Find a General Formula of a Linear Transformation From $\R^2$ to $\R^3$

Ohio State University exam problems and solutions in mathematics

Problem 353

Suppose that $T: \R^2 \to \R^3$ is a linear transformation satisfying
\[T\left(\, \begin{bmatrix}
1 \\
2
\end{bmatrix}\,\right)=\begin{bmatrix}
3 \\
4 \\
5
\end{bmatrix} \text{ and } T\left(\, \begin{bmatrix}
0 \\
1
\end{bmatrix} \,\right)=\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}.\] Find a general formula for
\[T\left(\, \begin{bmatrix}
x_1 \\
x_2
\end{bmatrix} \,\right).\]

(The Ohio State University, Linear Algebra Math 2568 Exam Problem)

 
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We give two solutions.

Solution 1. (Using linear combination)

Note that the set $B:=\left\{\, \begin{bmatrix}
1 \\
2
\end{bmatrix}, \begin{bmatrix}
0 \\
1
\end{bmatrix} \,\right\}$ form a basis of the vector space $\R^2$.


To find a general formula, we first express the vector $\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}$ as a linear combination of the basis vectors in $B$.
Namely, we find scalars $c_1, c_2$ satisfying
\[\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=c_1\begin{bmatrix}
1 \\
2
\end{bmatrix}+c_2\begin{bmatrix}
0 \\
1
\end{bmatrix}.\] This can be written as the matrix equation
\[\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=P\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}, \tag{*}\] where we have put
\[P=\begin{bmatrix}
1 & 0\\
2& 1
\end{bmatrix}.\] The $2 \times 2$ matrix $P$ is invertible as its determinant is $\det(P)=(1)(1)-(0)(2)=1\neq 0$, and the inverse matrix is given by
\[P^{-1}=\begin{bmatrix}
1 & 0\\
-2& 1
\end{bmatrix}.\]
Then we multiply $P^{-1}$ on the left of (*) and obtain
\begin{align*}
\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}&=P^{-1}\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
1 & 0\\
-2& 1
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
x_1 \\
-2x_1+x_2
\end{bmatrix}.
\end{align*}

Thus, we have determined the scalars
\[c_1=x_1 \text{ and } c_2=-2x_1+x_2,\] and the linear combination becomes
\[\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=x_1\begin{bmatrix}
1 \\
2
\end{bmatrix}+(-2x_1+x_2)\begin{bmatrix}
0 \\
1
\end{bmatrix}.\]


Now we compute $T\left(\, \begin{bmatrix}
x_1 \\
x_2
\end{bmatrix} \,\right)$ as follows.
\begin{align*}
T\left(\, \begin{bmatrix}
x_1 \\
x_2
\end{bmatrix} \,\right) &=T\left(\, x_1\begin{bmatrix}
1 \\
2
\end{bmatrix}+(-2x_1+x_2)\begin{bmatrix}
0 \\
1
\end{bmatrix} \,\right)\\[6pt] &=x_1T\left(\, \begin{bmatrix}
1 \\
2
\end{bmatrix} \,\right)+(-2x_1+x_2)T\left(\,\begin{bmatrix}
0 \\
1
\end{bmatrix} \,\right) && \text{by linearity of $T$}\\[6pt] &=x_1\begin{bmatrix}
3 \\
4 \\
5
\end{bmatrix}+(-2x_1+x_2)\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
3x_1 \\
4x_1 \\
3x_1 + x_2
\end{bmatrix}.
\end{align*}
Therefore, the general formula is given by
\[T\left(\, \begin{bmatrix}
x_1 \\
x_2
\end{bmatrix} \,\right)=\begin{bmatrix}
3x_1 \\
4x_1 \\
3x_1 + x_2
\end{bmatrix}.\]  

Solution 2. (Using the matrix representation of the linear transformation)

The second solution uses the matrix representation of the linear transformation $T$.
Let $A$ be the matrix for the linear transformation $T$.

Then by definition, we have
\[T(\mathbf{x})=A\mathbf{x}, \tag{**}\] for every $\mathbf{x}\in \R^2$.
(Note that the size of $A$ is $3\times 2$ because $T:\R^2\to \R^3$.)


We determine the matrix $A$ as follows.
We compute
\begin{align*}
A\begin{bmatrix}
1 & 0\\
2& 1
\end{bmatrix}&=\begin{bmatrix}
A\begin{bmatrix}
1 \\
2
\end{bmatrix} & A \begin{bmatrix}
0 \\
1
\end{bmatrix}
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
T\left(\, \begin{bmatrix}
1 \\
2
\end{bmatrix} \,\right) & T\left(\, \begin{bmatrix}
0 \\
1
\end{bmatrix} \,\right)
\end{bmatrix} && \text{by (**)} \\[6pt] &=\begin{bmatrix}
3 & 0 \\
4 & 0 \\
5 & 1
\end{bmatrix}.
\end{align*}

The $2\times 2$ matrix on the left hand side is invertible, and the inverse is
\[\begin{bmatrix}
1 & 0\\
2& 1
\end{bmatrix}^{-1}
=\begin{bmatrix}
1 & 0\\
-2& 1
\end{bmatrix}.\] (Remark: This is the matrix $P$ in the first solution.)

Multiplying by this inverse matrix on the right, we obtain
\begin{align*}
A&=\begin{bmatrix}
3 & 0 \\
4 & 0 \\
5 &1
\end{bmatrix}\begin{bmatrix}
1 & 0\\
2& 1
\end{bmatrix}^{-1}\\[6pt] &=\begin{bmatrix}
3 & 0 \\
4 & 0 \\
5 &1
\end{bmatrix}\begin{bmatrix}
1 & 0\\
-2& 1
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
3 & 0\\
4& 0 \\
3 & 1
\end{bmatrix}.
\end{align*}


Now that we have obtained the matrix $A$ for $T$, we can find the general formula as follows.
We have
\begin{align*}
T\left(\, \begin{bmatrix}
x_1 \\
x_2
\end{bmatrix} \,\right)&=A\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix} && \text{by (**)}\\[6pt] &=\begin{bmatrix}
3 & 0\\
4& 0 \\
3 & 1
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
3x_1 \\
4x_1 \\
3x_1 + x_2
\end{bmatrix}.
\end{align*}


In summary, the general formula is
\[T\left(\, \begin{bmatrix}
x_1 \\
x_2
\end{bmatrix} \,\right)=\begin{bmatrix}
3x_1 \\
4x_1 \\
3x_1 + x_2
\end{bmatrix}.\]

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