Linear operator proof

**demode** · March 25th 2010, 01:48 AM

How do we prove this:

Btw, here are the properties of inner products:

<u,v>=<v,u> (symmetry property)
<u+v,w>=<u,w>+<v,w> (additivity property)
<ku,v>=k<u,v> (homogeneity property)
<v,v> ≥ 0 and <v,v>=0 if and only if v=0 (positivity property)

So, I think to show that a mapping is linear we must show that it satisfies the additivity and homogeneity conditions. But I don't know how to prove these conditions here but here's what I think so far

So, for vectors $x_1,x_2$ and scalars a and b, T(x)= <x,u> v becomes

$T(ax_1+bx_2)= \left\langle ax_1+bx_2,u \right\rangle v$

By the additivity property we get

$= (\left\langle ax_1,u \right\rangle + \left\langle bx_2,u \right\rangle )v$

I don't know what to do after this point. Can anyone help?

**Defunkt** · March 25th 2010, 03:01 AM

Originally Posted by demode

How do we prove this:

Btw, here are the properties of inner products:

<u,v>=<v,u> (symmetry property)
<u+v,w>=<u,w>+<v,w> (additivity property)
<ku,v>=k<u,v> (homogeneity property)
<v,v> ≥ 0 and <v,v>=0 if and only if v=0 (positivity property)

So, I think to show that a mapping is linear we must show that it satisfies the additivity and homogeneity conditions. But I don't know how to prove these conditions here but here's what I think so far

So, for vectors $x_1,x_2$ and scalars a and b, T(x)= <x,u> v becomes

$T(ax_1+bx_2)= \left\langle ax_1+bx_2,u \right\rangle v$

By the additivity property we get

$= (\left\langle ax_1,u \right\rangle + \left\langle bx_2,u \right\rangle )v$

I don't know what to do after this point. Can anyone help?

To show that a transformation $T:V \to W$ is linear, you need to show that for any scalar $\alpha \in F$ and any two vectors $v_1,v_2 \in V$ , $T(\alpha v_1 + v_2) = \alpha T(v_1) + T(v_2)$ .

Now in your case, let $\alpha$ be a scalar and $x_1,x_2$ be vectors in V. Then, $T(\alpha x_1 + x_2) = <\alpha x_1 + x_2, u>v$ .

Now, you need to show that $T(\alpha x_1 + x_2) = \alpha T(x_1) + T(x_2)$ . Note that $\alpha T(x_1) + T(x_2) = \alpha <x_1,u>v ~ + <x_2,u>v$ . Can you show the equality now?

**demode** · March 25th 2010, 11:51 AM

Originally Posted by Defunkt

To show that a transformation $T:V \to W$ is linear, you need to show that for any scalar $\alpha \in F$ and any two vectors $v_1,v_2 \in V$ , $T(\alpha v_1 + v_2) = \alpha T(v_1) + T(v_2)$ .

Now in your case, let $\alpha$ be a scalar and $x_1,x_2$ be vectors in V. Then, $T(\alpha x_1 + x_2) = <\alpha x_1 + x_2, u>v$ .

Now, you need to show that $T(\alpha x_1 + x_2) = \alpha T(x_1) + T(x_2)$ . Note that $\alpha T(x_1) + T(x_2) = \alpha <x_1,u>v ~ + <x_2,u>v$ . Can you show the equality now?

$T(\alpha x_1+x_2) = \left\langle \alpha x_1+x_2,u \right\rangle v$

By the homogeneity property:

$=\alpha \left\langle x_1+x_2,u \right\rangle v$

By the additivity property:

$\alpha ( \left\langle x_1,u \right\rangle + \left\langle x_2,u \right\rangle)v$

Now what do I need to do? Do I just distribute the "v"?

**Defunkt** · March 25th 2010, 02:28 PM

Originally Posted by demode

$T(\alpha x_1+x_2) = \left\langle \alpha x_1+x_2,u \right\rangle v$

By the homogeneity property:

$=\alpha \left\langle x_1+x_2,u \right\rangle v$

This part is not correct. You know that $<\alpha v_1, v_2> = \alpha<v_1, v_2>$ but this is not the case here. You need to use additivity first.

By the additivity property:

$\alpha ( \left\langle x_1,u \right\rangle + \left\langle x_2,u \right\rangle)v$

Now what do I need to do? Do I just distribute the "v"?

.

**demode** · March 26th 2010, 10:34 PM

Is the following right?

$T(\alpha x_1,x_2)= \left\langle \alpha x_1 + x_2, u \right\rangle v$

$= ( \left\langle \alpha x_1,u \right\rangle + \left\langle x_2,u \right\rangle )v$

$= \left\langle \alpha x_1,u \right\rangle v+ \left\langle x_2,u \right\rangle v$

$= \alpha \left\langle x_1,u \right\rangle v + \left\langle x_2,u \right\rangle v$

**Defunkt** · March 27th 2010, 04:16 AM

Originally Posted by demode

Is the following right?

$T(\alpha x_1,x_2)= \left\langle \alpha x_1 + x_2, u \right\rangle v$

$= ( \left\langle \alpha x_1,u \right\rangle + \left\langle x_2,u \right\rangle )v$

$= \left\langle \alpha x_1,u \right\rangle v+ \left\langle x_2,u \right\rangle v$

$= \alpha \left\langle x_1,u \right\rangle v + \left\langle x_2,u \right\rangle v$

Yes, that is correct, and it is exact the equality you were asked to show.

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