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# Classroom Tips and Techniques: Locus of Eigenvalues

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Classroom Tips and Techniques: Locus of Eigenvalues

Robert J. Lopez

Emeritus Professor of Mathematics and Maple Fellow

Maplesoft

Monagan's MapleTech Article

Given that the earlier editions of [3] are dated 1980, 1986, 1990, and 1994, it must be obvious that Mike Monagan's MapleTech article was based on one of the first three editions of Steve Leon's text. Mike's article used matrices that differed slightly from those in Exercise 4, page 360, in [3].

The following is a paraphrase of the exercise, and consequently, of the MapleTech article.

 If , examine the locus of its eigenvalues as  varies linearly from  to 5.

Solution

The eigenvalues of  are  whereas the eigenvalues of  are . Because  is not symmetric, its eigenvalues can be, and actually do become, complex. Indeed, the eigenvalues of  are . Figure 1 is a graph of the real values of these eigenvalues; Figure 2 shows the eigenvalues in the complex plane. In each figure, , with  drawn in black; and , in red.

 > evs := [2+(1/2)*s+(1/2)*sqrt(s^2-8*s-8), 2+(1/2)*s-(1/2)*sqrt(s^2-8*s-8)]: plot(evs,s=-5..5,color=[black,red]);

 >

Figure 1   Eigenvalues of  in the real plane

 > evs := [2+(1/2)*s+(1/2)*sqrt(s^2-8*s-8), 2+(1/2)*s-(1/2)*sqrt(s^2-8*s-8)]: plots:-complexplot(evs,s=-5..5,color=[black,red],scaling=constrained,thickness=2);

 >

Figure 2   Eigenvalues of  in the complex plane

Figures 3 and 4 repeat Figures 1 and 2, respectively, but for . Figure 3 makes it clear that the analytic expressions giving  are discontinuous as real functions, but not as complex-valued functions. On the other hand, Figure 4 shows that in the complex plane, the loci of  do not have continuously turning tangents. Although  will be symmetric in the remaining examples in this article, Figures 3 and 4 portend some of the difficulties that will be encountered even for a restricted class of matrices.

 > evs := [2+(1/2)*s+(1/2)*sqrt(s^2-8*s-8), 2+(1/2)*s-(1/2)*sqrt(s^2-8*s-8)]: plot(evs,s=-5..15,color=[black,red]);

 >

Figure 3   Eigenvalues of  in the real plane

 > evs := [2+(1/2)*s+(1/2)*sqrt(s^2-8*s-8), 2+(1/2)*s-(1/2)*sqrt(s^2-8*s-8)]: plots:-complexplot(evs,s=-5..15,color=[black,red],scaling=constrained,thickness=2);

 >

Figure 4   Eigenvalues of  in the complex plane

Figure 5 contains an animation of the loci traced by  for . From this animation, or from an explicit calculation of the appropriate limits, the following can be deduced.

and

The characteristic polynomial for  is

and its discriminant, , has zeros .

 > evS := [2+(1/2)*S+(1/2)*sqrt(S^2-8*S-8), 2+(1/2)*S-(1/2)*sqrt(S^2-8*S-8)]: plots:-animate(plots:-complexplot,[evS,S=-5..s,color=[black,red],scaling=constrained,thickness=2],s=-5..15,frames=41,digits=3);

 >

Figure 5   Animation of  for

Hence, the loci of  bifurcate at  and . But the real question is, do the closed-form expressions  each define the locus of an eigenvalue (resulting in the black and red curves in Figure 4), or are the loci of the eigenvalues the connected curves in Figure 3?

Example 1

 Find the loci of the eigenvalues of  for .

Solution

The eigenvalues of the symmetric  are

Loci are graphed in Figure 6, where  is in black, and  is in red.

It might seem from Figure 6 that symmetry removes many of the difficulties posed by complex eigenvalues. However, for the 2 × 2 matrix  in Example 2,  are equal, so the loci of the eigenvalues will have a point in common. In the present example, the loci for  are separate and have continuously turning tangents.

 > evs1:=[1-(1/2)*s+(1/2)*sqrt(265*s^2-312*s+288), 1-(1/2)*s-(1/2)*sqrt(265*s^2-312*s+288)]: plot(evs1,s=0..1,color=[black,red]);

 >

Figure 6   Loci of eigenvalues ,

Example 2

 Find the loci of the eigenvalues of  for .

Solution

The eigenvalues of the symmetric  are

Loci are graphed in Figure 7, where  is in black, and  is in red.

Although the loci in Figure 7 intersect, each expression for an eigenvalue generates a unique locus with continuously turning tangent. Figure 7 raises the hope that perhaps tracking an eigenvalue from  to  might be a tractable task.

But alas, even though Example 3 might reinforce this belief, Examples 4 - 6 will prove this hope to be a chimera.

 > evs2:=[4*sqrt(5)*s-sqrt(5)-5*s+2, -4*sqrt(5)*s+sqrt(5)-5*s+2]: plot(evs2,s=0..1,color=[black,red]);

 >

Figure 7   Intersecting loci for

Example 3

 Find the loci of the eigenvalues of  for .

Solution

The eigenvalues , are obtained exactly with Maple's Eigenvectors command. The return is a list of length nearly 2000, and which would take two and a half pages to print.

Figure 8 contains a graph of the loci of the three eigenvalues, colored black, red, and green, respectively. The graph is drawn with increased precision; at standard precision, roundoff generates small imaginary parts that cause small gaps in the loci.

The loci in Figure 8 are separate and distinct, all with continuously turning tangents. For this  it is clearly a simple task to trace the eigenvalues from  to .

 > P:=Matrix([[-9+10*s,-4+5*s,1-8*s],[-4+5*s,6-9*s,-3+6*s],[1-8*s,-3+6*s, -3+9*s]]): evs3:=LinearAlgebra:-Eigenvalues(P,output=list): Digits:=25: plot(evs3,s=0..1,color=[black,red,green]); Digits:=10:

 >

Figure 8   Loci of

Example 4

 Find the loci of the eigenvalues of  for .

Solution

 • The eigenvalues of  are . Hence, the loci of the eigenvalues , and  have the point  in common. Just what this does to the loci remains to be seen.

 • The characteristic equation defines  implicitly. Maple's implicitplot command applied to this equation produces Figure 9 in which the curves threrefore represent the loci of the eigenvalues of .

 > CP:=x^3-(-11+4*s)*x^2-(1104*s^2-472*s+81)*x-10688*s^3+6192*s^2-1476*s-235: plots:-implicitplot(CP,s=0..1,x=-25..35,labels=[s,typeset(lambda)],gridrefine=5);

 >

Figure 9   Loci defined implicitly by the characteristic polynomial

 • In Figure 10, the loci are graphs of the exact eigenvalues obtained via Maple's Eigenvalues command.

 • The eigenvalues of  are approximately , and  and the loci emanating from these initial points are colored black, red, and green, respectively. The eigenvalues of  are approximately , and .

 • The red and green curves, each defined by an exact expression, do not have continuously turning tangents.

 > P:=Matrix([[3-8*s, 2+16*s, 4-4*s], [2+16*s, -9+20*s, 8-24*s], [4-4*s, 8-24*s, -5-8*s]]): Q:=LinearAlgebra:-Eigenvalues(P,output=list): plot(Q,s=0..1,color=[black,red,green],labels=[s,typeset(lambda)]);

 >

Figure 10   Loci via graph of exact expressions for the eigenvalues

The closed-form expressions for , are continuous, but not . (This can be established analytically by the calculations in Table 1.) So either the loci of the eigenvalues are defined by the closed-form expressions  and therefore do not have continuously turning tangents, or the loci are smooth curves and are only piecewise-defined by the analytic expressions  whose graphs appear in Figure 10. A precise definition of the locus of eigenvalues of a real, symmetric, matrix  is required.

 =  = =  = Table 1   Calculations showing that , do not define  curves

Example 5

 For the matrix  in Example 4, obtain the equivalent of Figure 10 and Table 1, using only numeric calculations.

Solution

Initializations

 • Define the matrix .

 • Obtain the characteristic polynomial.
 • Define  as a function of .

Numeric determination of loci of eigenvalues

 • The function  returns a list of numerically computed eigenvalues for each given value of the parameter .

 • The list  contains 101 equispaced values of .
 • Each  is a list of eigenvalues computed at .

 • Constructed from the lists  and , each  is a uniquely colored graph of the jth numerically calculated eigenvalue.

 • Figure 11 assembles the graphs , into a single graph via Maple's display command; it shows that the red and green curves share the common point .

 Figure 11   Numerically determined loci

Slopes on either side of  can be calculated numerically from  computed implicitly from the characteristic polynomial.

 =

Recall that  in  must be replaced by the appropriate eigenvalue  which is available only through numeric calculation via the function . The limiting process used in Table 1 can't be applied here; the requisite numeric calculations are summarized in Table 2.

 =  = =  = Table 2   Numeric calculation of slopes along the loci on either side of

Example 6

 Find the loci of the eigenvalues of  for .

Solution

Since  is a 5 × 5 matrix, only numeric techniques can be used to find its eigenvalues. However, note that the matrix has been chosen so that the eigenvalues of  are 1, 5, 10, 10, and 15.

In Table 3 the matrix  is defined, and the characteristic polynomial is defined as the function .

 Table 3   The matrix  and the characteristic polynomial as the function

The characteristic equation, (here, ) defines the loci of the eigenvalues of  implicitly. Figure 12 implements this insight via Maple's implicitplot command applied to the characteristic equation.

 Figure 12   Loci of eigenvalues defined implicitly by the characteristic equation

Table 4 shows the calculations needed to solve for the eigenvalues numerically, and to construct the separate loci based on these numeric data.

 • The function  returns a list of numerically computed eigenvalues for each given value of the parameter .

 • The list  contains 101 equispaced values of .
 • Each  is a list of eigenvalues computed at .

 • Constructed from the lists  and , each  is a uniquely colored graph of the jth numerically calculated eigenvalue.

Table 4   Numeric construction of the loci of eigenvalues

Figure 13 assembles the graphs , into a single graph via Maple's display command; it shows that the green and blue curves share the common point .

 Figure 13   Numerically calculated loci of eigenvalues

For the green and blue curves in Figure 13, slopes on either side of  can be calculated numerically from  computed implicitly from the characteristic polynomial. Table 5 contains the relevant calculations.

 • Obtain  implicitly with Maple's implicitdiff command.

 • Increase the number of working digits and define values of  on either side of .

Evaluate  on either side of  along the green and blue curves

=

=

=

=

Table 5   Numeric evaluation of slopes along the green and blue curves in Figure 13

There are no closed-form expressions for the eigenvalues of this 5 × 5 matrix . The eigenvalues are computed numerically by Maple's Eigenvalues or fsolve commands, each of which return a sorted list of eigenvalues. There is no user-control of this sort, but even if there were, what sorting rule could be invoked across an eigenvalue with algebraic multiplicity greater than 1? It would seem that the only way to define a unique locus of eigenvalues is to require that it be of class , that is, that it have a continuously turning tangent.

References

 [1] Mathematical Thoughts on the Root Locus, Robert J. Lopez, Tips & Techniques, Maple Reporter, July, 2013. [2] Using Computer Algebra to Help Understand the Nature of Eigenvalues and Eigenvectors, Michael Monagan, MapleTech, Issue 9, Spring 1993, Birkhäuser. [3] Linear Algebra with Applications, 5th ed., Steven J. Leon, 1998, Prentice Hall.

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