An Inquiry-Based Introduction to Linear Algebra and Applications

Subsection Application: Calculating Sums

Consider the problem of calculating sums of powers of integers. For example,

and so on. It is possible to prove these formulas if we have an idea of what the formula is, but how can one come up with these formulas in the first place? As we will see later in this section, we can make use of coordinate vectors to find and verify such formulas.

Subsection Introduction

In Section 16 we defined the coordinate vector of a vector \(\vx\) in \(\R^n\) with respect to a basis \(\CB = \vv_1, \vv_2, \vv_n\}\) of \(\R^n\) to be the vector \([x_1 \ x_2 \ \ldots \ x_n]^{\tr}\) if

We can define coordinate vectors in vector spaces in the same way. As we will see, this is a powerful idea as it allows us to translate problems in finite dimensional vector spaces to problems in \(\R^n\) where we have many tools at our disposal. First we define a coordinate vector.

Recall that there is exactly one way to write a vector as a linear combination of basis vectors, so there is only one coordinate vector of a given vector with respect to a basis. Therefore, the coordinate vector of any vector is well-defined.

Subsection The Coordinate Transformation

We have seen that both \(\R^2\) and \(\pol_1\) are vector spaces of dimension 2. Moreover, we can pair the polynomial \(a_0+a_1t\) in \(\pol_1\) with the vector \([a_0 \ a_1]\) in \(\R^2\) as the coordinate vector of \(a_0+a_1t\) with respect to the standard basis for \(\pol_1\text{.}\) In this way we can see that the vector space \(\pol_1\) “looks” like the vector space \(\R^2\text{.}\) In fact, using this same idea, we can show that any vector space of dimension \(n\) “looks” like \(\R^n\text{.}\) The key idea to make this work is the coordinate transformation.

As we saw in Preview Activity 34.1, the coordinate vector of a vector with respect to a basis depends on the basis we choose. In this activity we found that \([2+4t]_{\CB} = \left[ 1 \ \frac{1}{3} \right]^{\tr}\) and \([2+4t]_{\CC} = [ 3 \ -1]^{\tr}\text{,}\) where \(\CB = \{4+2t, -6+6t\}\) and \(\CC = \{1+t, 1-t\}\text{.}\) These coordinate vectors allow us to identify polynomials in \(\pol_1\) with vectors in \(\R^2\text{.}\) In this way, the coordinate vectors provide an identification between the vector space \(\pol_1\) and the vector space \(\R^2\) via a coordinate transformation.

Coordinate transformations have several useful properties that allow us to use them to compare the structure of vector spaces.

Activity 34.2 shows that coordinate transformations behave in ways similar to matrix transformations. Recall that if \(A\) is an \(m \times n\) matrix, then

for any vectors \(\vu\) and \(\vw\) in \(\R^n\) and any scalar \(c\text{.}\) Because of these properties, the transformation \(A\) preserves linear combinations. That is, if \(\vv_1\text{,}\) \(\vv_2\text{,}\) \(\ldots\text{,}\) and \(\vv_n\) are any vectors in \(\R^n\) and \(c_1\text{,}\) \(c_2\text{,}\) \(\ldots\text{,}\) \(c_n\) are any scalars, then

Activity 34.2 contains the essential ideas to show that if \(T\) is a coordinate transformation from a vector space \(V\) with basis \(\CB\) to \(\R^n\text{,}\) then

for any vectors \(\vu\) and \(\vw\) in \(V\) and any scalar \(c\text{.}\) The fact that any coordinate transformation satisfies these properties is contained in the following theorem.

Theorem 34.3 can be extended to any linear combination of vectors by mathematical induction. That is, if \(T : V \to \R^n\) is the coordinate mapping defined by \(T(\vx) = [\vx]_{\CB}\) for some basis \(\CB\text{,}\) and if \(\vu_1\text{,}\) \(\vu_2\text{,}\) \(\ldots\text{,}\) \(\vu_m\) are vectors in an \(n\)-dimensional vector space \(V\) with basis \(\CB\text{,}\) and \(x_1\text{,}\) \(x_2\text{,}\) \(\ldots\text{,}\) \(x_m\) are scalars, then \(T(x_1\vu_1+x_2\vu_2+\cdots +x_m\vu_m) = x_1T(\vu_1) + x_2T(\vu_2) + \cdots + x_mT(\vu_m)\text{.}\) In other words,

In other words, coordinate transformations take linear combinations of vectors in \(V\) to linear combinations of vectors in \(\R^n\text{.}\)

In essence, a coordinate transformation makes an identification of any vector space of dimension \(n\) with \(\R^n\text{.}\) One of the driving forces behind this identification is the fact that, just as in \(\R^n\text{,}\) if a vector space \(V\) has a basis of \(n\) elements, then any basis for \(V\) will have exactly \(n\) elements. The fact that any coordinate transformation from a vector space \(V\) to \(\R^n\) is linear means that elements in \(V\) behave the same way with respect to addition and multiplication by scalars in \(V\) as do their images in \(\R^n\) under the coordinate transformation. To make this identification complete, there are still two questions to address. First, given a coordinate transformation \(T\) from a vector space \(V\) to \(\R^n\text{,}\) is is possible for two vectors in \(V\) to have the same image in \(\R^n\) (in other words, is \(T\) one-to-one), and is every vector in \(\R^n\) the image of some vector in \(V\) under \(T\) (in other words, does \(T\) map \(V\) onto \(\R^n\))? If the coordinate transformations are one-to-one and onto, then, in essence, any vector space of dimension \(n\) is just a copy of \(\R^n\text{.}\) As a result, to understand any vector space with a basis of \(n\) vectors, it is enough to understand \(\R^n\text{.}\)

Activity 34.3 shows that a coordinate transformation maps an \(n\)-dimensional vector space \(V\) onto \(\R^n\text{.}\) What's more, any coordinate transformation is also one-to-one (the proof that \(T\) is one-to-one is left for the exercises). We summarize these results in the following theorem.

Theorem 34.3 and Theorem 34.4 show that a coordinate transformation from an \(n\)-dimensional vector space \(V\) with basis \(\CB\) to \(\R^n\) provides an identification of \(V\) with \(\R^n\text{,}\) where a vector \(\vv\) in \(V\) is identified with the vector \([\vv]_{\CB}\text{.}\) This identification allows us to transfer questions (linear dependence, independence, span) in \(V\) to \(\R^n\) where we can apply our knowledge of matrices to answer the questions.

Subsection Examples

What follows are worked examples that use the concepts from this section.

Subsection Summary

The key idea in this handout is the coordinate vector with respect to a basis.

• If \(\CB = \{\vv_1, \vv_2, \vv_3, \ldots, \vv_n\}\) is a basis for a vector space \(V\text{,}\) then the coordinate vector of \(\vx\) in \(V\) with respect to \(\CB\) is the vector

  \begin{equation*} [\vx]_{\CB} = [x_1, x_2, \ldots, x_n]^{\tr}\text{,} \end{equation*}

  where

  \begin{equation*} \vx = x_1 \vv_1 + x_2 \vv_2 + \cdots + x_n \vv_n\text{.} \end{equation*}

• The coordinate transformation \(\vx \mapsto [\vx]_{\CB}\) is a one-to-one and onto transformation from an \(n\)-dimensional vector space \(V\) to \(\R^n\) which preserves linear combinations.

• The coordinate transformation \(\vx \mapsto [\vx]_{\CB}\) allows us to translate problems in arbitrary vector spaces to \(\R^n\) where we have already developed tools to solve the problems.

Subsection Project: Finding Formulas for Sums of Powers

One way to derive formulas for sums of powers of whole numbers is to use different bases and coordinate vectors. One basis that will be useful is a basis of polynomials created by the binomial coefficients. Recall that the binomial coefficient \(\binom{n}{k}\) is equal to

with \(\binom{n}{k} = 0\) if \(n \lt k\text{.}\)

The binomial coefficient can be rewritten in a way to make it applicable to polynomials as

With this representation of \(\binom{n}{k}\text{,}\) we can define a new polynomial in \(t\) of degree \(k\) as

with \(p_0(t) = 1\text{.}\) For example,

These “generalized binomial coefficients” appear in Newton's generalized binomial theorem.

Two facts will make these polynomials useful for our sums.

The second useful fact about our polynomials \(p_k(t)\) is that they form a basis for \(\pol_n\text{.}\)

Now we have the tools we need to derive our formulas. To simplify computations, we will change coordinates to the standard basis \(\CS_n = \{1, t, t^2, \ldots, t^n\}\) for \(\pol_n\text{.}\)

We will illustrate the process of deriving formulas for our sums with the sum \(\sum_{t=0}^{n-1} t\text{.}\) We want to write \(t\) as a linear combination of the vectors in \(\CP_1\) so that we can utilize the hockey-stick identity. In this case, by definition we have \(t = p_1(t)\text{.}\) It follows by the hockey-stick identity and the fact that \(p_1(t) = \binom{t}{1}\) that

This is exactly the formula we saw at the beginning of this section. The cases for sums of higher powers work the same way, but we will need to do a little more work to write \(t^n\) in terms of the basis vectors in \(\CP_n\text{.}\)

Deriving formulas for higher powers involves the same process, just with more algebra.