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

Linear algebra problems and solutions

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.

 
LoadingAdd to solve later

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


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Linear Algebra
Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra
Union of Subspaces is a Subspace if and only if One is Included in Another

Let $W_1, W_2$ be subspaces of a vector space $V$. Then prove that $W_1 \cup W_2$ is a subspace of...

Close