An Inquiry-Based Introduction to Linear Algebra and Applications

Subsection Application: The Tennis Racket Effect

Try an experiment with a tennis racket (or a squash racket, or a ping pong paddle). Let us define a 3D coordinate system with the center of the racket as the origin with the head of the racket lying in the \(xy\)-plane. We let \(\vu_1\) be the vector in the direction of the handle and \(\vu_2\) the perpendicular direction (still lying in the plane defined by the head) as illustrated in Figure 28.1. We then let \(\vu_3\) be a vector perpendicular to the plane of the head. Hold the racket by the handle and spin it to make one rotation around the \(\vu_1\) axis. This is pretty easy. It is also not difficult to throw the racket so that it rotates around the \(\vu_3\text{.}\) Now toss the racket into the air to make one complete rotation around the axis of the vector \(\vu_2\) and catch the handle. Repeat this several times. You should notice that in most instances, the racket will also have made a half rotation around the \(\vu_1\) axis so that the other face of the racket now points up. This is quite different than the rotations around the \(\vu_1\) and \(\vu_3\) axes. A good video that illustrates this behavior can be seen at youtube.com/watch?v=4dqCQqI-Gis.

This effect is a result in classical mechanics that describes the rotational movement of a rigid body in space, called the tennis racket effect (or the Dzhanibekov effect, after the Russian cosmonaut Vladimir Dzhanibekov who discovered the theorem's consequences while in zero gravity in space — you can see an illustration of this in the video at youtube.com/watch?v=L2o9eBl_Gzw). The result is simple to see in practice, but is difficult to intuitively understand why the behavior is different around the intermediate axis. There is a story of a student who asked the famous physicist Richard Feynman if there is any intuitive way to understand the result; Feynman supposedly went into deep thought for about 10 or 15 seconds and answered, “no.” As we will see later in this section, we can understand this effect using the principal axes of a rigid body.

Subsection Introduction

We are familiar with quadratic equations in algebra. Examples of quadratic equations include \(x^2=1\text{,}\) \(x^2+y^2=1\text{,}\) and \(x^2+xy+y^2=3\text{.}\) We don't, however, have to restrict ourselves to two variables. A quadratic equation in \(n\) variables is any equation in which the sum of the exponents in any monomial term is 2. So a quadratic equation in the variables \(x_1\text{,}\) \(x_2\text{,}\) \(\ldots\text{,}\) \(x_n\) is an equation in the form

for some constant \(c\text{.}\) In matrix notation this expression on the left of this equation has the form

where \(\vx = \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right]\) and \(A\) is the \(n \times n\) matrix \(A = [a_{ij}]\text{.}\) For example, if \(A= \left[ \begin{array}{rcr} 1\amp 3\amp -2 \\ -1\amp 1\amp 2 \\ 0\amp 2\amp -2 \end{array} \right]\text{,}\) then we get the quadratic expression \(x_1^2+3x_1x_2-2x_1x_3-x_2x_1+x_2^2+2x_2x_3+2x_3x_2-2x_3^2\text{.}\) We should note here that the terms involving \(x_ix_j\) and \(x_jx_i\) are repeated in our sum, but

and so we could replace \(a_{ij}\) and \(a_{ji}\) both with \(\left(\frac{a_{ij} + a_{ji}}{2}\right)\) without changing the quadratic form. With this alteration in mind, we can then assume that \(A\) is a symmetric matrix. So in the previous example, the symmetric matrix \(A'= \left[ \begin{array}{rcr} 1\amp 1\amp -1 \\ 1\amp 1\amp 2 \\ -1\amp 2\amp -2 \end{array} \right]\) gives the same quadratic expression. This leads to the following definition.

As we show in Exercise 7, the symmetric matrix \(A\) is unique to the quadratic form, so we call the symmetric matrix \(A\) is the matrix of the quadratic form. It is these quadratic forms that we will study in this section.

Subsection Equations Involving Quadratic Forms in \(\R^2\)

When we consider equations of the form \(Q(\vx) = d\text{,}\) where \(Q\) is a quadratic form in \(\R^2\) and \(d\) is a constant, we wind up with old friends like \(x^2+y^2=1\text{,}\) \(2x^2+3y^2=2\text{,}\) or \(x^2-y^2=1\text{.}\) As we saw in Preview Activity 28.1 these equations are relatively easy to recognize. However, when we have cross-product terms, like in \(x^2-xy+y^2=1\text{,}\) it is not so easy to identify the curve the equation represents. If there was a way to eliminate the cross-product term \(xy\) from this form, we might be more easily able to recognize its graph. The discussion in this section will focus on quadratic forms in \(\R^2\text{,}\) but we will see later that the arguments work in any number of dimensions. While working in \(\R^2\) we will use the standard variables \(x\) and \(y\) instead of \(x_1\) and \(x_2\text{.}\)

In general, the equation of the form \(Q(\vx) = d\text{,}\) where \(Q\) is a quadratic form in \(\R^2\) defined by a matrix \(A = \left[ \begin{array}{cc} a\amp b/2\\b/2\amp c \end{array} \right]\) and \(d\) is a constant looks like

The graph of an equation like this is either an ellipse (a circle is a special case of an ellipse), a hyperbola, two non-intersecting lines, a point, or the empty set (see Exercise 5). The quadratic forms do not involve linear terms, so we don't consider the cases of parabolas. One way to see into which category one of these quadratic form equations falls is to write the equation in standard form.

The standard forms for quadratic equations in \(\R^2\) are as follows, where \(a\) and \(b\) are nonzero constants and \(h\) and \(k\) are any constants.

Lines:

\(\ds ax^2 = 1\) or \(\ds ay^2=1\) (\(a > 0\))

Ellipse:

\(\displaystyle \ds \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)

Hyperbola:

\(\ds \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) or \(\ds \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1\)

Preview Activity 28.1 contains the main tool that we need to convert a quadratic form into one of these standard forms. By this we mean that if we have a quadratic form \(Q\) in the variables \(x_1\text{,}\) \(x_2\text{,}\) \(\ldots\text{,}\) \(x_n\text{,}\) we want to find variables \(y_1\text{,}\) \(y_2\text{,}\) \(\ldots\text{,}\) \(y_n\) in terms of \(x_1\text{,}\) \(x_2\text{,}\) \(\ldots\text{,}\) \(x_n\) so that when written in terms of the variables \(y_1\text{,}\) \(y_2\text{,}\) \(\ldots\text{,}\) \(y_n\) the quadratic form \(Q\) contains no cross terms. In other words, we want to find a vector \(\vy = \left[ \begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \end{array} \right]\) so that \(Q(\vx) = \vy^{\tr} D \vy\text{,}\) where \(D\) is a diagonal matrix. Since every real symmetric matrix is orthogonally diagonalizable, we will always be able to find a matrix \(P\) that orthogonally diagonalizes \(A\text{.}\) The details are as follows.

Let \(Q\) be the quadratic form defined by \(Q(\vx) = \vx^{\tr} A \vx\text{,}\) where \(A\) is an \(n \times n\) symmetric matrix. As in Preview Activity 28.1, the fact that \(A\) is symmetric means that we can find an orthogonal matrix \(P = [ \vp_1 \ \vp_2 \ \vp_3 \ \cdots \ \vp_n ]\) whose columns are orthonormal eigenvectors of \(A\) corresponding to eigenvalues \(\lambda_1\text{,}\) \(\lambda_2\text{,}\) \(\ldots\text{,}\) \(\lambda_n\text{,}\) respectively. Letting \(\vy = P^{\tr} \vx\) give us \(\vx = P\vy\) and

where \(D\) is the diagonal matrix whose \(i\)th diagonal entry is \(\lambda_i\text{.}\)

Moreover, the set \(\B = \{\vp_1, \vp_2, \cdots, \vp_n\}\) is an orthonormal basis for \(\R^n\) and so defines a coordinate system for \(\R^n\text{.}\) Note that if \(y = [y_1 \ y_2 \ \cdots \ y_n]^{\tr}\text{,}\) then

Thus, the coordinate vector of \(\vx\) with respect to \(\B\) is \(\vy\text{,}\) or \([\vx]_{\B} = \vy\text{.}\) We summarize in Theorem 28.3.

The columns of the orthogonal matrix \(P\) in the Principal Axis Theorem form an orthogonal basis for \(\R^n\) and are called the principal axes for the quadratic form \(Q\text{.}\) Also, the coordinate vector of \(\vx\) with respect to this basis is \(\vy\text{.}\)

Subsection Classifying Quadratic Forms

If we draw graphs of equations of the type \(z=Q(\vx)\text{,}\) where \(Q\) is a quadratic form, we can see that a quadratic form whose matrix does not have 0 as an eigenvalue can take on all positive values (except at \(\vx=\vzero\)) as shown at left in Figure 28.4, all negative values (except at \(\vx=\vzero\)) as shown in the center of Figure 28.4, or both positive and negative values as depicted at right in Figure 28.4. We can see when these cases happen by analyzing the eigenvalues of the matrix that defines the quadratic form. Let \(A\) be a \(2 \times 2\) symmetric matrix with eigenvalues \(\lambda_1\) and \(\lambda_2\text{,}\) and let \(P\) be a matrix that orthogonally diagonalizes \(A\) so that \(P^{\tr}AP = D = \left[ \begin{array}{cc} \lambda_1 \amp 0 \\ 0 \amp \lambda_2 \end{array} \right]\text{.}\) If we let \(\vy = \left[ \begin{array}{c} w\\z \end{array} \right] = P^{\tr}\vx\text{,}\) then

Then \(Q(\vx) \geq 0\) if all of the eigenvalues of \(A\) are positive (with \(Q(\vx) > 0\) when \(\vx \neq \vzero\)) and \(Q(\vx) \leq 0\) if all of the eigenvalues of \(A\) are negative (with \(Q(\vx) \lt 0\) when \(\vx \neq \vzero\)). If one eigenvalue of \(A\) is positive and the other negative, then \(Q(\vx)\) will take on both positive and negative values. As a result, we classify symmetric matrices (and their corresponding quadratic forms) according to these behaviors.

For example, the quadratic form \(Q(\vx) = x^2+y^2\) at left in Figure 28.4 is positive definite (with repeated eigenvalue 1), the quadratic form \(Q(\vx) = -(x^2+y^2)\) at right in Figure 28.4 is negative definite (repeated eigenvalue \(-1\)), and the hyperbolic paraboloid \(Q(\vx) = x^2-y^2\) in the center of Figure 28.4 is indefinite (eigenvalues \(1\) and \(-1\)).

So we have argued that a quadratic form \(Q(\vx) = \vx^{\tr} A \vx\) is positive definite if \(A\) has all positive eigenvalues, negative definite if \(A\) has all negative eigenvalues, and indefinite if \(A\) has both positive and negative eigenvalues. Similarly, the quadratic form is positive semidefinite if \(A\) has all nonnegative eigenvalues and negative semidefinite if \(A\) has all nonpositive eigenvalues. Positive definite matrices are important, as we discuss in the next section.

Subsection Inner Products

We used the dot product to define lengths of vectors, to measure angles between vectors, and to define orthogonality in \(\R^n\text{.}\) We can generalize the notion of orthogonality by using different types of products called inner products that behave like the dot product.

Preview Activity 28.3 illustrates that there are functions from \(\R^n \times \R^n\) to \(\R\) other than the dot product that satisfy many of the properties in Theorem 23.6. Such functions allow us to broaden our ideas of what right angles look like. These functions are called inner products.

The dot product and the example in Preview Activity 28.3 provide two examples of inner products. The examples below provide two other important inner products on \(\R^n\text{.}\)

• If \(a_1\text{,}\) \(a_2\text{,}\) \(\ldots\text{,}\) \(a_n\) are positive scalars, then

  \begin{equation*} \langle [u_1 \ u_2 \ \cdots \ u_n]^{\tr}, [v_1 \ v_2 \ \cdots \ v_n]^{\tr} \rangle = a_1u_1v_1+a_2u_2v_2+ \cdots + a_nu_nv_n \end{equation*}

  defines an inner product on \(\R^n\text{.}\)

• Every invertible \(n \times n\) matrix \(A\) defines an inner product on \(\R^n\) by

  \begin{equation*} \langle \vu, \vv \rangle = (A\vu) \cdot (A\vv)\text{.} \end{equation*}

As Exercise 8 will demonstrate, every inner product on \(\R^n\) can be written in the form \(\langle \vu, \vv \rangle = \vu^{\tr} A \vv\) for some special type of matrix \(A\text{.}\)

Subsection Examples

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

Subsection Summary

• A quadratic form on \(\R^n\) is a function \(Q\) defined by

  \begin{equation*} Q(\vx) = \vx^{\tr} A \vx \end{equation*}

  for some \(n \times n\) symmetric matrix \(A\text{.}\)

• The Principal Axis Theorem tells us that there is a change of variable \(\vx = P\vy\) that will remove the cross-product terms from a quadratic form and allow us to identify the form and determine the principal axes for the form.

Subsection Project: The Tennis Racket Theorem

If a particle of mass \(m\) and velocity \(v\) is moving in a straight line, its kinetic energy \(KE\) is given by \(KE = \frac{1}{2}mv^2\text{.}\) If, instead, the particle rotates around an axis with angular velocity \(\omega\) (in radians per unit of time), its linear velocity is \(v = r \omega\text{,}\) where \(r\) is the radius of the particle's circular path. Substituting into the kinetic energy formula shows that the kinetic energy of the rotating particle is then \(KE = \frac{1}{2}\left(mr^2\right) \omega^2\text{.}\) The quantity \(mr^2\) is called the moment of inertia of the particle and is denoted by \(I\text{.}\) So \(KE = \frac{1}{2}I\omega^2\) for a rotating particle. Notice that the larger the value of \(r\text{,}\) the larger the inertia. You can imagine this with a figure skater. When a skater spins along their major axis with their arms outstretched, the speed at which they rotate is lower than when they bring their arms into their bodies. The moment of inertia for rotational motion plays a role similar to the mass in linear motion. Essentially, the inertia tells us how resistant the particle is to rotation.

To understand the tennis racket effect, we are interested in rigid bodies as they move through space. Any rigid body in three space has three principal axes about which it likes to spin. These axes are at right angles to each other and pass through the center of mass. Think of enclosing the object in an ellipsoid — the longest axis is the primary axis, the middle axis is the intermediate axis, and the third axis is the third axis. As a rigid body moves through space, it rotates around these axes and there is inertia along each axis. Just like with a tennis racket, if you were to imagine an axle along any of the principal axes and spin the object along that axel, it will either rotate happily with no odd behavior like flipping, or it won't. The former behavior is that of a stable axis and the latter an unstable axis. The Tennis Racket Theorem is a statement about the rotation of the body. Essentially, the Tennis Racket Theorem states that the rotation of a rigid object around its primary and third principal axes is stable, while rotation around its intermediate axis is not. To understand why this is so, we need to return to moments of inertia.

Assume that we have a rigid body moving through space. Euler's (rotation) equation describes the rotation of a rigid body with respect to the body's principal axes of inertia. Assume that \(I_1\text{,}\) \(I_2\text{,}\) and \(I_3\) are the moments of inertia around the primary, intermediate, and third principal axes with \(I_1 > I_2 > I_3\text{.}\) Also assume that \(\omega_1\text{,}\) \(\omega_2\text{,}\) and \(\omega_3\) are the components of the angular velocity along each axis. When there is no torque applied, using a principal orthogonal coordinates, Euler's equation tells us that

(The dots indicate a derivative with respect to time, which is common notation in physics.) We will use Euler's equations to understand the Tennis Racket Theorem.

Equation (28.10) shows that \(\omega_2\) is is bounded, so that any slight perturbations in angular velocity have a limited effect on \(\omega_2\text{.}\) A similar argument can be made for \(\omega_3\text{.}\) This implies that the rotation around the principal axes is stable — slight changes in angular velocity have limited effects on the rotations around the other axes.

We can make a similar argument for rotation around the third principal axes.

Now the issue is why is rotation around the second principal axis different.