Define two functions $T:\R^{2}\to\R^{2}$ and $S:\R^{2}\to\R^{2}$ by
\[
T\left(
\begin{bmatrix}
x \\ y
\end{bmatrix}
\right)
=
\begin{bmatrix}
2x+y \\ 0
\end{bmatrix}
,\;
S\left(
\begin{bmatrix}
x \\ y
\end{bmatrix}
\right)
=
\begin{bmatrix}
x+y \\ xy
\end{bmatrix}
.
\] Determine whether $T$, $S$, and the composite $S\circ T$ are linear transformations.

Solution.

We will prove that $T$ and $S\circ T$ are linear transformations, but $S$ is not.

$T$ is alinear transformation

To prove that $T$ is a linear transformation, note that for any $\mathbf{x},\mathbf{y}\in\R^{2}$, if we write
\[
\mathbf{x}
=
\begin{bmatrix}
x_{1} \\ x_{2}
\end{bmatrix}
,\;
\mathbf{y}
=
\begin{bmatrix}
y_{1} \\ y_{2}
\end{bmatrix}
,
\] then we have
\begin{align*}
T\left(\mathbf{x}+\mathbf{y}\right)
&=
T\left(
\begin{bmatrix}
x_{1}+y_{1} \\ x_{2}+y_{2}
\end{bmatrix}
\right)
=
\begin{bmatrix}
2(x_{1}+y_{1})+(x_{2}+y_{2}) \\ 0
\end{bmatrix}
\\
&=
\begin{bmatrix}
2x_{1}+x_{2} \\ 0
\end{bmatrix}
+
\begin{bmatrix}
2y_{1}+y_{2} \\ 0
\end{bmatrix}
=
T(\mathbf{x})+T(\mathbf{y})
.
\end{align*}

Next, for any scalar $r$, we have
\[
T(r\mathbf{x})
=
T\left(r
\begin{bmatrix}
x_{1} \\ x_{2}
\end{bmatrix}
\right)
=
T\left(
\begin{bmatrix}
rx_{1} \\ rx_{2}
\end{bmatrix}
\right)
=
\begin{bmatrix}
2rx_{1}+rx_{2} \\ 0
\end{bmatrix}
=r
\begin{bmatrix}
2x_{1}+x_{2} \\ 0
\end{bmatrix}
=
rT(\mathbf{x})
.
\] Hence $T$ is a linear transformation.

$S$ is not a linear transformation

To prove that $S$ is not a linear transformation, observe that
\[
S\left(
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
\right)
=
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
,
\quad
S\left(
\begin{bmatrix}
0 \\ 1
\end{bmatrix}
\right)
=
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
,
\quad
S\left(
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
\right)
=
\begin{bmatrix}
2 \\ 1
\end{bmatrix}
.
\] Therefore,
\begin{align*}
S\left(
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
+
\begin{bmatrix}
0 \\ 1
\end{bmatrix}
\right)
&=
S\left(
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
\right)
=
\begin{bmatrix}
2 \\ 1
\end{bmatrix}
% \\
% &
\neq
\begin{bmatrix}
2 \\ 0
\end{bmatrix}
=
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
+
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
\\
&=
S\left(
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
\right)
+
S\left(
\begin{bmatrix}
0 \\ 1
\end{bmatrix}
\right)
.
\end{align*}
Thus it is not the case that $S(\mathbf{x}+\mathbf{y})=S(\mathbf{x})+S(\mathbf{y})$ for all $\mathbf{x},\mathbf{y}\in\R^{2}$. It follows that $S$ cannot be a linear transformation.

The composite $S\circ T$ is a lineawr transformation

To prove that $S\circ T$ is linear, note that for any $\mathbf{x}\in\R^{2}$,
\[
S\circ T(\mathbf{x})
=
S\left(
T\left(
\begin{bmatrix}
x \\ y
\end{bmatrix}
\right)\right)
=
S\left(
\begin{bmatrix}
2x+y \\ 0
\end{bmatrix}
\right)
% =
% \begin{bmatrix}
% 2x+y+0 \\ (2x+y)\cdot 0
% \end{bmatrix}
=
\begin{bmatrix}
2x+y \\ 0
\end{bmatrix}
=
T(\mathbf{x})
.
\] Therefore, $S\circ T=T$. Since $T$ is a linear transformation, we can immediately conclude that $S\circ T$ is a linear transformation. Hence $T$ and $S\circ T$ are linear, while $S$ is not.

The post Are These Linear Transformations? first appeared on Problems in Mathematics.