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## Homework Statement

Today in my final i was given this exercise:

Given [itex]β_1=\{v_1,v_2,v_3\}[/itex] and [itex]β_2=\{u_1,u_2,u_3,u_4\}[/itex], basis of the vector spaces [itex]V[/itex] and [itex]U[/itex] respectively.

*a)*Find the linear transformation [itex]T:U\rightarrow V[/itex] so that [itex]T(v_i)≠T(v_j)[/itex] if [itex]i≠j[/itex], [itex]T(v_1)=u_1+u_2[/itex] and [itex]T[/itex] is injective

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

## Homework Equations

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

## The Attempt at a Solution

I thought that finding [itex][T]_{β_1 \rightarrow β_2}[/itex] first would be easier, or at least it made more sense for me. I found this matrix

[tex] \begin{pmatrix}

1&0&0\\

1&0&0\\

0&1&0\\

0&0&1\\

\end{pmatrix} [/tex]

Where the first column, [itex]\begin{pmatrix}

1\\

1\\

0\\

0\\

\end{pmatrix}[/itex]

comes from [itex]T(v_1)=u_1+u_2[/itex], and the other two were chosen so the nullspace of [itex][T]_{β_1 \rightarrow β_2}[/itex] is in fact, [itex]\{0\}[/itex]

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?

[tex] \begin{pmatrix}

1&0&0\\

1&0&0\\

0&1&0\\

0&0&1\\

\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}a\\b\\c\\d\end{pmatrix}[/tex], where [itex](x,y,z) \in V[/itex] and [itex](a,b,c,d) \in U[/itex]. So i end with [itex] a=x, b=x, c=y, d=z [/itex], but this is highly dependant on the basis chosen, right? shouldn't be independant?

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

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