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
\[\left[\,
[T(\mathbf{u})]_B, [T(\mathbf{v})]_B \,\right] =\begin{bmatrix}
1 & 2\\
4& 5
\end{bmatrix}.\] Click here if solved 19

The post Restriction of a Linear Transformation on the x-z Plane is a Linear Transformation first appeared on Problems in Mathematics.