# How to show that $M_{2\times 2}(\mathbb{R})=W_1\oplus W_2$ based on the following assumption?

Let the subspaces $W_1=\{\begin{pmatrix}a&b\\-b&a \end{pmatrix}|a, b\in \mathbb{R}\}$ and $W_2=\{\begin{pmatrix}c&d\\d&-c \end{pmatrix}|c, d\in \mathbb{R}\}$ of $M_{2\times 2}(\mathbb{R})$,how to show that $M_{2\times 2}(\mathbb{R})=W_1\oplus W_2$?

By the definition of direct sum, $V= W_1\oplus W_2 \iff V=W_1+W_2$ and $W_1\cap W_2=\{0\}$ .

Since both $W$ are subspaces of $V$, we can write $v=w_1+w_2$ where $w_1\in W_1$ and $w_2\in W_2$. Because every $v$ can be represented as $e\begin{pmatrix}a&b\\-b&a \end{pmatrix}+ f\begin{pmatrix}c&d\\d&-c \end{pmatrix}$, $V=W_1+W_2$. Since each $w_1\in W_1$ cannot linearly combines to form vectors in $W_2$, $W_1\cap W_2=\{0\}$ .

I think my proof is totally useless and does not show anything, could anyone revise it?

\begin{align*}&\bullet\;\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}\frac{a+d}2&\frac{b-c}2\\-\frac{b-c}2&\frac{a+d}2\end{pmatrix}+\begin{pmatrix}\frac{a-d}2&\frac{b+c}2\\\frac{b+c}2&-\frac{a-d}2\end{pmatrix}\;\implies\;V=W_1+W_2\\{}\\ &\bullet\;\begin{pmatrix}a&b\\c&d\end{pmatrix}\in W_1\cap W_2\implies\begin{cases}a=d=-d\\b=-c=c\\\end{cases}\;\implies a=b=c=d=0\end{align*}
and thus $\;V=W_1\oplus W_2\;$