Linear Algebra

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

Linear Transformation problems and solutions

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.

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Solution.

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

Recall that every linear transformation must map the zero vector to the zero vector.

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.

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