# Matrix Representation of a Linear Transformation of the Vector Space $R^2$ to $R^2$

## Problem 255

Let $B=\{\mathbf{v}_1, \mathbf{v}_2 \}$ be a basis for the vector space $\R^2$, and let $T:\R^2 \to \R^2$ be a linear transformation such that
$T(\mathbf{v}_1)=\begin{bmatrix} 1 \\ -2 \end{bmatrix} \text{ and } T(\mathbf{v}_2)=\begin{bmatrix} 3 \\ 1 \end{bmatrix}.$

If $\mathbf{e}_1=\mathbf{v}_1+2\mathbf{v}_2 \text{ and } \mathbf{e}_2=2\mathbf{v}_1-\mathbf{u}_2$, where $\mathbf{e}_1, \mathbf{e}_2$ are the standard unit vectors in $\R^2$, then find the matrix of $T$ with respect to the basis $\{\mathbf{e}_1, \mathbf{e}_2\}$.

## Solution.

The matrix representation $A$ of the linear transformation $T$ with respect to the basis $\{\mathbf{e}_1, \mathbf{e}_2 \}$ is given by
$A=\begin{bmatrix} T(\mathbf{e}_1) & T(\mathbf{e}_2) \end{bmatrix}. \tag{*}$

To find the vectors $T(\mathbf{e}_1), T(\mathbf{e}_2)$ we use the linearity of $T$.
We have
\begin{align*}
T(\mathbf{e}_1)&=T(\mathbf{v}_1+2\mathbf{v}_2)\\
&=T(\mathbf{v}_1)+2T(\mathbf{v}_2) \qquad \text{ (by the linearity of $T$)}\\
&=\begin{bmatrix}
1 \\
-2
\end{bmatrix} +2\begin{bmatrix}
3 \\
1
\end{bmatrix}\\
&=\begin{bmatrix}
7 \\
0
\end{bmatrix}.
\end{align*}

Also we have
\begin{align*}
T(\mathbf{e}_2)&=T(2\mathbf{v}_1-\mathbf{u}_2)\\
&=2T(\mathbf{v}_1)-T(\mathbf{u}_2)\\
&=2\begin{bmatrix}
1 \\
-2
\end{bmatrix}-\begin{bmatrix}
3 \\
1
\end{bmatrix}\\
&=\begin{bmatrix}
-1 \\
-5
\end{bmatrix}.
\end{align*}

Therefore the matrix $A=\begin{bmatrix} T(\mathbf{e}_1) & T(\mathbf{e}_2) \end{bmatrix}$ of the linear transformation $T$ is given by
$A=\begin{bmatrix} 7 & -1\\ 0& -5 \end{bmatrix}.$

Let $\mathbf{a}$ and $\mathbf{b}$ be vectors in $\R^n$ such that their length are $\|\mathbf{a}\|=\|\mathbf{b}\|=1$ and the inner product $\mathbf{a}\cdot \mathbf{b}=\mathbf{a}^{\trans}\mathbf{b}=-\frac{1}{2}.$...