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Section 9.2 Discovery guide

Subsection General linear groups

Discovery 9.1.

The group \(\GL_1(\R)\) consists of all invertible \(1 \times 1\) matrices. Demonstrate that \(\GL_1(\R)\) is isomorphic to \(\multunits{\R}\text{,}\) the multiplicative group of non-zero real numbers, by providing an explicit description of an isomorphism from one onto the other.

Discovery 9.2.

For which values of \(n\) is \(\GL_n(\R)\) an Abelian group?

Discovery 9.3.

(a)

Verify that the collection of \(n \times n\) matrices of the form

\begin{equation*} \left[\begin{array}{@{}ccc|c@{}} \amp \amp \amp 0 \\ \amp A \amp \amp \vdots \\ \amp \amp \amp 0 \\ \hline 0 \amp \cdots \amp 0 \amp 1 \\ \end{array}\right]\text{,} \end{equation*}

where \(A\) is a matrix in \(\GL_{n-1}(\R)\text{,}\) is a subgroup of \(\GL_n(\R)\text{.}\)

(b)

Verify that the map

\begin{equation*} A \mapsto \left[\begin{array}{@{}ccc|c@{}} \amp \amp \amp 0 \\ \amp A \amp \amp \vdots \\ \amp \amp \amp 0 \\ \hline 0 \amp \cdots \amp 0 \amp 1 \\ \end{array}\right] \end{equation*}

is an isomorphism of \(\GL_{n-1}(\R)\) onto that subgroup of \(\GL_n(\R)\text{.}\)

(c)

Using Task a as inspiration, come up with another way to create an isomorphism of \(\GL_{n-1}(\R)\) onto a subgroup of \(\GL_n(\R)\text{.}\)

Discovery 9.4.

Consider \(m\) and \(n\) to be fixed choices of positive integers, with \(m < n\text{.}\)

(a)

Consider the collection of \(n \times n\) matrices of the form

\begin{equation*} \left[\begin{array}{@{}c|c@{}} A \amp \zerovec \\ \hline \zerovec \amp I \\ \end{array}\right]\text{,} \end{equation*}

where \(A\) is a matrix in \(\GL_m(\R)\text{,}\) \(I\) is the \((n - m) \times (n - m)\) identity matrix, and the \(\zerovec\) entries represent appropriately-sized (likely non-square) zero matrices. Verify that this collection is a subgroup of \(\GL_n(\R)\text{.}\)

(b)

Verify that the map

\begin{equation*} A \mapsto \left[\begin{array}{@{}c|c@{}} A \amp \zerovec \\ \hline \zerovec \amp I \\ \end{array}\right] \end{equation*}

is an isomorphism of \(\GL_m(\R)\) onto that subgroup of \(\GL_n(\R)\text{.}\)

(c)

Using Task a as inspiration, come up with another way to create an isomorphism of \(\GL_m(\R)\) onto a subgroup of \(\GL_n(\R)\text{.}\) (Actually, there are many, many other ways to do this, but there are at least \(m-n\) other “obvious” ways based on the pattern of Task a.)

Subsection Special linear group

Discovery 9.5.

Use your preferred version of the Subgroup Test to verify that \(\SL_n(\R)\) is a subgroup of \(\GL_n(\R)\text{.}\)

Subsection Orthogonal group

Discovery 9.6.

By definition, a matrix \(A\) is orthogonal when it satisfies

\begin{equation*} \trans{A} A = I \text{.} \end{equation*}

Take the determinant of both sides of this equality, and then use the properties of the determinant to come to the conclusion that there are only two possible values that the determinant of an orthogonal matrix can take.

Now let's use the geometry of \(\R^2\) to try to understand the simplest (but non-trivial) case of orthogonal group: \(\Or_2(\R)\text{.}\) Suppose \(A\) is a matrix in \(\Or_2(\R)\text{.}\) From Fact 9.1.8, the result of computing \(A \vec{e}_1\) must also be on the unit circle. Write

\begin{gather} A \vec{e}_1 = (\cos \theta, \sin \theta) \text{,}\tag{✶} \end{gather}

where \(\theta\) is the angle from the positive \(x\)-axis to the radius through \(A \vec{e}_1\) (counter-clockwise). Then we already know

\begin{align} A = \begin{bmatrix} \cos \theta \amp ? \\ \sin \theta \amp ? \end{bmatrix} \text{.}\tag{✶✶} \end{align}

Discovery 9.7.

Under assumption (✶), what are the possibilities for the result of \(A \vec{e}_2\text{?}\)

Hint.

There are two possibilities. Consider Fact 9.1.8 and Figure 9.1.13.

Your results from Discovery 9.7 give you two different ways to fill in the second column of \(A\) in (✶✶).

Discovery 9.8.

Of the two possibilities for \(A\text{,}\) one represents a rotation of \(\R^2\text{.}\) (Remember that \(A\) permutes \(\R^2\text{.}\))

(a)

Which possibility for \(A\) is a rotation?

Hint.

If the two standard basis vectors are rotated together (without changing the right angle between them), this means the \(xy\)-axes are rotated, and so all of \(\R^2\) is rotated.

(b)

What is the value of the determinant of this rotation matrix?

Discovery 9.9.

Now consider the second possibility for \(A\) (i.e. the one not already considered in Discovery 9.8).

(a)

Plot \(\vec{e}_1\) and \(A \vec{e}_1\) on a set of \(xy\)-axes. Now draw the line through the origin at angle \(\theta/2\) to the positive \(x\)-axis (halfway between \(\vec{e}_1\) and \(A \vec{e}_1\)). If you reflect \(\vec{e}_1\) in this line, where does it land?

(b)

Now plot \(\vec{e}_2\) and \(A \vec{e}_2\) on the diagram you started in Task a. If you reflect \(\vec{e}_2\) in this line, where does it land?

(c)

What geometric transformation name would you assign to this second possibility for \(A\text{?}\)

(d)

What is the value of the determinant of this second possibility for \(A\text{?}\)

So now we have discovered two distinct possibilities for the form of elements in \(\Or_2(\R)\text{,}\) and geometrically these possibilities correspond either to a rotation of \(\R^2\) around the origin by some fixed angle, or a reflection of \(\R^2\) in a line through the origin. But we can be even more precise about how exactly \(\Or_2(\R)\) splits into these two different types. First, having considered the determinant of each of these possibilities, we find that \(\SO_2(\R)\) is precisely the collection of rotations in \(\Or_2(\R)\text{.}\) And second, we can map from rotation to reflection and back again using one fixed choice of reflection matrix.

Discovery 9.10.

(a)

Compute the \(2 \times 2\) orthogonal matrix \(A_{xy}\) corresponding to reflection in the line \(y = x\text{.}\)

Hint.

Remember that \(\theta\) represents the (counter-clockwise) angle between \(\vec{e}_1\) and \(A \vec{e}_1\text{,}\) and for a reflection the line of reflection is at the angle halfway between those two vectors.

(b)

Compute \(A_{xy}^2\text{.}\) Give a geometric reason for this result.

(c)

Determine \(\inv{A}_{xy}\text{.}\) (You can compute it if you like, but Task b tells you the answer.)

Give a geometric reason for this result.

(d)

Verify that if \(A\) is a \(2 \times 2\) rotation matrix, then the result of the product \(A_{xy} A\) is a reflection matrix.

Hint.

Use the determinant instead of trying to figure out the geometry. (This works because \(\Or_2(\R)\) is a group and so is closed under group algebra, and because of the results of our previous explorations of the possibilities for elements of \(\Or_2(\R)\) above.)

(e)

Now do the reverse of Task d: verify that if \(A'\) is a \(2 \times 2\) reflection matrix, then there is some \(2 \times 2\) rotation matrix so that

\begin{equation*} A' = A_{xy} A \text{.} \end{equation*}
Hint.

Again, no need for messing about with geometric arguments and calculation of angles and trig values — instead, use some algebra and the results of Task d combined with Task b and/or Task c.

Discovery 9.10 tells us that the group \(\Or_2(\R)\) can be broken up into two disjoint collections: the subgroup \(\SO_2(\R)\) (i.e. the subgroup of \(2 \times 2\) rotation matrices) and the collection

\begin{equation*} \begin{bmatrix} 0 \amp 1 \\ 1 \amp 0 \end{bmatrix} \cdot \SO_2(\R) \text{,} \end{equation*}

where this notation means “take each element of \(\SO_2(\R)\) and multiply it by \(\left[\begin{smallmatrix} 0 \amp 1 \\ 1 \amp 0 \end{smallmatrix}\right]\text{.}\)” This second subcollection is the collection of \(2 \times 2\) reflection matrices, but note that it is not a subgroup.

Discovery 9.11.

Verify that Discovery 9.10 can be repeated with \(A_{xy}\) replaced by any fixed choice of reflection matrix.

Discovery 9.11 tells us that we can use any fixed choice of \(2 \times 2\) reflection matrix \(A_0\) to create the subcollection of all \(2 \times 2\) reflection matrices as the collection

\begin{equation*} A_0 \cdot \SO_2(\R) \text{.} \end{equation*}

This pattern of “multiply each element in a subgroup by a fixed choice of element that is not in the subgroup” is one that we will see again in a more abstract fashion later in the course.