# Are These Linear Transformations?

## Problem 717

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.

Using Gram-Schmidt orthogonalization, find an orthogonal basis for the span of the vectors $\mathbf{w}_{1},\mathbf{w}_{2}\in\R^{3}$ if \[ \mathbf{w}_{1} = \begin{bmatrix} 1...