Orthogonality and inner product space

Question

The question relates to orthogonality and inner product spaces. If $ W,W_{1}, W_{2} $ are subspaces of finite dimensional vector space V. We have to prove the following

(a) $ W_{1} \subset W_{2} \implies W_{2}^\bot \subset W_{1}^\bot $

(b) $ (W_{1} + W_{2})^\bot = W_{1}^\bot \cap W_{2}^\bot $

(c) $ (W_{1} \cap W_{2})^\bot = W_{1}^\bot + W_{2}^\bot $

where $ W_{1}^\bot $ refers to the vector space orthogonal to $ W_{1} $

I have tried to solve the second part by the logic that $w\in W_{1} + W_{2} $ is expressed as

$$ w = w_{1} + w_{2} $$

where $ w_{1} \in W_{1} $ and $ w_{2} \in W_{2}$. A vector orthogonal to $w$ will exist in $ (W_{1} + W_{2})^\bot $ and let $ w^\bot $ be that vector. But since $ w = w_{1} + w_{2} $, $ w^\bot $ is orthogonal to $ w_{1} $ and $ w_{2} $ also. It implies that $ w^\bot $ element of $ W_{1}^\bot $ and $ W_{2}^\bot $ also. Hence $ w^\bot \subset W_{1}^\bot \cap W_{2}^\bot $.

Please correct me if I'm wrong and kindly help out with the other solutions.

Answer 1

The language makes things a little cumbersome, so let's try to avoid it as possible:

$$x\in\left(W_1+W_2\right)^\perp\iff \langle x,t\rangle=0\;,\;\forall\,t\in W_1+W_2\implies \forall\,w_i\in W_i\,,\,i=1,2\;:$$

$$0=\langle x,w_1+w_2\rangle\implies \;\text{choosing}\;\;w_2=0\;,\;\text{we get}\;\;\langle x,w_1\rangle=0\implies x\in W_1^\perp$$

and likewise $\,x\in W_2^\perp\;$ . The other direction is almost immediate (almost, though!)

Answer 2

(1) Assume $W_1 \subset W_2$. Now, let

$$ w \in W_2^\bot \implies w \notin W_{2} \implies w \notin W_1 \implies w \in W_1^{\bot} \implies W_2^{\bot} \subset W_1^{\bot}.$$

Note:

If $A\subset B$ and $x\in A$, then $x\in B$.