Example

Suppose $M$ is a linear transformation operating on 2-dimensional vectors.

If $M \begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}4\\5\end{pmatrix}$ and $M \begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}-3\\7\end{pmatrix}$, find $M \begin{pmatrix}-6\\8\end{pmatrix}$


Appealing to the linearity of $M$, we see that

$$\begin{align} M\begin{pmatrix}-6\\8\end{pmatrix} &= M \left( \begin{pmatrix}-6\\0\end{pmatrix} + \begin{pmatrix}0\\8\end{pmatrix} \right) \\ &= M\begin{pmatrix}-6\\0\end{pmatrix} + M\begin{pmatrix}0\\8\end{pmatrix} \\ &= M\left( -6 \begin{pmatrix}1\\0\end{pmatrix} \right) + M \left( 8 \begin{pmatrix}0\\1\end{pmatrix} \right) \\ &= -6 M\begin{pmatrix}1\\0\end{pmatrix} + 8 M \begin{pmatrix}0\\1\end{pmatrix} \\ &= -6 \begin{pmatrix}4\\5\end{pmatrix} + 8 \begin{pmatrix}-3\\7\end{pmatrix} \\ &= \begin{pmatrix}(-6) \cdot 4\\ (-6) \cdot 5\end{pmatrix} + \begin{pmatrix} 8 \cdot (-3)\\8 \cdot 7\end{pmatrix} \\ &= \begin{pmatrix}(-6) \cdot 4 + 8 \cdot (-3)\\ (-6) \cdot 5 + 8 \cdot 7\end{pmatrix} \\ &= \begin{pmatrix}-48\\26\end{pmatrix} \end{align}$$

We can apply this same process to find the output of any linear transformation that operates on two-dimensional vectors, provided we know what the linear transformation does to $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$.

As such, we can abbreviate the process above by writing $M$ as a matrix, and evaluating $M\begin{pmatrix}-6\\8\end{pmatrix}$ in the following way:

$$M\begin{pmatrix}-6\\8\end{pmatrix} = \begin{bmatrix}4 & -3\\5 & 7\end{bmatrix} \begin{pmatrix}-6\\8\end{pmatrix} = \begin{pmatrix}4 \cdot (-6) + (-3) \cdot 8\\5 \cdot (-6) + 7 \cdot 8\end{pmatrix} = \begin{pmatrix}-48\\26\end{pmatrix}$$