# Thread: Proof with Transformations and Subspaces

1. ## Proof with Transformations and Subspaces

I have this proof that I cannot figure out. If anyone could give me a little headstart that would be great. I am still struggling to fully grasp the concepts so maximum explanation is useful please.

Also I dont know how to make the perp symbol that looks like an upside down T so ill use an L instead.
V^L=V perp
W c V means W is a subspace of V

Suppose that A is a real symmetric n x n matrix. Show that if V is
a subspace of Rn and that A(V) c V, then A(V^L) c V^L.

Any help??!?!
Thanks

2. If $\displaystyle y\in V^{\perp}$ then, for all $\displaystyle x\in V$ we have:

$\displaystyle <Ay,x>=(Ay)^tx=y^tA^tx=y^tAx=<y,Ax>$

But by hypothesis $\displaystyle Ax\in V$, so:

$\displaystyle <y,Ax>=0=<Ay,x>\;\forall x\in V$

That is, $\displaystyle Ay\in V^{\perp}$ .

Regards.

Fernando Revilla

3. Originally Posted by BrianMath
I have this proof that I cannot figure out. If anyone could give me a little headstart that would be great. I am still struggling to fully grasp the concepts so maximum explanation is useful please.

Also I dont know how to make the perp symbol that looks like an upside down T so ill use an L instead.
V^L=V perp
W c V means W is a subspace of V

Suppose that A is a real symmetric n x n matrix. Show that if V is
a subspace of Rn and that A(V) c V, then A(V^L) c V^L.

Any help??!?!
Thanks

Denote by $\displaystyle \langle,\rangle$ the usual inner product in $\displaystyle \mathbb{R}^n$ , so that $\displaystyle y\in L^\bot\Longleftrightarrow \langle x,y\rangle=0\,\,\,\forall x\in L$

Since $\displaystyle A$ is symmetric this means it is self-adjoint , so $\displaystyle \langle Av,u\rangle=\langle v,Au\rangle,\,\,\,\forall u,v\in\mathbb{R}^n$

Let now $\displaystyle x\in L^\bot\Longrightarrow \langle x,v\rangle=0\,\,\,\forall v\in L$ , and thus for any $\displaystyle v\in L$ we get:

$\displaystyle \langle Ax,v\rangle=\langle x,Av\rangle = 0$ , since $\displaystyle Av\in L$ by assumption.

Tonio

4. ## Thanks

Damn, this is a lot simpler than I thought. I should have thought of this myself. I was trying things with the basis of V and some other things that were just unnecessary haha.
Thanks a bunch.