Given $V=M_{3}, \ U=\{A \in M_{3}; \ A^t=A \}$ subspace of $V$, find the supplementary subspace $W$ of $U$ ($W\leq V, \ V=U\oplus W$) I can't think in a general approach to solve this problem. I know that there must be a surjection in $M_{3}(\mathbb{R})$ from $U$ and $V$, and the diagonal of the elements from $U$ already satisfy the condition for each element diagonal from $M_{3}$, so $W$ elements diagonal can be all zeros. But I don't know how to solve the restriction imposed by the rest of the elements entries from $U$, also keeping $U\cap W = \{0_{V}\}$. I would like someone could show a solution to this problem.
 A: I will prove that, in general, if $U_n(\mathbb{R}):=\{A\in M_n(\mathbb{R}):A^t=A\}$ and $W_n(\mathbb{R}):=\{A\in M_n(\mathbb{R}):A^t=-A\}$ are the set of symmetric and antisymmetric $n\times n$ matrices respectively, then $U_n(\mathbb{R})\oplus W_n(\mathbb{R})=M_n(\mathbb{R})$. Your question is the particular case $n=3$.
Clearly, $U_n(\mathbb{R}),W_n(\mathbb{R})\leq M_n(\mathbb{R})$ are subspaces. And, if $A\in U_n(\mathbb{R})\cap W_n(\mathbb{R})$, then $A^t=A=-A^t\Rightarrow A=-A\Rightarrow A=0_{n\times n}$ which implies that $U_n(\mathbb{R})\cap W_n(\mathbb{R})=\{0_{n\times n}\}$. Now, given $A\in M_n(\mathbb{R})$, define $U_A=\frac{1}{2}(A+A^t)$ and $W_A=\frac{1}{2}(A-A^t)$. Then,
$U_A^t=[\frac{1}{2}(A+A^t)]^t=\frac{1}{2}(A^t+A)=U_A$ and $W_A^t=[\frac{1}{2}(A-A^t)]^t=\frac{1}{2}(A^t-A)=-W_A$ which means that $U_A\in U_n(\mathbb{R})$ and $W_A\in W_n(\mathbb{R})$ are matrices such that $A=U_A+W_A\in U_n(\mathbb{R})+W_n(\mathbb{R})$. Hence, since $A$ was arbitrary, we can also conclude that $M_n(\mathbb{R})=U_n(\mathbb{R})+W_n(\mathbb{R})$. Thus, $U_n(\mathbb{R})$ and $W_n(\mathbb{R})$ are supplementary subspaces of $M_n(\mathbb{R})$.
