Proof for $(U+W)^\perp=U^\perp \cap W^\perp$ if $U$ and $W$ are both subspaces.

Question

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?

Answer 1

$U+W$ is not the union of the two subspaces, which is not in general a linear subspace by the way, but the sum.

$$(U+W)^\perp=\{\varphi \in E^* : \varphi (u+w)=0 , \forall (u,w) \in U \times W\}$$

By the way the proof of the equality of your question is not trivial and requires the axiom of choice for infinite dimensional subspaces.

Answer 2

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 $(U+V)^⊥ = U^⊥ \cap V^⊥$.

Answer 3

Let $U$ and $W$ be subspaces of a vector space $V$ over a field $K$. In general, for a subset $S$ of $V$, $$S^\perp:=\big\{f\in V^*\,\big|\,f(s)=0\text{ for every }s\in S\big\}$$ is a vector subspace of the algebraic dual $V^*=\text{Hom}_K(V,K)$ of $V$ (and $S^\perp$ does not sit within $V$, at least not canonically). Then, $$(U+W)^\perp = \big\{f\in V^*\,\big|\,f(u+w)=0\text{ for all }u\in U\text{ and }w\in W\big\}\,.$$ If $f\in (U+W)^\perp$, then $f(u)=f(u+0)=0$ and $f(w)=f(0+w)=0$ for all $u\in U$ and $w\in W$. This shows that $f\in U^\perp \cap W^\perp$.

On the other hand, if $f\in U^\perp \cap W^\perp$, then by linearity, $f(u+w)=f(u)+(w)=0+0=0$ for all $u\in U$ and $w\in W$. Thus, $f\in (U+W)^\perp$. This shows that $(U+W)^\perp = U^\perp \cap W^\perp$.

However, if $V$ is endowed with a nondegenerate bilinear form $\langle\_,\_\rangle: V\times V\to K$ (this seems to be the case for the OP's question), then we can also define $$S_\perp:=\big\{v\in V\,\big|\,\langle v,s\rangle =0\text{ for every }s\in S\big\}\,.$$ Using a similar argument as above, we have $$(U+W)_\perp = U_\perp \cap W_\perp\,.$$

P.S. We also have $(U\cap W)^\perp = U^\perp +W^\perp$ and, if $V$ is endowed with a nondegenerate bilinear form, $(U\cap W)_\perp = U_\perp +W_\perp$. This is the part where the Axiom of Choice is needed.