Let $A$ be a $2 \times 3$ matrix, say \begin{align*} A = \left[ \begin{array}{rrr} 1 & 0 & -1\\ 3 & 1 & 2 \end{array} \right]. \end{align*} What do you get if you multiply $A$ by the vector $\vc{x}=(x,y,z)$? Remembering matrix multiplication, we see that \begin{align*} A\vc{x} = \left[ \begin{array}{rrr} 1 & 0 & -1\\ 3 & 1 & 2 \end{array} \right] \left[ \begin{array}{c} x\\ y\\ z \end{array} \right] = \left[ \begin{array}{c} x - z\\ 3x + y +2z \end{array} \right] =(x-z,3x+y+2z). \end{align*}

If we define a function $\vc{f}(\vc{x}) = A\vc{x}$, we have created a function of three variables $(x,y,z)$ whose output is a two-dimensional vector $(x-z,3x+y+2z)$. Using function notation, we can write $\vc{f} : \R^3 \to \R^2$. We have created a vector-valued function of three variables. So, for example, $\vc{f}(1,2,3) = (1-3,3\cdot 1 + 2 + 2 \cdot 3) = (-2, 11)$.

Given any $m \times n$ matrix $B$, we can define a function $\vc{g}: \R^n \to \R^m$ (note the order of $m$ and $n$ switched) by $\vc{g}(\vc{x}) = B \vc{x}$, where $\vc{x}$ is an $n$-dimensional vector. As another example, if \begin{align*} C = \left[ \begin{array}{rr} 5 & -3\\ 1 & 0\\ -7 & 4\\ 0 & -2 \end{array} \right], \end{align*} then the function $\vc{h}(\vc{y}) = C \vc{y}$, where $y=(y_1, y_2)$, is $\vc{h}(\vc{y}) = (5y_1-3y_2,y_1,-7y_1+4y_2,-2y_2)$.

In this way, we can associate with every matrix a function. What about going the other way around? Given some function, say $\vc{g}: \R^n \to \R^m$, can we associate with $\vc{g}(\vc{x})$ some matrix? We can only if $\vc{g}(\vc{x})$ is a special kind of function called a linear transformation. The function $\vc{g}(\vc{x})$ is a linear transformation if each term of each component of $\vc{g}(\vc{x})$ is a number times one of the variables. So, for example, the functions $\vc{f}(x,y)=(2x+y,y/2)$ and $\vc{g}(x,y,z)=(z,0,1.2 x)$ are linear transformation, but none of the following functions are: $\vc{f}(x,y)=(x^2,y,x)$, $\,\vc{g}(x,y,z)=(y,xyz)$, or $\vc{h}(x,y,z)=(x+1,y,z)$. Note that both functions we obtained from matrices above were linear transformations.

Let's take the function $\vc{f}(x,y)=(2x+y,y,x-3y)$, which is a linear transformation from $\R^2$ to $\R^3$. The matrix $A$ associated with $\vc{f}$ will be a $3 \times 2$ matrix, which we'll write as \begin{align*} A = \left[ \begin{array}{cc} a_{11} & a_{12}\\ a_{21} & a_{22}\\ a_{31} & a_{32} \end{array} \right]. \end{align*} We need $A$ to satisfy $\vc{f}(\vc{x})=A\vc{x}$, where $\vc{x}=(x,y)$.

The easiest way to find $A$ is the following. If we let $\vc{x}=(1,0)$, then $f(\vc{x})= A\vc{x}$ is the first column of $A$. (Can you see that?) So we know the first column of $A$ is simply \begin{align*} f(1,0)=(2,0,1) = \left[ \begin{array}{r} 2\\0\\1 \end{array} \right]. \end{align*}

Similarly, if $\vc{x}=(0,1)$, then $f(\vc{x})=A\vc{x}$ is the second column of $A$, which is \begin{align*} f(0,1) = (1,1,-3) = \left[ \begin{array}{r} 1\\1\\-3 \end{array} \right]. \end{align*}

Putting these together, we see that the linear transformation $\vc{f}(\vc{x})$ is associated with the matrix \begin{align*} A= \left[ \begin{array}{rr} 2 & 1\\ 0 & 1\\ 1 & -3 \end{array} \right]. \end{align*}

The important conclusion is that every linear transformation is associated with a matrix and vice versa.