Is this a valid linear transformation? (1) Which of the following is a linear transformation?
$ \quad $ (a) $ T ( \begin{pmatrix} x \\ y \end{pmatrix} )= \begin{pmatrix}  0 \\ x \\y \end{pmatrix}$, $ \quad $ (b) $ T ( \begin{pmatrix} x \\ y \\ z \end{pmatrix} ) = \begin{pmatrix}  y \\ x \end{pmatrix}$ $ \quad $ (c) $ T (\begin{pmatrix} x \\ y \end{pmatrix} )= \begin{pmatrix}  x^2 \\ y^2 \\ 0 \end{pmatrix}$
$ \quad $ (d) $ T (\begin{pmatrix} x \\ y \end{pmatrix}) = \begin{pmatrix}  1 \\ x \\y \end{pmatrix}$ $ \quad $ (e) $ T ( \begin{pmatrix}\ x \\ y  \\ z \end{pmatrix} ) = x \begin{pmatrix}  1 \\ 1 \\ z \end{pmatrix}$
I am new to this topic, am I correct in saying that only (a) is a linear transformation?
(2) $ T (\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}) = \begin{pmatrix}  2 \\ 2 \\2 \end{pmatrix}$ , $ \quad $ $ T (\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}) = \begin{pmatrix}  1 \\ -1 \\ -1 \end{pmatrix}$, $ \quad $ $ T (\begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}) = \begin{pmatrix}  0 \\ 0 \\ 0 \end{pmatrix}$ $ \quad $
$ \quad $$ M = \begin{pmatrix}1 & 1 & 0 \\ 2 & 1 & 1 \\ 3 & 2 & -2  \end{pmatrix}. $
$ \quad $ Then,  $ T (\begin{pmatrix} x \\ y \\ z \end{pmatrix}) = \begin{pmatrix}  2 &  1 & 0 \\ 2 & -1 & 0 \\ 2 & -1 & 0 \end{pmatrix} M^{-1}  \begin{pmatrix} x \\ y \\ z \end{pmatrix} $  ?
 A: $(a)$ and $(b)$ are.  The others do not possess the properties that $T(cv)=cT(v)$ and $T(v+w)=T(v)+T(w)$.
A: I think you need to understand the definition of linear transformation. Note that:

Let $V$ and $W$ vector spaces over field $F$. Let $T: V\to W$ a function, so $T$ is said linear transformation if, and only if,
$1.$ $\forall v_{1},v_{2}\in V: T(v_{1}+v_{2})=T(v_{1})+T(v_{2})$.
$2.$ $\forall \alpha \in F, \forall v \in V: T(\alpha v)=\alpha T(v)$.

Now, you can prove in this sense that
\begin{eqnarray}
T \text{is a linear transformation}  \iff \forall \alpha, \beta\in F, \forall v_{1},v_{2} \in V: T(\alpha v_{1}+\beta v_{2})=\alpha T(v_{1})+\beta T(v_{2})
\end{eqnarray}

Now, for your question you need to see if $T$ is a linear transformation, so:
b) $T$ Is a linear transformation, because
\begin{eqnarray}
T\left(\alpha \begin{pmatrix} x \\ y \\ z \end{pmatrix} +\beta \begin{pmatrix} a \\ b \\ c \end{pmatrix} \right)
&=& T\begin{pmatrix} \alpha x+\beta a \\ \alpha y +\beta b \\ \alpha z +\beta c \end{pmatrix}\\
&=&\begin{pmatrix} \alpha y +\beta b \\ \alpha x +\beta a \end{pmatrix}\\
&=&\alpha \begin{pmatrix} y \\ x \end{pmatrix} +\beta \begin{pmatrix} b \\ a \end{pmatrix}\\
&=&\alpha T\begin{pmatrix} x \\ y \\ z \end{pmatrix} +\beta T\begin{pmatrix} a \\ b \\ c \end{pmatrix}
\end{eqnarray}
a) Also $T$ is linear transformation (I think you can do this part).
c) $T4 is not linear transformation (I think you can do this part).
d) $T$ is not linear transformation, because
\begin{eqnarray}
T\left( \alpha \begin{pmatrix} x \\ y \end{pmatrix} +\beta\begin{pmatrix} a \\ b \end{pmatrix} \right)&=& T\begin{pmatrix} \alpha x+\beta a \\ \alpha y +\beta b \end{pmatrix}\\
&=&\begin{pmatrix} 1 \\ \alpha x+\beta a \\ \alpha y +\beta b \end{pmatrix}\\
&\not=& \alpha \begin{pmatrix} 1 \\ x \\ y \end{pmatrix}+\beta \begin{pmatrix} 1 \\ a \\ b \end{pmatrix}\\
&=& \alpha T \begin{pmatrix} x\\ y \end{pmatrix} +\beta T\begin{pmatrix} a \\ b \end{pmatrix} 
\end{eqnarray}
e) $T$ is not linear transformation ( I think you can do this part).
