Multivariable calculus and vector analysis

Matrices and linear functions

Let $A$ be some $2\times 3$ matrix, say

$$\begin{array}{lll}\hfill A=\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill -1\\ \hfill 3& \hfill 1& \hfill 2\end{array}\right].& & \hfill \end{array}$$

What do you get if you multiply $A$ by the vector $\mathbf{x}=\left(x,y,z\right)$? Remembering our matrix multiplication, we see that

$$\begin{array}{lll}\hfill A\mathbf{x}=\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill -1\\ \hfill 3& \hfill 1& \hfill 2\end{array}\right]\left[\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \\ \hfill z\hfill \end{array}\right]=\left[\begin{array}{c}\hfill x-z\hfill \\ \hfill 3x+y+2z\hfill \end{array}\right]=\left(x-z,3x+y+2z\right).& & \hfill \end{array}$$

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

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

$$\begin{array}{lll}\hfill C=\left[\begin{array}{cc}\hfill 5& \hfill -3\\ \hfill 1& \hfill 0\\ \hfill -7& \hfill 4\\ \hfill 0& \hfill -2\end{array}\right],& & \hfill \end{array}$$

then the function $\mathbf{h}\left(\mathbf{y}\right)=C\mathbf{y}$, where $y=\left(y_1,y_2\right)$, is $\mathbf{h}\left(\mathbf{y}\right)=\left(5y_1-3y_2,y_1,-7y_1+4y_2,-2y_2\right)$.

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

Let’s take the function $\mathbf{f}\left(x,y\right)=\left(2x+y,y,x-3y\right)$, which is a linear function from ${\mathbf{R}}^2$ to ${\mathbf{R}}^3$. The matrix $A$ associated with $\mathbf{f}$ will be a $3\times 2$ matrix, which we’ll write as

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

We need $A$ to satisfy $\mathbf{f}\left(\mathbf{x}\right)=A\mathbf{x}$, where $\mathbf{x}=\left(x,y\right)$.

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

$$\begin{array}{lll}\hfill f\left(1,0\right)=\left(2,0,1\right)=\left[\begin{array}{c}\hfill 2\\ \hfill 0\\ \hfill 1\end{array}\right].& & \hfill \end{array}$$

Similarly, if $\mathbf{x}=\left(0,1\right)$, then $f\left(\mathbf{x}\right)=A\mathbf{x}$ is the second column of $A$, which is

$$\begin{array}{lll}\hfill f\left(0,1\right)=\left(1,1,-3\right)=\left[\begin{array}{c}\hfill 1\\ \hfill 1\\ \hfill -3\end{array}\right].& & \hfill \end{array}$$

Putting these together, we see that the linear function $\mathbf{f}\left(\mathbf{x}\right)$ is associated with the matrix

$$\begin{array}{lll}\hfill A=\left[\begin{array}{cc}\hfill 2& \hfill 1\\ \hfill 0& \hfill 1\\ \hfill 1& \hfill -3\end{array}\right].& & \hfill \end{array}$$

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.

Contents by topic

1. Geometry in higher dimensions

1a. Preliminaries

1b. Lines and planes

1c. Matrices

• Matrices and determinants
• Multiplying matrices and vectors
• Dot product in matrix notation
• Matrices and linear functions
• The relationship between determinants and area or volume
• Matrix and vector multiplication examples

1d. Visualizing functions in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

6. Review material