Linear transformations question

[Rackhir]

Homework Statement

Today in my final i was given this exercise:
Given $β_1=\{v_1,v_2,v_3\}$ and $β_2=\{u_1,u_2,u_3,u_4\}$, basis of the vector spaces $V$ and $U$ respectively.
a) Find the linear transformation $T:U\rightarrow V$ so that $T(v_i)≠T(v_j)$ if $i≠j$, $T(v_1)=u_1+u_2$ and $T$ is injective

b) Find the transformation matrix from $β_1$ to $β_2$, $[T]_{β_1 \rightarrow β_2}$

Homework Equations

If $T$ is injective if and only if $Kernel(T)=\{0\}$, that means that the nullspace of the transformation matrix is $\{0\}$

The Attempt at a Solution

I thought that finding $[T]_{β_1 \rightarrow β_2}$ first would be easier, or at least it made more sense for me. I found this matrix
$$\begin{pmatrix}<br /> 1&0&0\\<br /> 1&0&0\\<br /> 0&1&0\\<br /> 0&0&1\\<br /> \end{pmatrix}$$

Where the first column, $\begin{pmatrix}<br /> 1\\<br /> 1\\<br /> 0\\<br /> 0\\<br /> \end{pmatrix}$
comes from $T(v_1)=u_1+u_2$, and the other two were chosen so the nullspace of $[T]_{β_1 \rightarrow β_2}$ is in fact, $\{0\}$
My big question, is this right? or it's horribly wrong and i should feel ashamed when i look myself in the mirror?

Then, for a) find the transformation per se, should i solve the following system?
$$\begin{pmatrix}<br /> 1&0&0\\<br /> 1&0&0\\<br /> 0&1&0\\<br /> 0&0&1\\<br /> \end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}a\\b\\c\\d\end{pmatrix}$$, where $(x,y,z) \in V$ and $(a,b,c,d) \in U$. So i end with $a=x, b=x, c=y, d=z$, but this is highly dependent on the basis chosen, right? shouldn't be dependent?

Any help will be highly appreciated, i'd love to know how this is actually solved.

 

[vela]

Your matrix is fine. Feel free to look in the mirror.

For part (a), all you probably needed to do was say what T does to the basis vectors in β₁, and then prove that T is injective.

 

[Rackhir]

Your matrix is fine. Feel free to look in the mirror.

For part (a), all you probably needed to do was say what T does to the basis vectors in β₁, and then prove that T is injective.

YEY, i feel relieved :) and what do you mean with "all you probably needed to do was say what T does to the basis vectors in β₁"? is $a=x, b=x, c=y, d=z \ \forall \ (x,y,z)∈V \ and \ (a,b,c,d)∈U$ wrong? as you can see, i have more problems finding out what i need to do, rather than doing it
I just hope my teachers dion't mark me wrong because i found the matrix first :P
Santiago

 

[vela]

Like you said, the calculation you did depends on the basis, but saying that T(v₁) = u₁+u₂ is independent of the basis. Whatever representation you use, T will always map v₁ to u₁+u₂.

Because T is linear, if you specify what it does to a set of basis vectors, you've determined what T will do to any vector in the space, so simply saying what T(v₁), T(v₂), and T(v₃) are is enough to fully describe T.

 

[Rackhir]

Like you said, the calculation you did depends on the basis, but saying that T(v₁) = u₁+u₂ is independent of the basis. Whatever representation you use, T will always map v₁ to u₁+u₂.

Because T is linear, if you specify what it does to a set of basis vectors, you've determined what T will do to any vector in the space, so simply saying what T(v₁), T(v₂), and T(v₃) are is enough to fully describe T.

Oh i see, so, with the matrix i chose, $T(v_1)=u_1+u_2$, $T(v_2)=u_3$ and $T(v_3)=u_4$. Well that does make sense now. And since they will be clearly linearly independent, the proof that T is injective is rather obvious.