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

## 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)

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}.

## Related Question.

Here are similar problems about linear transformations

##### Hyperplane in $n$-Dimensional Space Through Origin is a Subspace
A hyperplane in $n$-dimensional vector space $\R^n$ is defined to be the set of vectors \[\begin{bmatrix} x_1 \\ x_2 \\...