Understanding a basis in linear algebra


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Chapter 7 of Cherney, Denton, and Waldron begins by introducing the concept of a basis. The essence of a basis is that its simply a coordinate system with which you can describe a set of vectors, and consequently a vector space. A basis is one of those things that we take for granted until we discover the standard basis happens to be a special case. This is no different from first learning that the base 10 number system is one of many bases or that the integers are just one of many rings.

The first part of the section shows how to represent vectors in an alternate base. Then we look at how a linear operator can be represented as a matrix in terms of two bases. The most compelling example is Example 70, which shows how to represent the derivative operator as a matrix between two bases. In question 7.2.1, part (c) describes a similar problem:

Let L be the manufacturing process; it takes in supply packages and gives out two products (doors, and door frames) and it is linear in supplies. If Lf is 1 door and 2 frames and Lg is 3 doors and 1 frame, find a matrix for L.

This question is a little tricky because not only must you find the matrix representation of the linear operator L, but you must also define the bases. And since there are many bases to choose from, it can be difficult to know whether your solution is correct. So let’s think about how we solve this.

The input basis

First we need to define our input basis. Since the prior parts discuss the two functions f and g in terms of slabs of wood, fasteners, and brackets, it’s natural to assume that the basis is in terms of the ordered set (slab, fastener, bracket). If we follow Example 70 as a guide, we might be inclined to define a basis B = (slab, fastener, bracket). With an ordered triplet, our vector space is essentially \mathbb{W}^3. Then, given an arbitrary vector in this basis, we write \left[\begin{array}{c}a\\b\\c\end{array}\right]_B = a \cdot slab + b \cdot fastener + c \cdot bracket. Great, this is easy enough.

Not so fast, cowboy. Let’s verify with an example. Suppose a=2, b=3, c=2. Then \left[\begin{array}{c}a\\b\\c\end{array}\right]_B = 2 slabs + 3 fasteners + 2 brackets. What’s wrong with this picture? We know from the problem description that

A door factory can buy supplies in two kinds of packages, f and g. The package f contains 3 slabs of wood, 4 fasteners, and 6 brackets. The package g contains 5 fasteners, 3 brackets, and 7 slabs of wood.

So our chosen basis creates supplies (2 slabs + 3 fasteners + 2 brackets) that aren’t possible!

Before you leave a permanent impression of the wall on your forehead, let’s think about this some more. Our current basis creates linear combinations of slabs, fasteners, and brackets. However, it produces more combinations than we want. What we need is a way to control the supplies so that they are delivered in whole multiples of f and g. Indeed, what if we define a basis in terms of f and g? This basis looks like B = \left( \left( \begin{array}{c}3\\4\\6\end{array} \right), \left( \begin{array}{c}7\\5\\3\end{array} \right) \right). Supplies coming into the factory can now be written in terms of f and g: \left[\begin{array}{c}a\\b\end{array}\right]_B = af + bg.

The output basis

Okay, the hardest part is behind us. Now we do the same thing with the output basis. In this case, the output of the manufacturing process consists of door frames and doors, so again this makes a good candidate for the basis. Hence B' = \left( \left( \begin{array}{c}2\\1\end{array} \right), \left( \begin{array}{c}1\\3\end{array} \right) \right). Using these two bases, let’s look at the operator L. If we have one shipment of f, we get L\left(\begin{array}{c}1\\end{array}\right) = 1 \beta_1 + 0 \beta_2. Similarly, if we only have a shipment of g, we get L\left(\begin{array}{c}0\\1\end{array}\right) = 0 \beta_1 + 1 \beta_2. This implies that L = \left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right) .

While this answer may seem anticlimatic, the result is exactly what we expect: that a combination of shipments in f and g should result in a commensurate number of door frames and doors.

An alternative basis

Since this linear operator is a bit boring, let’s see if we can spice it up a little. Suppose we choose a different output basis, say B' = \left( \left( \begin{array}{c}1\\end{array} \right), \left( \begin{array}{c}0\\1\end{array} \right) \right). This time, the basis is simply the unit vectors for door frames and doors. Unlike with the input basis, we’re allowed to do this since the outputs are predefined. This gives us Lf = 2 \beta_1 + 1 \beta_2 and Lg = 1 \beta_1 + 3 \beta_2, so that L = \left(\begin{array}{cc}2 & 1 \\ 1 & 3\end{array}\right) . Again, this works since 1 shipment of f still yields 1 door frame and 2 doors.


What we’ve seen is that a basis is a coordinate system used to describe vectors. When modeling real-world processes, we can’t take standard bases for granted, since they may not exist in our non-numeric domain. In these situations it’s important to think about what the valid inputs and outputs are. Once we come up with a reasonable representation for these, then the matrix form of the linear operator should fall out naturally. The value is that then the calculations are trivial to perform.


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