# Linear Transformations

Algebra

### Linear Transformations

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.

transfnob.PNG (20.36 KiB) Viewed 377 times
Guest

### Re: Linear Transformations

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

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

Do the same for the others.
Guest