# Find a Value of a Linear Transformation From $\R^2$ to $\R^3$

## Problem 142

Let $T:\R^2 \to \R^3$ be a linear transformation such that $T(\mathbf{e}_1)=\mathbf{u}_1$ and $T(\mathbf{e}_2)=\mathbf{u}_2$, where $\mathbf{e}_1=\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \mathbf{e}_2=\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ are unit vectors of $\R^2$ and
$\mathbf{u}_1= \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}, \quad \mathbf{u}_2=\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}.$ Then find $T\left(\begin{bmatrix} 3 \\ -2 \end{bmatrix}\right)$.

Contents

## Hint.

A linear transformation from a vector space $V$ to a vector space $W$ is a map $f:V \to W$ satisfying the following linearity properties:

1. $f(u+v)=f(u)+f(v)$ for any vectors $u, v \in V$, and
2. $f(rv)=rf(v)$ for any vector $v \in V$ and any scalar $r$.

Note that the set $\{\mathbf{e}_1, \mathbf{e}_2\}$ is a basis for the vector space $\R^2$.
Thus the vector $\begin{bmatrix} 3 \\ -2 \end{bmatrix}$ can be written as a linear combination of the basis vectors $\mathbf{e}_1, \mathbf{e}_2$.

## Solution.

We first express the vector $\begin{bmatrix} 3 \\ -2 \end{bmatrix}$ as a linear combination of $\mathbf{e}_1$ and $\mathbf{e}_2$:
$\begin{bmatrix} 3 \\ -2 \end{bmatrix}=3\mathbf{e}_1-2\mathbf{e}_2.$ Then we have
\begin{align*}
T\left(\begin{bmatrix}
3 \\
-2
\end{bmatrix}\right)
&=T(3\mathbf{e}_1-2\mathbf{e}_2)\\
&=3T(\mathbf{e}_1)-2T(\mathbf{e}_2) \text{ by the linearity of } T\\
&=3\mathbf{u}_1-2\mathbf{u}_2\\
&=3\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}-2\begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}\\
&=\begin{bmatrix}
-7 \\
-2 \\
3
\end{bmatrix}.
\end{align*}

Thus, we found
$T\left(\begin{bmatrix} 3 \\ -2 \end{bmatrix}\right) =\begin{bmatrix} -7 \\ -2 \\ 3 \end{bmatrix}.$

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