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Prove that symmetric matrices of order $n$ form the subspace of $M_n(\Bbb F)$ with dimension $\frac{n(n+1)}{2}$.

What I know:

For $M\subseteq V$: $$M\leq V \Leftrightarrow \alpha x+\beta y \in M $$

$\forall \alpha, \beta \in \Bbb F, \forall x,y \in M$.

What I did: I took two arbitrary symmetric matrices $X$ and $Y$ and then I saw that $\alpha x+\beta y$ also gives me a symmetric matrix. That's how I show that it's a subspace, right? But how do I now find the dimension?

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lmc
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  • It is shown here: http://math.stackexchange.com/questions/12531/show-that-the-set-of-all-symmetric-real-matrices-is-a-subspace-determine-the-d?rq=1. – StubbornAtom Dec 18 '16 at 12:03

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Any matrix can be expressed as the sum of symmetric and alternating parts (rather like even/odd decomposition of functions). Define $$ S(A) = (A + A^T)/2 $$ and $$\Lambda(A) = (A - A^T)/2 $$ The sets $$ \{S(A): A \in M_n(V) \} \quad \{\Lambda(A): A \in M_n(V) \} $$ are clearly complementary subspaces, and have dimension $n(n+1)/2$, $n(n-1)/2$ repectively. Counting the dimension is easy. All terms in the upper right triangle are linearly independent. For the symmetric subspace we also get the terms on the diagonal. For the alternating subspace the diagonal is all-$0$, hence the difference of $n$ in their dimension.

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