1. ## Linear operator proof

How do we prove this:

Btw, here are the properties of inner products:

<u,v>=<v,u> (symmetry 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?

2. Originally Posted by demode
How do we prove this:

Btw, here are the properties of inner products:

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

3. 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 $\alpha T(x_1) + T(x_2) = \alpha v ~ + v$. 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:

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

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

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

4. 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 $<\alpha v_1, v_2> = \alpha$ but this is not the case here. You need to use additivity first.

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

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

5. Is the following right?

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

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

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

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

6. 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.