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**Showcase_22** Let $\displaystyle \tau: W \times V \rightarrow \mathbb{F}$ be a bilinear map, where $\displaystyle V$ and $\displaystyle W$ are vector spaces over a field $\displaystyle \mathbb{F}$. Recall that the dual vector space is denoted by $\displaystyle V^*$. Let $\displaystyle U, \ U_i$ be subspaces of V.

i). Prove that $\displaystyle U^{\perp}=\{w \in W | \tau(w,u)=0 \ \forall u \in U \}$

Prove what? You only wrote the definition of orthogonal complement. Perhaps it was to prove that this complement is a subspace?

My answer: $\displaystyle w= \alpha_1 w_1+\alpha_2 w_2)$.

Hence $\displaystyle \tau(\alpha_1 w_1+\alpha_2 w_2,u)=0$ by the definition of $\displaystyle U^{\perp}$.

We also have that $\displaystyle \alpha_1 \tau (w_1,u)+\alpha_2 \tau(w_2,u)=0$ from the definition of $\displaystyle U^{\perp}$.

Hence we have that $\displaystyle \tau(\alpha_1 w_1+ \alpha_2 w_2,u)=\alpha_1 \tau (w_1,u)+ \alpha_2 \tau (w_2,u) $ so $\displaystyle U^{perp}$ is a subspace of W.

ii). Prove that $\displaystyle U \subset (U^{\perp})^{\perp}$.

This is where things start to go a bit wrong. I have no idea what $\displaystyle (U^{\perp})^{\perp}$ is defined to be!

== $\displaystyle \left(U^\perp \right)^\perp:=\left\{v\in V\;\slash \;\tau(v,w)=0\;\;\forall w\in U^\perp\right\}$==

iii). Prove that $\displaystyle U_1 \subset U_2 \Rightarrow U_2^{\perp} \subset U_1^{\perp}$

I'm also a little stuck on this as well. I decided to start by letting $\displaystyle u_1 \in U_1 \Rightarrow u_1 \in U_2$.

I also know that $\displaystyle \tau(w,u_1)=0$.

However, I wasn't sure where this would take me =S

$\displaystyle \mbox{ Let} w_2\in U_2^\perp \Longrightarrow \forall u_2\in U_2\,,\;\tau(u_2,w_2)=0$ , and this is true is particular for all $\displaystyle u_1\in U_1$ since $\displaystyle U_1 \subset U_2$ !

iv). Given that $\displaystyle T_U: W \rightarrow U^*$ defined by $\displaystyle T_U(w)(u)=\tau(w,u)$ for $\displaystyle w \in W, \ u \in U$ is a linear map from $\displaystyle W$ to $\displaystyle U^*$, and that $\displaystyle ker(T_U)=U^{\perp}$, deduce that $\displaystyle dim(U)+dim(U^{\perp}) \geq dim(W)$

I figure this has something to do with rank+ nullity....

Of course! And it also has to do with the important fact that $\displaystyle U^*\cong U$ for finite dimensional spaces, so by the rank+nullity theorem:

$\displaystyle \dim W = \dim(\ker T_U=U^\perp)+\dim (Im\,\, T_U) \geq \dim U^\perp +\dim U^*$

Tonio

I'd really appreciate any help with this. I'm finding the concept of bilinear maps and dual vector spaces quite hard =S