Week 9 Activity

Section 9.4 Week 9 Activity

Subsection 9.4.1 Dihedral Groups

Activity 9.4.1.

Make a multiplication table for the dihedral group \(D_5\text{.}\) (Label the rotations \(R_i\text{;}\) the index \(i\) indicates how many fifth of a full turn for the rotation. Label the reflections \(F_i\text{;}\) the index indicates reflection over a line through the \(i\) vertex, counting counterclockwise. Then start calculating the compositions using vertex opertaions. One you have done a number of these, you can start looking for patterns in the multiplication table and potentially use those patterns to finish the table.)

Solution.

There are many multiplications to calculate. I recommend calculating matrices by vertex operations. If I number the vertices of the pentagon counterclockwise from 1 to 5, then \(R_2\) (rotation by \(\frac{2}{5}\) of a turn) can be represented by this movement of vertices.

\begin{align*} 1 \amp \mapsto 3\\ 2 \amp \mapsto 4\\ 3 \amp \mapsto 5\\ 4 \amp \mapsto 1\\ 5 \amp \mapsto 2 \end{align*}

Reflection \(F_3\) (reflection over the line through vertex 3 and the middle of the opposite edge)) can be represented by this action on the vertices.

\begin{align*} 1 \amp \mapsto 5\\ 2 \amp \mapsto 4\\ 3 \amp \mapsto 3\\ 4 \amp \mapsto 2\\ 5 \amp \mapsto 1 \end{align*}

To calculate \(R_2 \circ F_3\text{,}\) I can use these vertex changes. In composition, I do the right transformation first. Therefore, starting at vertex 1, \(F_3\) send it to vertex 5, and then \(R_2\) sends vertex 5 to vertex 2. Therefore, the whole composition sends vertex 1 to vertex 2. I calculate the movement of the other four vertices similarly.

\begin{align*} 1 \amp \mapsto 2\\ 2 \amp \mapsto 1\\ 3 \amp \mapsto 5\\ 4 \amp \mapsto 4\\ 5 \amp \mapsto 3 \end{align*}

Then I have to identify this transformation, since is still should be something in the group. It cannot be a rotation, since vertex 4 is fixed. The only rotation that fixes vertex 4 is \(F_4\text{.}\) Therefore, I have \(R_2 \circ F_3 = F_4\text{.}\) Note that the multiplication is not commutative. If I did \(F_3 \circ R_2\text{,}\) the result would be these vertex transformations.

\begin{align*} 1 \amp \mapsto 3\\ 2 \amp \mapsto 2\\ 3 \amp \mapsto 1\\ 4 \amp \mapsto 5\\ 5 \amp \mapsto 4 \end{align*}

This is the reflection \(F_2\text{,}\) so \(F_3 \circ R_2 = F_2\text{.}\) This is just a demonstration calculation: you’ll have to many of these to produce the whole table. Between calculating compositions and using the patterns of the table to fill in remaining entries, you should end up with this multiplication for \(D_5\text{.}\)

Table 9.4.1.

\(\circ\)	\(\Id\)	\(R_1\)	\(R_2\)	\(R_3\)	\(R_4\)	\(F_1\)	\(F_2\)	\(F_3\)	\(F_4\)	\(F_5\)
\(\Id\)	\(\Id\)	\(R_1\)	\(R_2\)	\(R_3\)	\(R_4\)	\(F_1\)	\(F_2\)	\(F_3\)	\(F_4\)	\(F_5\)
\(R_1\)	\(R_1\)	\(R_2\)	\(R_3\)	\(R_4\)	\(\Id\)	\(F_4\)	\(F_5\)	\(F_1\)	\(F_2\)	\(F_3\)
\(R_2\)	\(R_2\)	\(R_3\)	\(R_4\)	\(\Id\)	\(R_1\)	\(F_2\)	\(F_3\)	\(F_4\)	\(F_1\)	\(F_2\)
\(R_3\)	\(R_3\)	\(R_4\)	\(\Id\)	\(R_1\)	\(R_2\)	\(F_5\)	\(F_1\)	\(F_2\)	\(F_3\)	\(F_4\)
\(R_4\)	\(R_4\)	\(\Id\)	\(R_1\)	\(R_2\)	\(R_3\)	\(F_3\)	\(F_4\)	\(F_5\)	\(F_1\)	\(F_2\)
\(F_1\)	\(F_1\)	\(F_3\)	\(F_5\)	\(F_2\)	\(F_4\)	\(\Id\)	\(R_3\)	\(R_1\)	\(R_4\)	\(R_2\)
\(F_2\)	\(F_2\)	\(F_4\)	\(F_1\)	\(F_3\)	\(F_5\)	\(R_2\)	\(\Id\)	\(R_3\)	\(R_1\)	\(R_4\)
\(F_3\)	\(F_3\)	\(F_5\)	\(F_2\)	\(F_4\)	\(F_1\)	\(R_4\)	\(R_2\)	\(\Id\)	\(R_3\)	\(R_1\)
\(F_4\)	\(F_4\)	\(F_1\)	\(F_3\)	\(F_5\)	\(F_2\)	\(R_1\)	\(R_4\)	\(R_2\)	\(\Id\)	\(R_3\)
\(F_5\)	\(F_5\)	\(F_2\)	\(F_4\)	\(F_1\)	\(F_3\)	\(R_3\)	\(R_1\)	\(R_4\)	\(R_2\)	\(\Id\)

Subsection 9.4.2 Linear Algebra of Polynomials

Activity 9.4.2.

Write \(5x^4 + 2x^3\) as a linear combination of \(x^4 - x^3 + 7x\) and \(x^3 -5x\text{.}\) (Write the linear combination with unknown coefficients and work with the polynomials to try to figure out what the two unknown coefficients should be.)

Solution.

I need to find two constants for this equation.

\begin{equation*} 5x^4 + 2x^3 = a(x^4 - x^3 + 7x) + b(x^3 - 5x) \end{equation*}

The only power of \(x^4\) in either of the two polynomials is in the first. Therefore, the only way to get 5x^4 in the result is for \(a=5\text{.}\) Let me put that value in and see what the situation is.

\begin{align*} 5x^4 + 2x^3 \amp = 5(x^4 - x^3 + 7x) + b(x^3 - 5x) = \\ \amp = 5x^4 + (-5 + b)x^3 + (35 - 5b)x \end{align*}

The coefficient of \(x^3\) needs to be 2. The only way to do this is to make \(b = 7\text{.}\) Let me check if this works.

\begin{align*} 5x^4 + 2x^3 \amp = 5(x^4 - x^3 + 7x) + (-5 + 7)x^3 + (35 - 5(7))x \\ \amp = 5x^4 + 2x^3 + 0x = 5x^4 + 2x^3 \end{align*}

This works.

Activity 9.4.3.

Write \(7x^5 - 7x^2 + 14\) as a linear combination of \(4x^6 - x^3 + 1\text{,}\) \(x^5 + x^3\) and \(x^3 + x^2 - 2\text{.}\) (Write the linear combination with unknown coefficients and work with the polynomials to try to figure out what the two unknown coefficients should be.)

Solution.

I need to find three constants for this equation.

\begin{equation*} 7x^5 - 7x^2 + 14 = a(4x^6 - x^3 + 1) + b(x^5 + x^3) + c(x^3 + x^2 - 2) \end{equation*}

There is no \(x^6\) in the polynomial I want to build, not in any of the other terms. Therefore, \(a=0\) is the only possibility to get ride of the \(x^6\text{.}\)

\begin{equation*} 7x^5 - 7x^2 + 14 = b(x^5 + x^3) + c(x^3 + x^2 - 2) \end{equation*}

Then to match the power of \(x^5\text{,}\) I need \(b=7\text{.}\) Let me put that in and see what it looks like.

\begin{equation*} 7x^5 - 7x^2 + 14 = 7x^5 + (7 + c)x^3 + cx^2 - 2c \end{equation*}

Then to get \(0\) as the coefficient of \(x^3\text{,}\) I need c = -7

\begin{equation*} 7x^5 - 7x^2 + 14 = 7x^5 + (7 - 7)x^3 - 7x^2 + 14 = 7x^5 - 7x^2 + 14 \end{equation*}

The remaining coefficients all work out, so \(a = 0\text{,}\) \(b = 7\) and \(c = -7\) works.

Activity 9.4.4.

What is the dimension of \(\Span \{ x^2 - 4, x, 4, x-1, x^2 + 4x + 3 \}\text{.}\) (You can do this algebraically, but it’s better to try to argue from the standard basis of polynomials that I talked about in the notes and the video.)

Solution.

The dimension is the number of linear independent polynomials that I need to make this span. I can use the fact, from the notes and the videos, that the monomials are enough to span the polynomials. I have \(4\text{,}\) \(x\) and \(x^2 - 4\text{,}\) which gets me the first three monoials (I can get \(x^2\) by subtracting \(4\) from \(x^2 - 4\text{.}\)) Between the first three monomials, nothing else is needed to make any degree 2 or less polynomials. Therefore, this span has dimension 3.

Activity 9.4.5.

What is the dimension of the set of all polynomials with degree 6 or less?

Solution.

Again, I’m going to use the monomials from the video. To get degree 6 polynomials, I need these monomials: \(\{ 1, x, x^2, x^3, x^4, x^5, x^6 \}\text{.}\) These are all linearly independent and their are seven of them. Therefore, the dimension of all polynomials with degree 6 or less is 7.

Activity 9.4.6.

What’s the dimension of the set of all odd-degree polynomials?

Solution.

Since the odd degree are unbounded, this degree can be as high as I want. It includes all the monomials \(\{x, x^3, x^5, x^7 \ldots \}\text{.}\) Since there are infinitely many of these monomials and they are all linearly independent, the dimension of the odd-degree polynomials is infinite.

Activity 9.4.7.

If the set \(\{x^6, x^4, x^3, x^2, 4x^6 + 3x^4 - 9x^3 - 4x^2 - 15\}\) linearly independent?

Solution.

The first four terms are different monomials, so they are independent. The real question here is the last polynomial. It has all the same monomials as the first four. The only different is the constant, \(-15\text{.}\) Does that make it independent?

It does make the last polynomials independent. Using the first first, I can build \(4x^6 + 3x^4 - 9x^3 - 4x^2\text{,}\) but there is no way that I can make a linear combination of the monomials and get the \(-15\text{.}\) I only get multiples of the powers -- linear combinations cannot add another constant. Therefore, the fifth polynomials is independent, and this whole set is a linearly independent set.

Subsection 9.4.3 Conceptual Review Questions

What is a group? What is a matrix group? What is a dihedral group?
What does it mean to make a multiplication table?
What is an abstract vector spaces?
How to the definitions and ideas of linear algebra extend from vectors to other environments?
Why do polynomials, series and functions form abstract vector spaces?

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