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.

Sponsored Links

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.

Sponsored Links

More from my site

You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

If the Sum of Entries in Each Row of a Matrix is Zero, then the Matrix is Singular

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

Close