# Restriction of a Linear Transformation on the x-z Plane is a Linear Transformation

## Problem 428

Let $T:\R^3 \to \R^3$ be a linear transformation and suppose that its matrix representation with respect to the standard basis is given by the matrix
$A=\begin{bmatrix} 1 & 0 & 2 \\ 0 &3 &0 \\ 4 & 0 & 5 \end{bmatrix}.$

(a) Prove that the linear transformation $T$ sends points on the $x$-$z$ plane to points on the $x$-$z$ plane.

(b) Prove that the restriction of $T$ on the $x$-$z$ plane is a linear transformation.

(c) Find the matrix representation of the linear transformation obtained in part (b) with respect to the standard basis
$\left\{\, \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \,\right\}$ of the $x$-$z$ plane.

## Proof.

### (a) Prove that the linear transformation $T$ sends points on the $x$-$z$ plane to points on the $x$-$z$ plane.

Each point on the $x$-$z$ plane is of the form
$\begin{bmatrix} x \\ 0 \\ z \end{bmatrix}$ for some $x, z \in \R$.
We have
\begin{align*}
T\left(\,\begin{bmatrix}
x \\
0 \\
z
\end{bmatrix}\,\right)&=A\begin{bmatrix}
x \\
0 \\
z
\end{bmatrix}\6pt] &=\begin{bmatrix} 1 & 0 & 2 \\ 0 &3 &0 \\ 4 & 0 & 5 \end{bmatrix}\begin{bmatrix} x \\ 0 \\ z \end{bmatrix}\\[6pt] &=\begin{bmatrix} x+2z \\ 0 \\ 4x+5z \end{bmatrix}. \end{align*} Since the y-coordinate of the last vector is 0, and thus the output vector lies in the x-z plane. ### (b) Prove that the restriction of T on the x-z plane is a linear transformation. Let V be the x-z plane in \R^3. Then V is a subspace of the vector space \R^3. In part (a), we showed that the restriction of T on V is given by the formula \begin{align*} T\left(\,\begin{bmatrix} x \\ 0 \\ z \end{bmatrix}\,\right)=\begin{bmatrix} x+2z \\ 0 \\ 4x+5z \end{bmatrix}. \tag{*} \end{align*} We abuse the notation and write this restriction as T: V\to V. (The precise notation is T|_{V}:V\to V.) Let \[\begin{bmatrix} x_1 \\ 0 \\ z_1 \end{bmatrix}, \begin{bmatrix} x_2 \\ 0 \\ z_2 \end{bmatrix} be arbitrary vectors in $V$ and let $r\in \R$ be an arbitrary real number.
Then we have
\begin{align*}
T\left(\,\begin{bmatrix}
x_1 \\
0 \\
z_1
\end{bmatrix}+\begin{bmatrix}
x_2 \\
0 \\
z_2
\end{bmatrix}\,\right)&=T\left(\,\begin{bmatrix}
x_1+x_2 \\
0 \\
z_1 +z_2
\end{bmatrix}\,\right)\6pt] &=\begin{bmatrix} (x_1+x_2)+2(z_1+z_2) \\ 0 \\ 4(x_1+x_2)+5(z_1+z_2)\\ \end{bmatrix}\\ &=\begin{bmatrix} x_1+2z_1 \\ 0 \\ 4x_1+5z_1 \end{bmatrix}+\begin{bmatrix} x_2+2z_2 \\ 0 \\ 4x_2+5z_2 \end{bmatrix}\\[6pt] &=T\left(\,\begin{bmatrix} x_1 \\ 0 \\ z_1 \end{bmatrix}\,\right)+T\left(\,\begin{bmatrix} x_2 \\ 0 \\ z_2 \end{bmatrix}\,\right) \end{align*} and \begin{align*} T\left(\,r\begin{bmatrix} x_1 \\ 0 \\ z_1 \end{bmatrix}\,\right)&=T\left(\,\begin{bmatrix} rx_1 \\ 0 \\ rz_1 \end{bmatrix}\,\right)\\[6pt] &=\begin{bmatrix} (rx_1)+2(rz_1) \\ 0 \\ 4(rx_1)+5(z_1) \end{bmatrix}\\[6pt] &=r\begin{bmatrix} x_1+2z_1 \\ 0 \\ 4x_1+5z_1 \end{bmatrix}\\[6pt] &=r T\left(\, \begin{bmatrix} x_1 \\ 0 \\ z_1 \end{bmatrix} \,\right). \end{align*} It follows that the restriction T:V\to V is a linear transformation. ### (c) Find the matrix representation of the linear transformation obtained in part (b) with respect to the standard basis Let B=\{\mathbf{u}, \mathbf{v}\} be the standard basis of the x-z plane: \[\mathbf{u}=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \mathbf{v}=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}. It follows from formula (*) that we have
\begin{align*}
T(\mathbf{u})=T\left(\, \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix} \,\right)=\begin{bmatrix}
1 \\
0 \\
4
\end{bmatrix}=\mathbf{u}+4\mathbf{v}\end{align*}
and
\begin{align*}
T(\mathbf{v})=T\left(\, \begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}\,\right)=\begin{bmatrix}
2 \\
0 \\
5
\end{bmatrix} =2\mathbf{u}+5\mathbf{v}.
\end{align*}
Thus the coordinate vectors with respect to the basis $B$ are
$[T(\mathbf{u})]_B=\begin{bmatrix} 1\\ 4 \end{bmatrix}_{B} , [T(\mathbf{v})]_B=\begin{bmatrix} 2 \\ 5 \end{bmatrix}_{B},$ and the matrix representation of the linear transformation $T:V\to V$ with respect to the standard basis $B=\{\mathbf{u}, \mathbf{v}\}$ is
$\left[\, [T(\mathbf{u})]_B, [T(\mathbf{v})]_B \,\right] =\begin{bmatrix} 1 & 2\\ 4& 5 \end{bmatrix}.$

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