For an assignment in class, I have the following question.
Let n $\geq 1$ and let W be a subspace of $Mn\times n(K)$ such that $dim(W)>\frac{n^2-n}{2}$. Prove that W contains a non-zero matrix with the property $A^T=A$.
I am not sure how to start this; I found the dimension of the vector spaces to be between $n^2 \geq dim(W) \geq \frac{n^2-n}{2}+1$.
One way I thought about approaching the problem was that whatever the dimension of the vector space, its basis spans the set so it should be possible to represent a transpose matrix by a linear combination of the matrices in the basis.
So far I have found out the various possibilities of the dimension of W depending on the value of n, and how a transpose matrix looks like for various values of n, but I am kind of stuck on the proof side of the question.
Best Answer
As OP mentions in the comments, the dimension of the space $S_n(K)$ of symmetric $n \times n$ matrices over $K$ is $\frac{1}{2}(n^2 + n)$.
Hint The dimension of the sum of two vector subspaces $S, W$ of a vector space $M$ is $$\dim (S + W) = \dim S + \dim W - \dim (S \cap W) .$$ On the other hand, since $S + W$ is a subspace of $M$, $\dim(S + W) \leq \dim M$.