Give a Formula For a Linear Transformation From $\R^2$ to $\R^3$

Problem 339

Let $\{\mathbf{v}_1, \mathbf{v}_2\}$ be a basis of the vector space $\R^2$, where
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
1
\end{bmatrix} \text{ and } \mathbf{v}_2=\begin{bmatrix}
1 \\
-1
\end{bmatrix}.\] The action of a linear transformation $T:\R^2\to \R^3$ on the basis $\{\mathbf{v}_1, \mathbf{v}_2\}$ is given by
\begin{align*}
T(\mathbf{v}_1)=\begin{bmatrix}
2 \\
4 \\
6
\end{bmatrix} \text{ and } T(\mathbf{v}_2)=\begin{bmatrix}
0 \\
8 \\
10
\end{bmatrix}.
\end{align*}

Find the formula of $T(\mathbf{x})$, where
\[\mathbf{x}=\begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2.\]

 
Add to solve later

Sponsored Links

Solution.

We give two solutions.

Solution 1 (linear combination)

Since we know the values of $T$ on the basis vectors $\mathbf{v}_1, \mathbf{v}_2$, if we express the vector $\mathbf{x}$ as a linear combination of $\mathbf{v}_1, \mathbf{v}_2$, we can find $F(\mathbf{x})$ by the linearity of the linear transformation $T$.

So let us find the scalars $c_1, c_2$ such that
\[\mathbf{x}=c_1\mathbf{v}_1+c_2\mathbf{v}_2.\] We write this as
\begin{align*}
\begin{bmatrix}
x \\
y
\end{bmatrix}=c_1\begin{bmatrix}
1 \\
1
\end{bmatrix}+c_2\begin{bmatrix}
1 \\
-1
\end{bmatrix}
=
\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}.
\end{align*}

The matrix $\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}$ is invertible (as its determinant is $-2$) and its inverse matrix is
\[\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}^{-1}=\frac{1}{2}\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}.\]

Thus, we have
\begin{align*}
\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}&=\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}^{-1}\begin{bmatrix}
x \\
y
\end{bmatrix}\\[6pt] &=\frac{1}{2}\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix}\\[6pt] &=\frac{1}{2}\begin{bmatrix}
x+y \\
x-y
\end{bmatrix}
\end{align*}
Therefore, we obtain the linear combination
\[\mathbf{x}=\frac{1}{2}(x+y)\mathbf{v}_1+\frac{1}{2}(x-y)\mathbf{v}_2.\]

Now we compute $T(\mathbf{x})$ as follows.
We have
\begin{align*}
T(\mathbf{x})&=T\left(\frac{1}{2}(x+y)\mathbf{v}_1+\frac{1}{2}(x-y)\mathbf{v}_2 \right)\\[6pt] &=\frac{1}{2}(x+y)T(\mathbf{v}_1)+\frac{1}{2}(x-y)T(\mathbf{v}_2) && \text{by linearity of $T$}\\[6pt] &=\frac{1}{2}(x+y)\begin{bmatrix}
2 \\
4 \\
6
\end{bmatrix}+\frac{1}{2}(x-y)\begin{bmatrix}
0 \\
8 \\
10
\end{bmatrix} \\[6pt] &=\begin{bmatrix}
x+y \\
6x-2y \\
8x-2y
\end{bmatrix}.
\end{align*}
Hence the formula is
\[T(\mathbf{x})=\begin{bmatrix}
x+y \\
6x-2y \\
8x-2y
\end{bmatrix}.
\]

Solution 2 (Matrix representation)

In the second solution, we use the matrix representation for the linear transformation $T$.
Let $A$ be the matrix of $T$ with respect to the standard basis $\{\begin{bmatrix}
1 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
1
\end{bmatrix}\}$ of $\R^2$.
Thus, we have $T(\mathbf{x})=A\mathbf{x}$ by definition.

To find the matrix $A$, we compute
\begin{align*}
A\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}&=A\begin{bmatrix}
\mathbf{v}_1 & \mathbf{v}_2 \\
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
A\mathbf{v}_1 & A\mathbf{v}_2
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
2 & 0 \\
4 & 8 \\
6 &10
\end{bmatrix}
\end{align*}
It follows that we have
\begin{align*}
A&=\begin{bmatrix}
2 & 0 \\
4 & 8 \\
6 &10
\end{bmatrix}\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}^{-1}\\[6pt] &=\begin{bmatrix}
2 & 0 \\
4 & 8 \\
6 &10
\end{bmatrix}\frac{1}{2}\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
1 & 1 \\
6 & -2 \\
8 &-2
\end{bmatrix}.
\end{align*}

We now be able to find $T(\mathbf{x})$ as follows.
We have
\begin{align*}
T(\mathbf{x})&=A\mathbf{x}\\[6pt] &=\begin{bmatrix}
1 & 1 \\
6 & -2 \\
8 &-2
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
x+y \\
6x-2y \\
8x-2y
\end{bmatrix},
\end{align*}
which is, of course, the same formula that we obtained in solution 1.

Related Question.

A similar problem for a linear transformation from $\R^3$ to $\R^3$ is given in the post “Determine linear transformation using matrix representation“.

Instead of finding the inverse matrix in solution 1, we could have used the Gauss-Jordan elimination to find the coefficients.
See the post “Give a formula for a linear transformation if the values on basis vectors are known” for a similar problem and its solution using this alternative method.

Click here if solved 44
Add to solve later

Sponsored Links