Coordinate Vectors and Coordinate Transformations

Section 34 Coordinate Vectors and Coordinate Transformations

Focus Questions

By the end of this section, you should be able to give precise and thorough answers to the questions listed below. You may want to keep these questions in mind to focus your thoughts as you complete the section.

How do we find the coordinate vector of a vector $x$ with respect to a basis $B = {v_{1}, v_{2}, \dots, v_{n}} ?$
How can we visualize coordinate systems?
How do we identify a vector space of dimension $n$ with $R^{n} ?$
What properties does the coordinate transformation $x \mapsto [x]_{B}$ have?
How are the coordinates of a vector with respect to a basis related to its coordinates with respect to the standard basis?

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Subsection Application: Calculating Sums

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

\begin{aligned} \sum_{t = 0}^{n - 1} t & = \frac{1}{2} (n^{2} - n) \\ \sum_{t = 0}^{n - 1} t^{2} & = \frac{1}{6} (2 n^{3} - 3 n^{2} + 1) \\ \sum_{t = 0}^{n - 1} t^{3} & = \frac{1}{4} (n^{4} - 2 n^{3} + n^{2}) \end{aligned}

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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.

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Subsection Introduction

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In Section 16 we defined the coordinate vector of a vector

x

R^{n}

with respect to a basis

B = v_{1}, v_{2}, v_{n}}

R^{n}

to be the vector

[x_{1} x_{2} \dots x_{n}]^{T}

x = x_{1} v_{1} + x_{2} v_{2} + \dots + x_{n} v_{n} .

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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.

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Definition 34.1.

Let $B = {v_{1}, v_{2}, \dots, v_{n}}$ be an ordered basis for a vector space $V .$ For any vector $x$ in $V,$ the coordinate vector of $x$ with respect to $B$ is the vector

[x]_{B} = [\begin{matrix} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \end{matrix}],

where

x = x_{1} v_{1} + x_{2} v_{2} + \dots + x_{n} v_{n} .

The scalars $x_{1},$ $x_{2},$ $\dots,$ $x_{n}$ are the coordinates of the vector $x$ with respect to the basis.

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Preview Activity 34.1.

Let $b_{1} (t) = 4 + 2 t,$ $b_{2} (t) = - 6 + 6 t,$ $c_{1} (t) = 1 + t,$ $c_{2} (t) = 1 - t,$ and let $B = {b_{1} (t), b_{2} (t)}$ and $C = {c_{1} (t), c_{2} (t)} .$

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(a)

Show that $B$ and $C$ are bases for $P_{1} .$

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(b)

Let $p (t) = 3 b_{1} (t) + 2 b_{2} (t) .$ What is $[p (t)]_{B} ?$

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(c)

Find the coordinate vector of $f (t) = 2 + 4 t$ with respect to both $B$ and $C .$

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(d)

Find a polynomial $r (t)$ such that $[r (t)]_{B} = [1 3]^{T} .$ Determine exactly how many such polynomials $s (t)$ have the property that $[s (t)]_{B} = [1 3]^{T} .$ Explain your reasoning.

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(e)

Find a polynomial $q (t)$ such that $[q (t)]_{C} = [1 3]^{T} .$ Determine exactly how many such polynomials $s (t)$ have the property that $[s (t)]_{C} = [1 3]^{T} .$ Explain your reasoning.

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(f)

What specific property does a basis have that ensures the responses to parts (c) and (d) will always be the same?

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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.

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Subsection The Coordinate Transformation

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We have seen that both

R^{2}

and

P_{1}

are vector spaces of dimension 2. Moreover, we can pair the polynomial

a_{0} + a_{1} t

P_{1}

with the vector

[a_{0} a_{1}]

R^{2}

as the coordinate vector of

a_{0} + a_{1} t

with respect to the standard basis for

P_{1} .

In this way we can see that the vector space

P_{1}

“looks” like the vector space

R^{2} .

In fact, using this same idea, we can show that any vector space of dimension

n

“looks” like

R^{n} .

The key idea to make this work is the coordinate transformation.

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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 + 4 t]_{B} = {[1 \frac{1}{3}]}^{T}

and

[2 + 4 t]_{C} = [3 - 1]^{T},

where

B = {4 + 2 t, - 6 + 6 t}

and

C = {1 + t, 1 - t} .

These coordinate vectors allow us to identify polynomials in

P_{1}

with vectors in

R^{2} .

In this way, the coordinate vectors provide an identification between the vector space

P_{1}

and the vector space

R^{2}

via a coordinate transformation.

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Definition 34.2.

Let $V$ be a vector space of dimension $n$ with basis $B .$ The coordinate transformation $T$ from $V$ to $R^{n}$ with respect to the ordered basis $B$ is the mapping defined by

T (x) = [x]_{B}

for any vector $x$ in $V .$

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

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Activity 34.2.

Let $a (t) = 8 + 2 t$ and $b (t) = - 5 + t$ in $P_{1} .$ Suppose we know $C = {1 + t, 2 - t}$ is a basis of $P_{1} .$ Let $T (x) = [x]_{C}$ for $x$ in $P_{1} .$

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(a)

What are $T (a (t))$ and $T (b (t)) ?$

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(b)

Find $T (a (t)) + T (b (t)) .$

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(c)

What is $T (a (t) + b (t)) = [a (t) + b (t)]_{C} ?$

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(d)

What is the relationship between $T (a (t)) + T (b (t))$ and $T (a (t) + b (t)) ?$

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(e)

Show that if $c$ is any scalar, then $T (c a (t)) = c T (a (t)) .$

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(f)

Where have we seen functions with these properties before?

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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

A (u + w) = A u + A w and A (c u) = c A u

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for any vectors

u

and

w

R^{n}

and any scalar

c .

Because of these properties, the transformation

A

preserves linear combinations. That is, if

v_{1},

v_{2},

\dots,

and

v_{n}

are any vectors in

R^{n}

and

c_{1},

c_{2},

\dots,

c_{n}

are any scalars, then

A (c_{1} v_{1} + c_{2} v_{2} + \dots + c_{n} v_{n}) = c_{1} A v_{1} + c_{2} A v_{2} + \dots + c_{n} A v_{n} .

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Activity 34.2 contains the essential ideas to show that if

T

is a coordinate transformation from a vector space

V

with basis

B

R^{n},

then

T (u + w) = T (u) + T (w) and T (c u) = c T (u)

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for any vectors

u

and

w

V

and any scalar

c .

The fact that any coordinate transformation satisfies these properties is contained in the following theorem.

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Theorem 34.3.

If a vector space $V$ has a basis $B$ of $n$ vectors, then the coordinate mapping $T : V \to R^{n}$ defined by $T (x) = [x]_{B}$ satisfies

$T (u + w) = T (u) + T (w)$ and
$T (c u) = c T (u)$

for any vectors $u$ and $w$ in $V$ and any scalar $c .$

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Proof.

Let $v_{1},$ $v_{2},$ $\dots,$ and $v_{n}$ be vectors that form a basis $B$ for a vector space $V .$ To verify the first property, let $u$ and $w$ be two arbitrary vectors from $V .$ We will show that $T (u) + T (w) = T (u + w)$ for these two vectors. Consider $T (u) .$ If $T (u) = [u_{1} u_{2} \dots u_{n}]^{T},$ then the fact that $T (u) = [u]_{B}$ implies that

u = u_{1} v_{1} + u_{2} v_{2} + \dots + u_{n} v_{n} .

Similarly, if $T (w) = [w_{1} w_{2} \dots w_{n}]^{T},$ then $T (w) = [w]_{B}$ implies that

w = w_{1} v_{1} + w_{2} v_{2} + \dots + w_{n} v_{n} .

We then obtain

\begin{aligned} u + w & = (u_{1} v_{1} + u_{2} v_{2} + \dots + u_{n} v_{n}) + (w_{1} v_{1} + w_{2} v_{2} + \dots + w_{n} v_{n}) \\ = (u_{1} + w_{1}) v_{1} + (u_{2} + w_{2}) v_{2} + \dots + (u_{n} + w_{n}) v_{n} . \end{aligned}

Thus, by definition of $T$ again,

T (u + w) = [u + w]_{B} = [(u_{1} + w_{1}) (u_{2} + w_{2}) \dots (u_{n} + w_{n})]^{T} .

To show that $T (u + w) = T (u) + T (w),$ note that

\begin{aligned} T (u) + T (w) & = [u_{1} u_{2} \dots u_{n}]^{T} + [w_{1} w_{2} \dots w_{n}]^{T} \\ = [(u_{1} + w_{1}) (u_{2} + w_{2}) \dots (u_{n} + w_{n})]^{T} \\ = [u + w]_{B} \\ = T (u + w) . \end{aligned}

The proof of the second property is left for the exercises.

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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 (x) = [x]_{B}

for some basis

B,

and if

u_{1},

u_{2},

\dots,

u_{m}

are vectors in an

n

-dimensional vector space

V

with basis

B,

and

x_{1},

x_{2},

\dots,

x_{m}

are scalars, then

T (x_{1} u_{1} + x_{2} u_{2} + \dots + x_{m} u_{m}) = x_{1} T (u_{1}) + x_{2} T (u_{2}) + \dots + x_{m} T (u_{m}) .

In other words,

[x_{1} u_{1} + x_{2} u_{2} + \dots + x_{m} u_{m}]_{B} = x_{1} [u_{1}]_{B} + x_{2} [u_{2}]_{B} + \dots + x_{m} [u_{m}]_{B} .

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In other words, coordinate transformations take linear combinations of vectors in

V

to linear combinations of vectors in

R^{n} .

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In essence, a coordinate transformation makes an identification of any vector space of dimension

n

with

R^{n} .

One of the driving forces behind this identification is the fact that, just as in

R^{n},

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

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

R^{n},

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} .

As a result, to understand any vector space with a basis of

n

vectors, it is enough to understand

R^{n} .

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Activity 34.3.

Let $V$ be a vector space with an ordered basis $B = {v_{1}, v_{2}, \dots, v_{n}} .$ Then $T$ maps $V$ into $R^{n} .$ We want to show that $T$ maps $V$ onto $R^{n} .$ Recall that a function $f$ from a set $X$ to a set $Y$ is onto if for any element $y$ in $Y,$ there is an element $x$ in $X$ such that $f (x) = y .$ Let $b = [b_{1} b_{2} \dots b_{n}]^{T}$ be a vector in $R^{n} .$ Must there be a vector $v$ in $V$ so that $T (v) = b ?$ If so, find such a vector. If not, explain why not.

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Activity 34.3 shows that a coordinate transformation maps an

n

-dimensional vector space

V

onto

R^{n} .

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.

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Theorem 34.4.

If a vector space $V$ has a basis $B$ of $n$ vectors, then the coordinate mapping $T : V \to R^{n}$ defined by $T (x) = [x]_{B}$ is both one-to-one and onto.

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Theorem 34.3 and Theorem 34.4 show that a coordinate transformation from an

n

-dimensional vector space

V

with basis

B

R^{n}

provides an identification of

V

with

R^{n},

where a vector

v

V

is identified with the vector

[v]_{B} .

This identification allows us to transfer questions (linear dependence, independence, span) in

V

R^{n}

where we can apply our knowledge of matrices to answer the questions.

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Activity 34.4.

Let $V = P_{3}$ and let $B = {1, t, t^{2}, t^{3}} .$ Let $S = {1 + t + t^{2} + t^{3}, t - t^{3}, 1 + 2 t^{2}, 1 + 5 t - t^{3}} .$

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(a)

Find each of $[1 + t + t^{2} + t^{3}]_{B},$ $[t - t^{3}]_{B},$ $[1 + 2 t^{2}]_{B},$ and $[1 + 5 t - t^{3}]_{B} .$

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(b)

Are the vectors $[1 + t + t^{2} + t^{3}]_{B},$ $[t - t^{3}]_{B},$ $[1 + 2 t^{2}]_{B},$ and $[1 + 5 t - t^{3}]_{B}$ linearly independent or dependent? Explain. If the vectors are linearly dependent, write one of the vectors as a linear combination of the others.

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(c)

The coordinate transformation identifies the vectors in $S = {1 + t + t^{2} + t^{3}, t - t^{3}, 1 + 2 t^{2}, 1 + 5 t - t^{3}}$ with their coordinate vectors in $R^{4} .$ Use that information to determine if $S$ is a linearly independent or dependent set. If dependent, write one of the vectors in $S$ as a linear combination of the others.

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Subsection Examples

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

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Example 34.5.

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(a)

Find the coordinate vector of $v$ with respect to the ordered basis $B$ in the indicated vector space.

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(i)

$B = {1 + t, 2 - t}$ in $P_{1}$ with $v = 4 + t$

Solution.

Find the coordinate vector of $v$ with respect to the ordered basis $B$ in the indicated vector space.

We need to write $v = 4 + t$ as a linear combination of $1 + t$ and $2 - t .$ If $4 + t = c_{1} (1 + t) + c_{2} (2 - t),$ then equating coefficients of like power terms yields the equations $4 = c_{1} + 2 c_{2}$ and $1 = c_{1} - c_{2} .$ The solution to this system is $c_{1} = 2$ and $c_{2} = 1,$ so $[4 + t]_{B} = [2 1]^{T} .$

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(ii)

$B = {[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}], [\begin{array}{rc} 1 & 0 \\ - 1 & 1 \end{array}], [\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}], [\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}]}$ in $M_{2 \times 2}$ with $v = [\begin{array}{cc} 2 & 3 \\ 1 & 0 \end{array}]$

Solution.

Find the coordinate vector of $v$ with respect to the ordered basis $B$ in the indicated vector space.

We need to write $v$ as a linear combination of the vectors in $B .$ If

\begin{aligned} [\begin{array}{cc} 2 & 3 \\ 1 & 0 \end{array}] & = c_{1} [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] + c_{2} [\begin{array}{rc} 1 & 0 \\ - 1 & 1 \end{array}] \\ + c_{3} [\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}] + c_{4} [\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}], \end{aligned}

equating corresponding components produces the system

\begin{aligned} c_{1} & + & c_{2} & + & c_{3} & = & 2 \\ c_{3} & = & 3 \\ - c_{2} & = & 1 \\ c_{1} & + & c_{2} & + & c_{3} & + & c_{4} & = & 0 & . \end{aligned}

The solution to this system is $c_{1} = 0,$ $c_{2} = - 1,$ $c_{3} = 3,$ and $c_{4} = - 2,$ so $[v]_{B} = [0 - 1 3 - 2]^{T} .$

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(b)

Find the vector $v$ in the indicated vector space $V$ given the basis $B$ of $V$ and $[v]_{B} .$

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(i)

$V = P_{2},$ $B = {1 + t^{2}, 1 + t, t + t^{2}},$ $[v]_{B} = [2 1 3]^{T}$

Solution.

Find the vector $v$ in the indicated vector space $V$ given the basis $B$ of $V$ and $[v]_{B} .$

Since $[v]_{B} = [2 1 3]^{T},$ it follows that

v = 2 (1 + t^{2}) + 1 (1 + t) + 3 (t + t^{2}) = 3 + 4 t + 5 t^{2} .

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(ii)

$B = {\cos (x), \frac{1}{1 + x^{2}}},$ $V = Span B$ as a subspace of $F,$ $[x]_{B} = [2 - 1]^{T}$

Solution.

Find the vector $v$ in the indicated vector space $V$ given the basis $B$ of $V$ and $[v]_{B} .$

Since $[x]_{B} = [2 - 1]^{T},$ it follows that

v = 2 \cos (x) - \frac{1}{1 + x^{2}} .

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Example 34.6.

Let $p_{1} (t) = 1,$ $p_{2} (t) = 2 - t,$ $p_{3} (t) = 3 + t - t^{2},$ $p_{4} (t) = t + t^{3},$ and $p_{5} (t) = 2 t + t^{2} + t^{4}$ be in $P_{4} .$ Also, let $B = {1, t, t^{2}, t^{3}, t^{4}}$ be the standard basis for $P_{4} .$

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(a)

Find $[p_{1} (t)]_{B},$ $[p_{2} (t)]_{B},$ $[p_{3} (t)]_{B},$ $[p_{4} (t)]_{B},$ and $[p_{5} (t)]_{B} .$

Solution.

The coordinate vectors of a polynomial with respect to the standard basis in $P_{4}$ is found by just reading off the coefficients of the polynomial. So

\begin{aligned} [p_{1} (t)]_{B} & = [1 0 0 0 0]^{T} \\ [p_{2} (t)]_{B} & = [2 - 1 0 0 0]^{T} \\ [p_{3} (t)]_{B} & = [3 1 - 1 0 1]^{T} \\ [p_{4} (t)]_{B} & = [0 1 0 1 0]^{T} \\ [p_{5} (t)]_{B} & = [0 2 1 0 1]^{T} . \end{aligned}

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(b)

Use the result of part (a) to explain why ${p_{1} (t), p_{2} (t), p_{3} (t), p_{4} (t), p_{5} (t)}$ is a basis for $P_{4} .$

Solution.

Let

\begin{aligned} C & = {p_{1} (t), p_{2} (t), p_{3} (t), p_{4} (t), p_{5} (t)} and \\ S & = {[p_{1} (t)]_{B}, [p_{2} (t)]_{B}, [p_{3} (t)]_{B}, [p_{4} (t)]_{B}, [p_{5} (t)]_{B}} . \end{aligned}

Since the coordinate transformation is one-to-one and onto, the two sets $C$ and $S$ are either both linearly dependent or linearly independent. Technology shows that the reduced row echelon form of

[[p_{1} (t)]_{B} [p_{2} (t)]_{B} [p_{3} (t)]_{B} [p_{4} (t)]_{B} [p_{5} (t)]_{B}]

is $I_{5},$ so the sets $C$ and $S$ are both linearly linearly independent. Since $\dim (P_{4}) = 5,$ it follows that any linearly independent set of five vectors in $P_{4}$ is a basis for $P_{4} .$ Therefore, the set $C$ is a basis for $P_{4} .$

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(c)

Let $p (t) = 2 - t + t^{2} - 3 t^{3} + 4 t^{4} .$ Find $[p (t)]_{B} .$

Solution.

Similar to part (a), we have $[p (t)]_{B} = [2 - 1 1 - 3 4]^{T} .$

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(d)

Use the coordinate vectors in parts (a) and (c) to write $p (t)$ as a linear combination of $p_{1} (t),$ $p_{2} (t),$ $p_{3} (t),$ $p_{4} (t),$ and $p_{5} (t)$

Solution.

Technology shows that the reduced row echelon form of the augmented matrix

[[p_{1} (t)]_{B} [p_{2} (t)]_{B} [p_{3} (t)]_{B} [p_{4} (t)]_{B} [p_{5} (t)]_{B} | [p (t)]_{B}]

[\begin{array}{cccccr} 1 & 0 & 0 & 0 & 0 & - 25 \\ 0 & 1 & 0 & 0 & 0 & 9 \\ 0 & 0 & 1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 0 & - 3 \\ 0 & 0 & 0 & 0 & 1 & 4 \end{array}] .

[p (t)]_{B} = - 25 [p_{1} (t)]_{B} + 9 [p_{2} (t)]_{B} + 3 [p_{3} (t)]_{B} - 3 [p_{4} (t)]_{B} + 4 [p_{5} (t)]_{B}

and

p (t) = - 25 p_{1} (t) + 9 p_{2} (t) + 3 p_{3} (t) - 3 p_{4} (t) + 4 p_{5} (t) .

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Subsection Summary

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

If $B = {v_{1}, v_{2}, v_{3}, \dots, v_{n}}$ is a basis for a vector space $V,$ then the coordinate vector of $x$ in $V$ with respect to $B$ is the vector

$[x]_{B} = [x_{1}, x_{2}, \dots, x_{n}]^{T},$

where

$x = x_{1} v_{1} + x_{2} v_{2} + \dots + x_{n} v_{n} .$
The coordinate transformation $x \mapsto [x]_{B}$ 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 $x \mapsto [x]_{B}$ allows us to translate problems in arbitrary vector spaces to $R^{n}$ where we have already developed tools to solve the problems.

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Exercises Exercises

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1.

Let $B = {[\begin{array}{r} 0 \\ 1 \\ - 1 \end{array}], [\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}]}$ be a basis of the subspace defined by the equation $y - 4 x + z = 0 .$ Find the coordinates of the vector $b = [\begin{matrix} 3 \\ 4 \\ 2 \end{matrix}]$ with respect to the basis $B .$

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2.

Given basis $B = {1 + t, 2 + t^{2}, t + t^{2}}$ of $P_{2},$

🔗

(a)

For which $p (t)$ in $P_{2}$ is $[p (t)]_{B} = [1 - 1 3]^{T} ?$

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(b)

Determine coordinates of $q (t) = - 1 + t + 2 t^{2}$ in $P_{2}$ with respect to basis $B .$

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3.

Find two different bases $B_{1}$ and $B_{2}$ of $R^{2}$ so that $[b]_{B_{1}} = [b]_{B_{2}} = [\begin{matrix} 2 \\ 1 \end{matrix}],$ where $b = [\begin{matrix} 5 \\ 3 \end{matrix}] .$

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4.

If $[b_{1}]_{B} = [\begin{matrix} 1 \\ 2 \end{matrix}]$ and $[b_{2}]_{B} = [\begin{matrix} 2 \\ 1 \end{matrix}]$ with respect to some basis $B,$ where $b_{1} = [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$ and $b_{2} = [\begin{matrix} 2 \\ 1 \\ 3 \end{matrix}],$ what are the coordinates of $[\begin{array}{r} - 2 \\ 3 \\ 1 \end{array}] ?$

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5.

If $[b_{1}]_{B} = [\begin{matrix} 1 \\ 1 \end{matrix}]$ and $[b_{2}]_{B} = [\begin{matrix} 2 \\ 1 \end{matrix}]$ with respect to some basis $B,$ where $b_{1} = [\begin{matrix} 3 \\ 1 \\ 3 \end{matrix}]$ and $b_{2} = [\begin{matrix} 4 \\ 1 \\ 5 \end{matrix}],$ what are the vectors in $B ?$

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6.

Let $B = {v_{1}, v_{2}, \dots, v_{n}}$ be a basis for a vector space $V .$ Describe how the coordinates of a vector with respect to $B$ will change if $v_{1}$ is replaced with $\frac{1}{2} v_{1} .$

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7.

Let $B = {1, t, 1 + t^{2}}$ in $P_{2} .$

🔗

(a)

Show that $B$ is a basis for $P_{2} .$

🔗

(b)

Let $p_{1} (t) = 1 + 2 t^{2},$ $p_{2} (t) = 1 + t + 2 t^{2},$ and $p_{3} (t) = 2 - t + t^{2}$ in $P_{2} .$

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(i)

Find $[p_{1} (t)]_{B},$ $[p_{2} (t)]_{B},$ and $[p_{3} (t)]_{B} .$

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(ii)

Use the coordinate vectors in part i. to determine if the set ${p_{1} (t), p_{2} (t), p_{3} (t)}$ is linearly independent or dependent.

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8.

Let $W = Span {2 + 4 t + 6 t^{3}, 3 - t^{2}, 3 - t^{2} + 9 t^{3}}$ in $P_{3} .$ Let $B = {1, t, t^{2}, t^{3}}$ be the standard basis for $P_{3} .$

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(a)

Calculate $[2 + 4 t + 6 t^{3}]_{B},$ $[3 - t^{2}]_{B}$ and $[3 - t^{2} + 9 t^{3}]_{B} .$

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(b)

Use the coordinate vectors from part (a) to determine if the polynomials $2 + 4 t + 6 t^{3},$ $3 - t^{2},$ and $3 - t^{2} + 9 t^{3}$ are linearly independent or dependent.

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(c)

Let $p (t) = 4 + 2 t - t^{2} + 9 t^{3} .$ Find $[p (t)]_{B} .$

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(d)

Use the calculations from parts (a) and (c) to determine if $p (t)$ is in $W .$ If so, write $p (t)$ as a linear combination of the polynomials $2 + 4 t + 6 t^{3},$ $3 - t^{2},$ and $3 - t^{2} + 9 t^{3} .$ If not, explain why not.

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9.

Let $B = {[\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}], [\begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array}], [\begin{array}{cc} 0 & 0 \\ 1 & 1 \end{array}], [\begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array}]}$ in $M_{2 \times 2} .$

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(a)

Show that $B$ is a basis of $M_{2 \times 2} .$

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(b)

Let

A = [\begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array}], B = [\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}], C = [\begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array}], D = [\begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array}] .

Find $[A]_{B}, [B]_{B}, [C]_{B}, [D]_{B} .$

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(c)

Determine if the set ${A, B, C, D}$ is linearly dependent or independent.

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10.

Let $V$ be a vector space of dimension $n$ and let $B$ be a basis for $V .$ Show that the coordinate transformation $T$ from $V$ to $R^{n}$ defined by $T (x) = [x]_{B}$ satisfies $T (0_{V}) = 0,$ where $0_{V}$ is the additive identity in $V .$

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11.

Prove the second property of Theorem 34.3. That is, if a vector space $V$ has a basis of $n$ vectors, then the coordinate mapping $T : V \to R^{n}$ defined by $T (x) = [x]_{B}$ satisfies

T (c u) = c T (u)

for any vector $u$ in $V$ and any scalar $c .$

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12.

Prove Theorem 34.4 by demonstrating that if $V$ is a vector space with ordered basis $B = {v_{1}, v_{2}, \dots, v_{n}},$ then the coordinate mapping $T : V \to R^{n}$ defined by $T (x) = [x]_{B}$ is one-to-one.

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13.

The coordinate transformation $T$ is one-to-one, so it has an inverse $T^{- 1} .$ Let $V$ be an $n$ -dimensional vector space that has a basis $B,$ and let $T : V \to R^{n}$ be the coordinate transformation defined by $T (x) = [x]_{B} .$ Let $S = {u_{1}, u_{2}, \dots, u_{k}}$ be a subset of $V$ and let $R = {[u_{1}]_{B}, [u_{2}]_{B}, \dots, [u_{k}]_{B}}$ in $R^{n} .$

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(a)

Suppose $x$ is in $V$ with $x = x_{1} u_{1} + x_{2} u_{2} + \dots + x_{k} u_{k} .$ Write the vector $[x]_{B}$ as a linear combination of the vectors in $R .$ Explain your reasoning and explain how your result shows that $T$ preserves linear combinations.

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(b)

Now suppose $w$ is in $V$ so that $[w]_{B} = w_{1} [u_{1}]_{B} + w_{2} [u_{2}]_{B} + \dots + w_{k} [u_{k}]_{B} .$ Write the vector $w$ as a linear combination of the vectors in $S .$ Explain your reasoning and explain how your result shows that $T^{- 1}$ preserves linear combinations.

Hint.

Apply $T$ to an appropriate vector and use the fact that $T$ is one-to-one.

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14.

Let $S = {u_{1}, u_{2}, \dots, u_{k}}$ be a subset of an $n$ -dimensional vector space $V$ with basis $B .$ Let $R = {[u_{1}]_{B}, [u_{2}]_{B}, \dots, [u_{k}]_{B}}$ in $R^{n} .$

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(a)

Show that if $S$ is linearly independent in $V,$ then $R$ is linearly independent in $R^{n} .$

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(b)

Is the converse of part (a) true? That is, if $R$ is linearly independent in $R^{n},$ must $S$ be linearly independent in $V ?$ Justify your answer.

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(c)

Repeat parts (a) and (b), replacing “linearly independent” with “linearly dependent”.

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15.

Let $V$ be an $n$ -dimensional vector space with basis $B,$ and let $S = {w_{1}, w_{2}, \dots, w_{k}}$ be a subset of $V$ that spans $V .$ Prove that ${[w_{1}]_{B}, [w_{2}]_{B}, \dots, [w_{k}]_{B}}$ spans $R^{n} .$

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16.

Suppose $B_{1}$ is a basis of a vector space $V$ with $n$ elements and $B_{2}$ is a basis of $W$ with $n$ elements. Show that the map $T_{V W}$ which sends every $x$ in $V$ to the vector $y$ in $W$ such that $[x]_{B_{1}} = [y]_{B_{2}}$ is one-to-one and onto.

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17.

Label each of the following statements as True or False. Provide justification for your response.

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(a) True/False.

The coordinates of a non-zero vector cannot be the same in the coordinate systems defined by two different bases.

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(b) True/False.

The coordinate vector of the zero vector with respect to any basis is always the zero vector.

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(c) True/False.

If $W$ is a $k$ dimensional subset of an $n$ dimensional vector space $V,$ and $B$ is a basis of $W,$ then $[w]_{B}$ is a vector in $R^{n}$ for any $w$ in $W .$

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(d) True/False.

The order of vectors in a basis do not affect the coordinates of vectors with respect to this basis.

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(e) True/False.

If $T$ is a coordinate transformation from a vector space $V$ with basis $B$ to $R^{n},$ then the vector $[x]_{B}$ is unique to the vector $x$ in $V .$

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(f) True/False.

If $T$ is a coordinate transformation from a vector space $V$ with basis $B$ to $R^{n},$ then there is always a vector $x$ in $V$ that maps to any vector $b$ in $R^{n} .$

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(g) True/False.

A coordinate transformation from a vector space $V$ with basis $B$ to $R^{n}$ always maps the additive inverse of a vector $x$ in $V$ to the additive inverse of the vector $[x]_{B}$ in $R^{n} .$

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(h) True/False.

A coordinate transformation provides a unique identification of vectors in an $n$ -dimensional vector space with vectors in $R^{n}$ in a way that preserves the algebraic structure of the spaces.

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(i) True/False.

If the coordinate vector of $x$ in a vector space $V$ is $[\begin{array}{r} 1 \\ - 1 \\ 2 \end{array}],$ then the coordinate vector of $2 x$ is $[\begin{array}{r} 2 \\ - 2 \\ 4 \end{array}] .$

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Subsection Project: Finding Formulas for Sums of Powers

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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

(\binom{n}{k}) = \frac{n!}{k! (n - k)!},

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with

(\binom{n}{k}) = 0

n < k .

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

(\binom{n}{k}) = \frac{n (n - 1) (n - 2) \dots (n - k + 1)}{k!} = Π_{i = 1}^{k} \frac{1}{i} (n - i + 1) .

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With this representation of

(\binom{n}{k}),

we can define a new polynomial in

t

of degree

k

p_{k} (t) = \frac{1}{k!} (t) (t - 1) (t - 2) \dots (t - k + 1)

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with

p_{0} (t) = 1 .

For example,

\begin{aligned} p_{1} (t) & = \frac{1}{1!} t = t \\ p_{2} (t) & = \frac{1}{2} (t) (t - 1) = \frac{1}{2} (t^{2} - t) \\ p_{3} (t) & = \frac{1}{6} (t) (t - 1) (t - 2) = \frac{1}{6} (t^{3} - 3 t^{2} + 2 t) . \end{aligned}

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

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Two facts will make these polynomials useful for our sums.

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Project Activity 34.5.

Our polynomials $p_{k} (t)$ are defined in terms of binomial coefficients. A useful identity will relate sums of binomial coefficients to other binomial coefficients. This identity, called the hockey-stick identity after the way it can be visualized on Pascal's triangle, is as follows:

\sum_{k = 0}^{n - 1} (\binom{k}{r}) = (\binom{n}{r + 1})

for positive integers $r .$ Use the definition of the binomial coefficients and some algebra to verify the hockey-stick identity.

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The second useful fact about our polynomials

p_{k} (t)

is that they form a basis for

P_{n} .

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Project Activity 34.6.

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(a)

Let $k$ be a positive integer. Explain why $p_{k} (0) = p_{k} (1) = p_{k} (2) = \dots = p_{k} (k - 1) = 0$ and $p_{k} (k) = 1 .$

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(b)

Let $P_{n} = {p_{0} (t), p_{1} (t), p_{2} (t), \dots, p_{n} (t)} .$ Show that $P_{n}$ is a basis for $P_{n} .$ (Hint: Let $c_{1},$ $c_{2},$ $\dots,$ $c_{n}$ be scalars and consider the equation

c_{0} p_{0} (t) + c_{1} p_{1} (t) + c_{2} p_{2} (t) + \dots + c_{n} p_{n} (t) = 0 .

Evaluate this equation at $t = 0,$ $t = 1,$ $\dots,$ $t = n$ and use the result of part (a).)

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Now we have the tools we need to derive our formulas. To simplify computations, we will change coordinates to the standard basis

S_{n} = {1, t, t^{2}, \dots, t^{n}}

for

P_{n} .

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We will illustrate the process of deriving formulas for our sums with the sum

\sum_{t = 0}^{n - 1} t .

We want to write

t

as a linear combination of the vectors in

P_{1}

so that we can utilize the hockey-stick identity. In this case, by definition we have

t = p_{1} (t) .

It follows by the hockey-stick identity and the fact that

p_{1} (t) = (\binom{t}{1})

that

\begin{aligned} \sum_{t = 0}^{n - 1} t & = \sum_{t = 0}^{n - 1} p_{1} (t) \\ = \sum_{t = 0}^{n - 1} (\binom{t}{1}) \\ = (\binom{n}{2}) \\ = \frac{1}{2} (n^{2} - n) . \end{aligned}

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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

P_{n} .

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Project Activity 34.7.

Consider the sum $\sum_{t = 0}^{n - 1} t^{2} .$ We want to write $t^{2}$ as a linear combination of $p_{0} (t),$ $p_{1} (t)$ and $p_{2} (t) .$ To do so, we will use the coordinate vectors with respect to $S_{2}$ and do our work in $R^{3} .$

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(a)

Find $[p_{0} (t)]_{S_{2}},$ $[p_{1} (t)]_{S_{2}},$ $[p_{2} (t)]_{S_{2}},$ and $[t^{2}]_{S_{2}} .$

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(b)

Use the coordinate vectors from part (a) to write $t^{2}$ as a linear combination of the vectors in $P_{2} .$

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(c)

Use the result of part (b) and, the hockey-stick identity, and the fact that $p_{k} (t) = (\binom{t}{k})$ to find a formula for $\sum_{t = 0}^{n - 1} t^{2} .$

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

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Project Activity 34.8.

Use the process outlined in Project Activity 34.7 to derive formulas for the following sums.

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(a)

$\sum_{t = 0}^{n - 1} t^{3}$

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(b)

$\sum_{t = 0}^{n - 1} t^{4}$