Matrices and linear functions

Let A be some 2 x 3 matrix, say

$$\left[\vphantom{ \begin{array}{rrr} 1 & 0 & -1\ 3 & 1 & 2 \end{array} }\right. \begin{array}{rrr} 1 & 0 & -1\ 3 & 1 & 2 \end{array} \left.\vphantom{ \begin{array}{rrr} 1 & 0 & -1\ 3 & 1 & 2 \end{array} }\right]$$

What do you get if you multiply A by the vector x = (x, y, z)? Remembering our matrix multiplication, we see that

$$\left[\vphantom{ \begin{array}{rrr} 1 & 0 & -1\ 3 & 1 & 2 \end{array} }\right. \begin{array}{rrr} 1 & 0 & -1\ 3 & 1 & 2 \end{array} \left.\vphantom{ \begin{array}{rrr} 1 & 0 & -1\ 3 & 1 & 2 \end{array} }\right] \left[\vphantom{ \begin{array}{c} x\ y\ z \end{array} }\right. \begin{array}{c} x\ y\ z \end{array} \left.\vphantom{ \begin{array}{c} x\ y\ z \end{array} }\right] \left[\vphantom{ \begin{array}{c} x - z\ 3x + y +2z \end{array} }\right. \begin{array}{c} x - z\ 3x + y +2z \end{array} \left.\vphantom{ \begin{array}{c} x - z\ 3x + y +2z \end{array} }\right]$$

If we define a function f(x) = Ax, we have created a function of three variables (x, y, z) whose output is a two-dimensional vector (x - z, 3x + y + 2z). Using our function notation, we can write f : R³$\to$R². We have created a vector-valued function of three variables. So, for example, f(1, 2, 3) = (1 - 3, 3 ^. 1 + 2 - 2 ^. 3) = (- 2, - 1).

Given any m x n matrix B, we can define a function g : Rⁿ$\to$R^m (note the order of m and n switched) by g(x) = Bx, where x is an n-dimensional vector. As another example, if

$$\left[\vphantom{ \begin{array}{rr} 5 & -3\ 1 & 0\ -7 & 4\ 0 & -2 \end{array} }\right. \begin{array}{rr} 5 & -3\ 1 & 0\ -7 & 4\ 0 & -2 \end{array} \left.\vphantom{ \begin{array}{rr} 5 & -3\ 1 & 0\ -7 & 4\ 0 & -2 \end{array} }\right]$$

then the function h(y) = Cy, where y = (y₁, y₂), is h(y) = (5y₁ -3y₂, y₁, -7y₁ +4y₂, -2y₂).

In this way, we can associate with every matrix a function. What about going the other way around? Given some function, say g : Rⁿ$\to$R^m, can we associate with g(x) some matrix? We can only if g(x) is a special kind of function called a linear function. The function g(x) is linear if each term in g(x) is a number times one of the variables. So, for example, the functions f(x, y) = (2x + y, y/2) and g(x, y, z) = (z, 0, 1.2x) are linear, but none of the following functions are linear: f(x, y) = (x², y, x),  g(x, y, z) = (y, xyz), or h(x, y, z) = (x + 1, y, z). Note that both functions we obtained from matrices above were linear.

Let's take the function f(x, y) = (2x + y, y, x - 3y), which is a linear function from R² to R³. The matrix A associated with f will be a 3 x 2 matrix, which we'll write as

$$\left[\vphantom{ \begin{array}{cc} a_{11} & a_{12}\ a_{21} & a_{22}\ a_{31} & a_{32} \end{array} }\right. \begin{array}{cc} a_{11} & a_{12}\ a_{21} & a_{22}\ a_{31} & a_{32} \end{array} \left.\vphantom{ \begin{array}{cc} a_{11} & a_{12}\ a_{21} & a_{22}\ a_{31} & a_{32} \end{array} }\right]$$

We need A to satisfy f(x) = Ax, where x = (x, y).

The easiest way to find A is the following. If we let x = (1, 0), then f (x) = Ax is the first column of A. (Can you see that?) So we know the first column of A is simply

$$\left[\vphantom{ \begin{array}{r} 2\ 0\ 1 \end{array} }\right. \begin{array}{r} 2\ 0\ 1 \end{array} \left.\vphantom{ \begin{array}{r} 2\ 0\ 1 \end{array} }\right]$$

Similarly, if x = (0, 1), then f (x) = Ax is the second column of A, which is

$$\left[\vphantom{ \begin{array}{r} 1\ 1\ -3 \end{array} }\right. \begin{array}{r} 1\ 1\ -3 \end{array} \left.\vphantom{ \begin{array}{r} 1\ 1\ -3 \end{array} }\right]$$

Putting these together, we see that the linear function f(x) is associated with the matrix

$$\left[\vphantom{ \begin{array}{rr} 2 & 1\ 0 & 1\ 1 & -3 \end{array} }\right. \begin{array}{rr} 2 & 1\ 0 & 1\ 1 & -3 \end{array} \left.\vphantom{ \begin{array}{rr} 2 & 1\ 0 & 1\ 1 & -3 \end{array} }\right]$$

The important conclusion is that every linear function is associated with a matrix and vice versa. We will sometimes use this correspondence in this course because, as you soon will learn, the multivariable version of a derivative will be a matrix. We can also view the derivative as the linear function associated with that matrix.