# Proof with Transformations and Subspaces

• Nov 29th 2010, 07:30 PM
BrianMath
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.

Attachment 19900

Any help??!?!
Thanks
• Nov 30th 2010, 03:07 AM
FernandoRevilla
If $y\in V^{\perp}$ then, for all $x\in V$ we have:

$=(Ay)^tx=y^tA^tx=y^tAx=$

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

$=0=\;\forall x\in V$

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

Regards.

Fernando Revilla
• Nov 30th 2010, 03:18 AM
tonio
Quote:

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.

Attachment 19900

Any help??!?!
Thanks

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

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

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

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

Tonio
• Nov 30th 2010, 04:40 AM
BrianMath
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.