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&amp;0&amp;0\\<br /> 1&amp;0&amp;0\\<br /> 0&amp;1&amp;0\\<br /> 0&amp;0&amp;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&amp;0&amp;0\\<br /> 1&amp;0&amp;0\\<br /> 0&amp;1&amp;0\\<br /> 0&amp;0&amp;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.