$. Thus, $\displaystyle \text{Mat}_n\left(\mathbb{R}\right)= \text{Sym}_n \left(\mathbb{R}\right)\oplus\text{ASym}_n\left( \mathbb{R} \right)$ and so

$\displaystyle \text{codim}\left(\text{ASym}_n \mathbb{R}\right)=\dim\text{Sym}_n\left(\mathbb{R} \right)=\text{codim} \text{Sym}_n\left(\mathbb{R}\right)^\perp$

And so recalling that $\displaystyle \text{ASym}_n\left(\mathbb{R}\right)\subseteq\text {Sym}_n\left(\mathbb{R}\right)^\perp$ you may conclude with a dimension argument