Linear Transformation $T:\R^2 \to \R^2$ Given in Figure

Linear Transformation problems and solutions

Problem 610

Let $T:\R^2\to \R^2$ be a linear transformation such that it maps the vectors $\mathbf{v}_1, \mathbf{v}_2$ as indicated in the figure below.

linear transformation from R^2 to R^2

Find the matrix representation $A$ of the linear transformation $T$.

 
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Solution 1.

From the figure, we see that
\[\mathbf{v}_1=\begin{bmatrix}
-3\\ 1
\end{bmatrix} \text{ and } \mathbf{v}_2=\begin{bmatrix}
5\\ 2
\end{bmatrix},\] and
\[T(\mathbf{v}_1)=\begin{bmatrix}
2\\ 2
\end{bmatrix} \text{ and } T(\mathbf{v}_2)=\begin{bmatrix}
1\\ 3
\end{bmatrix}.\]


Let $A$ be the matrix representation of the linear transformation $T$. By definition, we have $T(\mathbf{x})=A\mathbf{x}$ for any $\mathbf{x}\in \R^2$.

We determine $A$ as follows.
We have
\begin{align*}
\begin{bmatrix}
2& 1 \\
2& 3
\end{bmatrix}
&=[T(\mathbf{v}_1), T(\mathbf{v}_2)]\\
&=[A\mathbf{v}_1, A\mathbf{v}_2]\\
&=A[\mathbf{v}_1, \mathbf{v}_2]\\
&=A\begin{bmatrix}
-3& 5 \\
1& 2
\end{bmatrix}.
\end{align*}


Note that the determinant of the leftmost matrix is $-11$, hence it is invertible and the inverse is given by
\[\begin{bmatrix}
-3& 5 \\
1& 2
\end{bmatrix}^{-1}=\frac{1}{11}\begin{bmatrix}
-2& 5 \\
1& 3
\end{bmatrix}.
\] Hence
\begin{align*}
A&= \begin{bmatrix}
2& 1 \\
2& 3
\end{bmatrix} \begin{bmatrix}
-3& 5 \\
1& 2
\end{bmatrix}^{-1}\\[6pt] &=\frac{1}{11}\begin{bmatrix}
2& 1 \\
2& 3
\end{bmatrix}\begin{bmatrix}
-2& 5 \\
1& 3
\end{bmatrix}. \\[6pt] &=\frac{1}{11}\begin{bmatrix}
-3& 13 \\
-1& 19
\end{bmatrix}.
\end{align*}

Therefore, the matrix representation of $T$ is
\[A=\frac{1}{11}\begin{bmatrix}
-3& 13 \\
-1& 19
\end{bmatrix}.\]

Solution 2.

Let $\{\mathbf{e}_1, \mathbf{e}_2\}$ be the standard basis for $\R^2$.
Then the matrix representation $A$ of the linear transformation $T$ is given by
\[A=[T(\mathbf{e}_1), T(\mathbf{e}_2)].\] From the figure, we see that
\[\mathbf{v}_1=\begin{bmatrix}
-3\\ 1
\end{bmatrix} \text{ and } \mathbf{v}_2=\begin{bmatrix}
5\\ 2
\end{bmatrix},\] and
\[T(\mathbf{v}_1)=\begin{bmatrix}
2\\ 2
\end{bmatrix} \text{ and } T(\mathbf{v}_2)=\begin{bmatrix}
1\\ 3
\end{bmatrix}.\]


The values of $T(\mathbf{e}_1)$ and $T(\mathbf{e}_2)$ are not indicated in the figure but we can determine these values as follows.
Note that $\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$ (as they are linearly independent from the figure).

Let us express $\mathbf{e}_1$ as a linear combination of $\mathbf{v}-1$ and $\mathbf{v}_2$.
Let
\[\mathbf{e}_1=c_1\mathbf{v}_1+c_2\mathbf{v}_2,\] where $c_1, c_2$ are scalars to be determined.

This is equivalent to
\[\begin{bmatrix}
1
\\ 0
\end{bmatrix}
=[\mathbf{v}_1, \mathbf{v}_2]\begin{bmatrix}
c_1\\ c_2
\end{bmatrix}
=\begin{bmatrix}
-3
& 5 \\
1& 2
\end{bmatrix}
\begin{bmatrix}
c_1\\ c_2
\end{bmatrix}.\]

As we have
\[\begin{bmatrix}
-3& 5 \\
1& 2
\end{bmatrix}^{-1}=\frac{1}{11}\begin{bmatrix}
-2& 5 \\
1& 3
\end{bmatrix},
\] it follows that
\begin{align*}
\begin{bmatrix}
c_1\\ c_2
\end{bmatrix}
=\begin{bmatrix}
-3& 5 \\
1& 2
\end{bmatrix}^{-1}\begin{bmatrix}
1\\ 0
\end{bmatrix}=
\frac{1}{11}\begin{bmatrix}
-2& 5 \\
1& 3
\end{bmatrix}
\begin{bmatrix}
1\\ 0
\end{bmatrix}
= \frac{1}{11}\begin{bmatrix}
-2
\\ 1
\end{bmatrix}.
\end{align*}
Thus, we obtain the linear combination
\[\mathbf{e}_1=-\frac{2}{11}\mathbf{v}_1+\frac{1}{11}\mathbf{v}_2,\]


By the linearity of $T$, we have
\begin{align*}
T(\mathbf{e}_1)&=-\frac{2}{11}T(\mathbf{v}_1)+\frac{1}{11}T(\mathbf{v}_2)\\
&=-\frac{2}{11}\begin{bmatrix}
2
\\ 2
\end{bmatrix}
+\frac{1}{11}\begin{bmatrix}
1\\ 3
\end{bmatrix}
\\
&=\frac{1}{11}\begin{bmatrix}
-3\\ -1
\end{bmatrix}.
\end{align*}


Similarly, let
\[\mathbf{e}_2=d_1\mathbf{v}_1+d_2\mathbf{v}_2,\] where $d_1, d_2$ are scalars to be determined.
Then we have
\[\begin{bmatrix}
0\\ 1
\end{bmatrix}
=\begin{bmatrix}
-3 & 5 \\
1& 2
\end{bmatrix}
\begin{bmatrix}
d_1\\ d_2
\end{bmatrix}.\] It follows that
\begin{align*}
\begin{bmatrix}
d_1\\ d_2
\end{bmatrix}
=\begin{bmatrix}
-3& 5 \\
1& 2
\end{bmatrix}^{-1}\begin{bmatrix}
0\\ 1
\end{bmatrix}=
\frac{1}{11}\begin{bmatrix}
-2& 5 \\
1& 3
\end{bmatrix}
\begin{bmatrix}
0\\ 1
\end{bmatrix}
= \frac{1}{11}\begin{bmatrix}
5
\\ 3
\end{bmatrix}.
\end{align*}
Thus, we have the linear combination
\[\mathbf{e}_2=\frac{5}{11}\mathbf{v}_1+\frac{3}{11}\mathbf{v}_2,\] and by linearity of $T$ we obtain
\begin{align*}
T(\mathbf{e}_2)&=\frac{5}{11}T(\mathbf{v}_1)+\frac{3}{11}T(\mathbf{v}_2)\\
&=\frac{5}{11}T(\mathbf{v}_1)+\frac{3}{11}T(\mathbf{v}_2)\\
&=\frac{5}{11}\begin{bmatrix}
2
\\ 2
\end{bmatrix}
+\frac{3}{11}\begin{bmatrix}
1\\ 3
\end{bmatrix}
\\
&=\frac{1}{11}\begin{bmatrix}
13\\ 19
\end{bmatrix}.
\end{align*}


In conclusion, the matrix representation for $T$ is
\[A=[T(\mathbf{e}_1), T(\mathbf{e}_2)]=\frac{1}{11}\begin{bmatrix}
-3& 13 \\
-1& 19
\end{bmatrix}.\]


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