Let $V=M_{n \times n} (F)$ and $S=\{ \mbox{symmetric matrices} \}$ and let $ S \leqslant V $. Find $ T \leqslant V $ such that $ V=S \oplus T $.
My general concept is first to find out $B$, the basis of the symmetric matrix $S$. And then extend it to make $B{1}$ the basis of V.The vectors that we add in $B$ will span the matrix T.
But I don't know how to put this thing together. Need Help!
First we assume that our field over which the matrices live has characteristics not $2$ (e.g. the real numbers). We say that a matrix is antisymmetric if $a_{ij}=-a_{ji}$. We denote by $T$ the vector space(!) anti-symmetric matrices. Our claim is that $T\oplus S=M_n$. We have to check that 1) every matrix can be written as the sum of a symmetric and anti-symmetric matrix and 2) The intersection of T with S is the null-space. The second is easy as from $a_{ij}=a_{ji}=-a_{ji}$ it follows that $a_{ij}=0$ (which is false in characteristic $2$). For the first you have to check that every matrix is the sum of a symmetric and an anti-symmetric matrix. Now check using a computation: $A=\frac 12(A+A^t)+\frac 12(A-A^t)$. The first summand is symmetric and the second is anti-symmetric ( here $A^t$ denotes the transpose of a matrix). Now you are done!
Hint : any matrix $A$ can be written $A = \frac{1}{2}(A + A^t) + \frac{1}{2}(A - A^t)$.
