How to show $\mathrm{Sym}_{n\times n}(\mathbb{R}) + \mathrm{Skew}_{n\times n}(\mathbb{R})= \mathrm{M}_{n\times n}(\mathbb{R})$? $$\mathrm{Sym}_{n\times n}(\mathbb{R}) + \mathrm{Skew}_{n\times n}(\mathbb{R}) \stackrel{?}{=} \mathrm{M}_{n\times n}(\mathbb{R})$$ Using the simplest core ideas of linear algebra, could someone perhaps give a hint as to where I might start to show this to be the case?
 A: As pointed out by Mariano Suárez-Alvarez, you can use dimension. There is a more elementary way, which generalizes exactly the same way to cases where the dimension argument fails, namely in $C^*$ and von Neumann algebras.
Here is the trick:

$$
M=\frac{M+M^t}{2}+\frac{M-M^t}{2}.
$$

Recall that $(M^t)^t=M$ and that $M$ is symmetric (resp. antisymmetric) if and only if $M^t=M$ (resp. $M^t=-M$).
I let you check it proves the harder direction of the equality you want, namely $M_n(\mathbb{R})\subseteq Sym+Skew$. The reverse inclusion is obvious.
A: Here I give another proof:
We can see easily that
$$f:\mathcal{M}_n(\mathbb{R})\to \mathcal{M}_n(\mathbb{R}),\quad A\mapsto A^T$$
is an endomorphism of vector spaces and that $f^2=\mathrm{Id}$ so $x^2-1$ is the minimal polynomial of $f$, moreover we have $\ker(f-\mathrm{Id})=\mathrm{Sym}_{n\times n}(\mathbb{R})$ and $\ker(f+\mathrm{Id})=\mathrm{Skew}_{n\times n}(\mathbb{R})$ hence
$$\ker(f-\mathrm{Id})\oplus\ker(f+\mathrm{Id})=\mathrm{Sym}_{n\times n}(\mathbb{R})\oplus \mathrm{Skew}_{n\times n}(\mathbb{R})=\mathcal{M}_n(\mathbb{R})$$
A: If $M_1 \in Sym_{n\times n}(\mathbb{R})$, then $a_{ij} = a_{ji}$, so $a_{ij}=\frac{1}{2}a_{ij}+\frac{1}{2}a_{ji}$. Similarly, if $M_2 \in Skew_{n\times n}(\mathbb{R})$, then $a_{ij} = -a_{ji}$, so $a_{ij}=\frac{1}{2}a_{ij}-\frac{1}{2}a_{ji}$, by the definition of symmetric and skew-symmetric matrices; thus, $\frac{1}{2}a_{ij}+\frac{1}{2}a_{ji}+\frac{1}{2}a_{ij}-\frac{1}{2}a_{ji}=a_{ij}$.

I think there is a clearer way...
