In this case $(U+W)^\perp$ means the orthogonal complement of the union of $U$ and $W$ (regardless of repetition, is it correct?). The vectors inside the orthogonal complement has to be orthogonal to every vector in $(U+W)$ and so it must be composed of the orthogonal complement for each $U$ and $W$.
Digitally, $(U+W)^\perp=${$x$ in $ℝ^n|x \cdot f_u =0$ $\iff x \cdot f_w =0$ }=U$^\perp \cap W^\perp$ .
But I felt the proof is either somewhat incorrect or incomplete, could anyone give some opinions?
Best Answer
Please excuse the poor formatting. Before I get into the proof, I'd like to explain the intuition. If you're orthogonal to any vector of the form $(c_1)u+(c_2)v$, then taking $c_2$ to be zero, you must be orthogonal to every $u$ and similarly every $v$. Thus being in the orthogonal compliment of $(U+V)$ implies that you're in the orthogonal compliment of U and in the orthogonal compliment of $V$. The reverse is obviously true. If you're orthogonal to $u$ and $v$ then you're clearly orthogonal to any linear combo of them.
If $ x$ is in $(U+V)^⊥$ then $ x \cdot ( u+ v)=0$ for any $ u$ in $U$ and $ v$ in $V$. Namely, since $ 0$ is in $U$, $x \cdot v=0$ for all $v$ in $V$. Similarly $ x \cdot u=0$ for all $ u$ in $V$. Thus, $ x$ is in $U^⊥$ and $ x$ is in $V^⊥$, so x is in the intersection of $U^⊥$ and $V^⊥$. So $(U+V)^⊥$ is a subset of $U^⊥ \cap V^⊥$. The reverse is true immediately. If $x$ is in $U^⊥$ and $V^⊥$, then clearly $x(u+v)= xu+xv=0$. Thus, $x$ is in $(U+V)^⊥$. So $U^⊥ \cap V^⊥$ is a subset of $(U+V)^⊥$. This means that $(U+V)^⊥ = U^⊥ \cap V^⊥$.