Consider the subspace $W:=\Bigl \{ \begin{bmatrix} a & b \\ b & a \end{bmatrix} : a,b \in \mathbb{R}\Bigr \}$ of $\mathbb{M}^2(\mathbb{R})$.
I have a few questions about how this can be decomposed.
1) Is there a subspace $V$ of $\mathbb{M}^2(\mathbb{R})$ such that $W\oplus V=\mathbb{M}^2(\mathbb{R})$? If so, what is one?
2) Further, is there a different (i.e., $\ne V$) subspace with the same property? And if not, is there a different proper subspace $U$ such that $W+U= \mathbb{M}^2(\mathbb{R})$?
Which examples, if any, would work for these questions? I haven't made much progress, so seeing explicit examples and explanations would be greatly appreciated.
$$W = \operatorname{span}\left\{\begin{bmatrix}1&0\\0&1\end{bmatrix}, \begin{bmatrix}0&1\\1&0\end{bmatrix}\right\}.$$ Can you find a basis for $\mathbb{M}^2(\mathbb{R})$ that contains the above two matrices? Then the span of the new basis elements is your desired $V$.
Note that your question is exactly the same if you view the matrices as vectors, i.e., $W=\{(a,b,b,a): a,b \in \mathbb{R}\}$ in $\mathbb{R}^4$.
Suppose $W = \operatorname{span}\{(1,0,0,1), (0,1,1,0)\}$. The two vectors are a basis for $W$. Can you extend $\{(1,0,0,1), (0,1,1,0)\}$ into a basis for $\mathbb{R}^4$? One way is $\{(1,0,0,1), (0,1,1,0), (0,0,0,1), (0,0,1,0)\}$.
I claim $V:= \operatorname{span}\{(0,0,0,1),(0,0,1,0)\}$ satisfies your condition 1). Indeed this follows from the definition of a basis.
This is not unique; there are other ways to extend into a basis for $\mathbb{R}^4$. One other way yields $V:= \operatorname{span}\{(1,0,0,-1), (0,1,-1,0)\}$.