# Is the Following Function $T:\R^2 \to \R^3$ a Linear Transformation?

## Problem 627

Determine whether the function $T:\R^2 \to \R^3$ defined by
$T\left(\, \begin{bmatrix} x \\ y \end{bmatrix} \,\right) = \begin{bmatrix} x_+y \\ x+1 \\ 3y \end{bmatrix}$ is a linear transformation.

## Solution.

The function $T:\R^2 \to \R^3$ is a not a linear transformation.

However, we have
$T\left(\, \begin{bmatrix} 0 \\ 0 \end{bmatrix} \,\right) =\begin{bmatrix} 0+0 \\ 0+1 \\ 3\cdot 0 \end{bmatrix}=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \neq \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$ So the function $T$ does not map the zero vector $\begin{bmatrix} 0 \\ 0 \end{bmatrix}$ to the zero vector $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$.
Thus, $T$ is not a linear transformation.

### Another solution

Another way to see this is, for example, as follows.
Let
$\mathbf{u}=\begin{bmatrix} 1 \\ 0 \end{bmatrix} \text{ and } \mathbf{v}=\begin{bmatrix} 0 \\ 1 \end{bmatrix}.$ (In fact, you may take any two vectors.)

Then we have
$T(\mathbf{u})+T(\mathbf{v})=T\left(\, \begin{bmatrix} 1 \\ 0 \end{bmatrix} \,\right)+T\left(\, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \,\right) =\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}+\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}=\begin{bmatrix} 2 \\ 3 \\ 3 \end{bmatrix}.$ On the other hand, we have
$T\left(\, \mathbf{u}+\mathbf{v} \,\right) =T\left(\, \begin{bmatrix} 1 \\ 1 \end{bmatrix} \,\right) =\begin{bmatrix} 2 \\ 2 \\ 3 \end{bmatrix}.$

Therefore, we see that
$T(\mathbf{u})+T(\mathbf{v}) \neq T\left(\, \mathbf{u}+\mathbf{v} \,\right),$ and hence $T$ is not a linear transformation.

Let $A$ be an $n\times n$ matrix. Suppose that the sum of elements in each row of $A$ is zero....