Linear Transformations

[Guest]

Good afternoon people.
So i have to demonstrate that the problems below are Linear Transformations, i have searched and i know i have to do it using a couple of "rules", it is a linear transformation if:
T(u+v) = T(u) + T(v) and T(Lu) = LT(u), the thing is that i really can't understand how to develop that and find the demonstration.
Thanks for reading.

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[Guest]

As you say, a transformation, T, is "linear" if and only if T(u+ v)= Tu+ Tv and T(au)= aTu. We can combine those into a single test: T(au+ bv)= aT(u)+ bTv.

The first transformation, [tex]T\begin{pmatrix}x \\ y \\ z \end{pmatrix}= \begin{pmatrix}1 \\ z \end{pmatrix}[/tex].
Let [tex]u= \begin{pmatrix}x \\ y \\ z \end{pmatrix}[/tex] and [tex]v= \begin{pmatrix}p \\ q \\ r \end{pmatrix}[/tex]. Then [tex]T(au+ bv)= T\left(\begin{pmatrix}ax \\ ay \\ az \end{pmatrix}+ \begin{pmatrix}bp \\ bq \\ br \end{pmatrix}\right)[/tex][tex]= T\left(\begin{pmatrix}ax+ bp \\ ay+ bq \\ az+ br \end{pmatrix}\right)[/tex][tex]= \begin{pmatrix}1 \\ az+ br \end{pmatrix}[/tex].
While [tex]aTu= aT\left(\begin{pmatrix}x \\ y \\ z \end{pmatrix}\right)= a\begin{pmatrix}1 \\ z \end{pmatrix}= \begin{pmatrix} a \\ az\end{pmatrix}[/tex]
and [tex]bTu= bT\left(\begin{pmatrix}p \\ q \\ r \end{pmatrix}\right)= b\begin{pmatrix}1 \\ r \end{pmatrix}= \begin{pmatrix} b \\ br\end{pmatrix}[/tex]
so [tex]aTu+ bTv= \begin{pmatrix}a+ b \\ az+ br \end{pmatrix}[/tex]

Those are not the same so this is NOT a linear transformation.

Do the same for the others.