Proof with Transformations and Subspaces

**BrianMath** · November 29th 2010, 07:30 PM

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.

Proof with Transformations and Subspaces-screen-shot-2010-11-29-11.32.53-pm.png

Any help??!?!
Thanks

**FernandoRevilla** · November 30th 2010, 03:07 AM

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

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

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

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

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

Regards.

Fernando Revilla

**tonio** · November 30th 2010, 03:18 AM

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.

Click image for larger version.

Name: Screen shot 2010-11-29 at 11.32.53 PM.png
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ID: 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

**BrianMath** · November 30th 2010, 04:40 AM

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.

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