Linear transformation image.
Question in mext post.

Question

Linear transformation image.
Question in mext post.

chillout · Answer

Let $T:U \rightarrow V$ be a linear transformation and $\{u_1,u_2,u_3,\dots,u_n\}$ a basis of U. Show that $Im(T)=[T(u_1),T(u_2),T(u_3),\dots,T(u_n)]$

chillout · Answer

There you go, @Kainui :)

kainui · Answer

What sorta definitions and stuff are you working with? I thought these two statements were pretty much interchangeable so I'm not really sure how I'd go about showing this, although I could do it in a kind of silly way.

chillout · Answer

U and V are 2 vector spaces over the same field, Im is the image and T holds the usual linear transform definitions $(T(u+v) = T(u)+T(v), T(ku)=kT(u),\  k \in \mathbb{R}$)
And $\{u_1,\dots,u_n\}$ a set of elements

chillout · Answer

Is this what you were asking for?

chillout · Answer

I had this start... Let $\omega=T(u_1),\dots,T(u_n)$ we can write

$\omega=\alpha T(u_1),\dots,\gamma T(u_n)$

kainui · Answer

Yeah that's what I was imagining something along those lines. I'd probably use those and write that every vector can be decomposed as a linear combination (is that allowed) of basis vectors and use linearity properties to separate it out just like that

chillout · Answer

in which $\alpha$ and $\gamma$ are scalars... I know I'm getting into a silly proof.

chillout · Answer

Yeah I'm allowed to use whatever is needed to prove it.

chillout · Answer

Using the axiom: $T(ku)=kT(u)$ I can just assume those scalars are equal to 1 and if $T(u_1),\dots,T(u_n)$ is the image of $\omega$, then $\omega$ is the image of $T$

chillout · Answer

Is it this really silly?

kainui · Answer

Well ok here's what I'd do, although I don't know. Maybe by show they just mean write the same thing but more explicitly.

$$u = a_1 u_1 + a_2 u_2 + \cdots + a_n u_n$$
Then
$$T(u) = T(a_1 u_1 + a_2 u_2 + \cdots + a_n u_n)$$$$ = T(a_1 u_1 )+ T(a_2 u_2) + \cdots + T(a_n u_n) $$$$= a_1 T( u_1) + a_2 T(u_2) + \cdots + a_n T( u_n)$$

I am not sure if this is really what they're looking for or not really but I think it's what ou're saying too so I'm really not quite sure what they expect unfortunately! :X

chillout · Answer

That's the line of reasoning I was going into! So yeah, one more reason to buy your answer too :D

chillout · Answer

Yeah your answer is more complete than mine. That's what the book says. Thanks for your time!

chillout · Answer

Wait! There's the second part of the proof... Almost missed it...

kainui · Answer

awesome, I haven't left haha

chillout · Answer

Well, I think the rest of the proof is really redundant and I've taken enough of your time. I think our proofs kind of complement each other.

chillout · Answer

It assumes another element and writes in terms of a known element

chillout · Answer

in this case, your "u" element

kainui · Answer

Oh ok well have fun