An Inquiry-Based Introduction to Linear Algebra and Applications

Subsection Application: Rotations in 3D

An aircraft in flight, like a plane or the space shuttle, can perform three independent rotations: roll, pitch, and yaw. Roll is a rotation about the axis through the nose and tail of the aircraft, pitch is rotation moving the nose of the aircraft up or down through the axis from wingtip to wingtip, and yaw is the rotation when the nose of the aircraft turns left or right about the axis though the plane from top to bottom. These rotations take place in \(3\)-space and the axes of the rotations change as the aircraft travels through space. To understand how aircraft maneuver, it is important to know about general rotations in space. These are more complicated than rotations in \(2\)-space, and, as we will see later in this section, involve orthogonal sets.

Subsection Introduction

If \(\CB = \{\vv_1, \vv_2, \ldots, \vv_m\}\) is a basis for a subspace \(W\) of \(\R^n\text{,}\) we know that any vector \(\vw\) in \(W\) can be written uniquely as a linear combination of the vectors in \(\CB\text{.}\) In the past, the way we have found the coordinates of \(\vx\) with respect to \(\CB\text{,}\) i.e. the weights needed to write a vector \(\vx\) as a linear combination of the elements in \(\CB\text{,}\) has been to row reduce the matrix \([\vv_1 \ \vv_2 \ \cdots \ \vv_m \ | \ \vx]\) to solve the corresponding system. This can be a cumbersome process, especially if we need to do it many times. This process also forces us to determine all of the weights at once. For certain types of bases, namely the orthogonal and orthonormal bases, there is a much easier way to find individual weights for this linear combination.

Recall that two nonzero vectors \(\vu\) and \(\vv\) in \(\R^n\) are orthogonal if \(\vu \cdot \vv = 0\text{.}\) We can extend this idea to an entire set. For example, the standard basis \(\CS = \{\ve_1, \ve_2, \ve_3\}\) for \(\R^3\) has the property that any two distinct vectors in \(\CS\) are orthogonal to each other. The basis vectors in \(\CS\) make a very nice coordinate system for \(\R^3\text{,}\) where the basis vectors provide the directions for the coordinate axes. We could rotate this standard basis, or multiply any of the vectors in the basis by a nonzero constant, and retain a basis in which all distinct vectors are orthogonal to each other (e.g., \(\{[2 \ 0 \ 0]^{\tr}, [0 \ 3 \ 0]^{\tr}, [0 \ 0 \ 1]^{\tr}\}\)). We define this idea of having all vectors be orthogonal to each other for sets, and then for bases.

Subsection Orthogonal Sets

We defined orthogonal sets in \(\R^n\) and bases of subspaces of \(\R^n\) in Definitions 24.1 and Definition 24.2. We saw that the standard basis in \(\R^3\) is an orthogonal set and an orthogonal basis of \(\R^3\) — there are many other examples as well.

The example from Activity 24.2 suggests that we can have three orthogonal nonzero vectors in \(\R^3\text{,}\) but no more. Orthogonal vectors are, in a sense, as far apart as they can be. So we might expect that there is no linear relationship between orthogonal vectors. The following theorem makes this clear.

Subsection Properties of Orthogonal Bases

Orthogonality is a useful and important property for a basis to have. In Preview Activity 24.1 we saw that if a vector \(\vx\) in the span of an orthogonal basis \(\{\vv_1, \vv_2, \vv_3\}\) could be written as a linear combination of the basis vectors as \(\vx = x_1 \vv_1 + x_2 \vv_2 + x_3 \vv_3\text{,}\) then \(x_1 = \frac{\vx \cdot \vv_1}{\vv_1 \cdot \vv_1}\text{.}\) If we continued that same argument we could show that

We can apply this idea in general to see how the orthogonality of an orthogonal basis allows us to quickly and easily determine the weights to write a given vector as a linear combination of orthogonal basis vectors. To see why, let \(\CB = \{\vv_1, \vv_2, \ldots, \vv_m\}\) be an orthogonal basis for a subspace \(W\) of \(\R^n\) and let \(\vx\) be any vector in \(W\text{.}\) We know that

for some scalars \(x_1\text{,}\) \(x_2\text{,}\) \(\ldots\text{,}\) \(x_m\text{.}\) Let \(1\leq k\leq m\text{.}\) Then, using orthogonality of vectors \(\vv_1, \vv_2, \ldots, \vv_m\text{,}\) we have

So

Thus, we can calculate each weight individually with two simple dot products. We summarize this discussion in the next theorem.

Subsection Orthonormal Bases

The decomposition (24.3) is even simpler if \(\vv_k \cdot \vv_k = 1\) for each \(k\text{,}\) that is, if \(\vv_k\) is a unit vector for each \(k\text{.}\) In this case, the denominators are all 1 and we don't even need to consider them. We have a familiar example of such a basis for \(\R^n\text{,}\) namely the standard basis \(\CS = \{\ve_1, \ve_2, \ldots, \ve_n\}\text{.}\)

Recall that

so the condition \(\vv \cdot \vv = 1\) implies that the vector \(\vv\) has norm 1. An orthogonal basis with this additional condition is a very nice basis and is given a special name.

In other words, an orthonormal basis is an orthogonal basis in which every basis vector is a unit vector. A good question to ask here is how we can construct an orthonormal basis from an orthogonal basis.

In general, we can construct an orthonormal basis \(\{\vu_1, \vu_2, \ldots, \vu_m\}\) from an orthogonal basis \(\CB = \{\vv_1, \vv_2, \ldots, \vv_m\}\) by normalizing each vector in \(\CB\) (that is, dividing each vector by its norm).

Subsection Orthogonal Matrices

We have seen in the diagonalization process that we diagonalize a matrix \(A\) with a matrix \(P\) whose columns are linearly independent eigenvectors of \(A\text{.}\) In general, calculating the inverse of the matrix whose columns are eigenvectors of \(A\) in the diagonalization process can be time consuming, but if the columns form an orthonormal set, then the calculation is very straightforward.

The result of Activity 24.5 is that if the columns of a square matrix \(P\) form an orthonormal set, then \(P^{-1} = P^{\tr}\text{.}\) This makes calculating \(P^{-1}\) very easy. Note, however, that this only works if the columns of \(P\) form an orthonormal basis for \(\Col P\text{.}\) You should also note that if \(P\) is an \(n \times n\) matrix satisfying \(P^{\tr}P = I_n\text{,}\) then the columns of \(P\) must form an orthonormal set. Matrices like this appear quite often and are given a special name.

Orthogonal matrices are useful because they satisfy some special properties. For example, if \(P\) is an orthogonal \(n \times n\) matrix and \(\vx, \vy \in \R^n\text{,}\) then

This property tells us that the matrix transformation \(T\) defined by \(T(\vx) = P\vx\) preserves dot products and, hence, orthogonality. In addition,

so \(||P\vx|| = ||\vx||\text{.}\) This means that \(T\) preserves length. Such a transformation is called an isometry and it is convenient to work with functions that don't expand or contract things. Moreover, if \(\vx\) and \(\vy\) are nonzero vectors, then

Thus \(T\) also preserves angles. Transformations defined by orthogonal matrices are very well behaved transformations. To summarize,

We have discussed orthogonal and orthonormal bases for subspaces of \(\R^n\) in this section. There are reasonable questions that follow, such as

• Can we always find an orthogonal (or orthonormal) basis for any subspace of \(\R^n\text{?}\)

• Given a vector \(\vv\) in \(W\text{,}\) can we find an orthogonal basis of \(W\) that contain \(\vv\text{?}\)

We will answer these questions in subsequent sections.

Subsection Examples

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

Subsection Summary

• A subset \(S\) of \(\R^n\) is an orthogonal set if \(\vu \cdot \vv = 0\) for every pair of distinct vector \(\vu\) and \(\vv\) in \(S\text{.}\)

• Any orthogonal set of nonzero vectors is linearly independent.

• A basis \(\CB\) for a subspace \(W\) of \(\R^n\) is an orthogonal basis if \(\CB\) is also an orthogonal set.

• An orthogonal basis \(\CB\) for a subspace \(W\) of \(\R^n\) is an orthonormal basis if each vector in \(\CB\) has unit length.

• If \(\CB = \{\vv_1, \vv_2, \ldots, \vv_m\}\) is an orthogonal basis for a subspace \(W\) of \(\R^n\) and \(\vx\) is any vector in \(W\text{,}\) then

  \begin{equation*} \vx = \sum_{i=1}^m c_i \vv_i \end{equation*}

  where \(c_i = \frac{\vx \cdot \vv_i}{\vv_i \cdot \vv_i}\text{.}\)

• An \(n \times n\) matrix \(P\) is an orthogonal matrix if \(P^{\tr}P = I_n\text{.}\) Orthogonal matrices are important, in part, because the matrix transformations they define are isometries.

Subsection Project: Understanding Rotations in 3-Space

Recall that a counterclockwise rotation of \(2\)-space around the origin by an angle \(\theta\) is accomplished by left multiplication by the matrix \(\left[ \begin{array}{cr} \cos(\theta)\amp -\sin(\theta) \\ \sin(\theta)\amp \cos(\theta) \end{array} \right]\text{.}\) Notice that the columns of this rotation matrix are orthonormal, so this rotation matrix is an orthogonal matrix. As the next activity shows, rotation matrices in 3D are also orthogonal matrices.

By Project Activity 24.7 we know that the determinant of any rotation matrix is either \(1\) or \(-1\text{.}\) Having a determinant of \(1\) preserves orientation, and we will identify these rotations as being counterclockwise, and we will identify the others with determinant of \(-1\) as being clockwise. We will set the convention that a rotation is always measured counterclockwise (as we did in \(\R^2\)), and so every rotation matrix will have determinant \(1\text{.}\)

Returning to the counterclockwise rotation of \(2\)-space around the origin by an angle \(\theta\) determined by left multiplication by the matrix \(\left[ \begin{array}{cr} \cos(\theta)\amp -\sin(\theta) \\ \sin(\theta)\amp \cos(\theta) \end{array} \right]\text{,}\) we can think of this rotation in \(3\)-space as the rotation that keeps points in the \(xy\) plane in the \(xy\) plane, but rotates these points counterclockwise around the \(z\) axis. In other words, in the standard \(xyz\) coordinate system, with standard basis \(\ve_1\text{,}\) \(\ve_2\text{,}\) \(\ve_3\text{,}\) our rotation matrix \(R\) has the property that \(R \ve_3 = \ve_3\text{.}\) Now \(R \ve_3\) is the third column of \(R\text{,}\) so the third column of \(R\) is \(\ve_3\text{.}\) Similarly, \(R \ve_1\) is the first column of \(R\) and \(R \ve_2\) is the second column of \(R\text{.}\) Since \(R\) is a counterclockwise rotation of the \(xy\) plane space around the origin by an angle \(\theta\) it follows that this rotation is given by the matrix

In this notation in (24.5), the subscript gives the direction of the line fixed by the rotation and the angle provides the counterclockwise rotation in the plane perpendicular to this vector. This vector is called a normal vector for the rotation. Note also that the columns of \(R_{\ve_3}(\theta)\) form an orthogonal set such that each column vector has norm \(1\text{.}\)

This idea describes a general rotation matrix \(R_{\vn}(\theta)\) in 3D by specifying a normal vector \(\vn\) and an angle \(\theta\text{.}\) For example, with roll, a normal vector points from the tail of the aircraft to its tip. It is our goal to understand how we can determine an arbitrary rotation matrix of the form \(R_{\vn}(\theta)\text{.}\) We can accomplish this by using the rotation around the \(z\) axis and change of basis matrices to find rotation matrices around other axes. Let \(\CS = \{\ve_1, \ve_2, \ve_3\}\) be the standard basis for \(\R^3\)

The difference between the two options is that, in the first we have \(\det\left(\underset{\CS \leftarrow \CB_1}{P}\right) = -1\) while \(\det\left(\underset{\CS \leftarrow \CB}{P} \right) = 1\) in the second. Using \(\CB_1\) will give clockwise rotations while \(\CB\) gives counterclockwise rotations (this is the difference between a left hand system and a right hand system). So it is important to ensure that our change of basis matrix is one with determinant \(1\text{.}\)

We do one more example to illustrate the process before tackling the general case.

In the next activity we summarize the general process to find a 3D rotation matrix \(R_{\vn}(\theta)\) for any normal vector \(\vn\text{.}\) There is a GeoGebra applet at geogebra.org/m/n9gbjhfx that allows you to visualize rotation matrices in 3D.