Bilinear maps and Dual Vector Spaces

**Showcase_22** · Oct 27th 2009, 01:08 PM

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

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

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

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

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

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

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

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

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

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

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

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

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

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

**tonio** · Oct 27th 2009, 02:33 PM

Originally Posted by Showcase_22

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Tonio

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

.

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