4 More About the Exponential Linear Transformation
Section 6.4 More About the Exponential Linear Transformation
In the case when there the eigenvalues are distinct, embedded in the calculations utilizing (6.3.28) and (6.3.32) is a somewhat more elegant way to calculate the exponential matrix. The essential fact that allows this more elegant formula to be obtained is a result known as the Lagrange Interpolation Formula. Lagrange interpolation was mentioned in Exercise 2.4.2 and, later, in Example 2.6.3. The essentials of the derivation of this formula begins with a polynomial of degree
N 1 that we shall write simply
2 N 1 fx
0 1 x 2 x N 1 x
The idea is to calculate the coefficients 0 , 1 , 2 ,..., N 1 from knowledge of the values fx 1 , fx 2 ,..., fx N at N distinct points xx 1 , 2 ,..., x N . If (6.4.1) is evaluated at these points, the
coefficients 0 , 1 , 2 ,..., N 1 are the solution of
1 x 1 x 1 0 fx 1
2 N 1
1 x 2 x 2 x 2
fx 1 2
3 x 3 x 3 2 fx 2
1 x N x N x N N 1 fx N
If we were actually to calculate the coefficients we are led again to the solution of an equation of the form (6.3.33). In this case, however, the polynomial is formally rearranged such that the polynomial (6.4.1) is written
2 N 1 fx
0 1 x 2 x N 1 x L j xfx j (6.4.3)
where the quantities L j x , for j 1, 2,..., N , are N 1 degree polynomials to be determined. One
obvious property that follows from (6.4.3) is
1 if j k
L j x k
0 if j k
These functions are solutions of
Chap. 6
• ADDITIONAL TOPICS EIGENVALUE PROBLEMS
1 1 1 1 Lx 1 1
x 1 x 2 x 3 x N L 2 x x
x 1 x 2 x 3 x N Lx 3 x
x 1 x 2 x 3 x N L N x x
Equation (6.4.5) follows from the multiplication of (6.4.2) by the matrix
Lx 1 L 2 x Lx 3 L N x , making use of (6.4.3) to eliminate L j xfx j from j 1
the result and forcing the result to hold as an identity in the coefficients 0 , 1 , 2 ,..., N 1 . Given
what we know about the determinant of the Vandermondian matrix, equation (6.3.34), and Cremers
rule, equation (1.11.6), we can write the solution for Lx 1 , for example, as
Lx 1
with similar formulas for the other functions L j x , for j 2,..., N . The two determinants in (6.4.6) can be evaluated by the formula (6.3.34) to yield
k 1 x x 2 x x 3 x x N
Lx 1 N
x 1 x 2 x 1 x 3 x 1 x N
Sec. 6.4 • More About the Exponential Linear Transformation
The general expression for the functions L j x , for j 1, 2,..., N follows by a similar argument and is
L j x N
Equations (6.4.8) and (6.4.3) 2 combine to yield the Lagrange Interpolation Formula
k j fx
L j xfx j fx j N
Equation (6.4.9) is an identity for the N 1 degree polynomial (6.4.1). Given the duality
between polynomials and polynomials of linear transformations, we can apply (6.4.9) to the polynomial (6.3.28) and write
Equation (6.4.10) is a special case of a result known as Sylvester’s Theorem. 9 Equation (6.3.37)
3 ,a
result that was derived without the use of (6.4.10), is of the same form. The point is that the results in Section 6.3 contain results like (6.4.10). They are simply not manipulated into the form (6.4.10)
9 See, for example, Elementary Matrices by Frazier, Duncan and Collar, Cambridge University Press, 1938, Section 3.9.
10 Sometimes equation (6.2.17) is used to write (6.4.10) in the equivalent form
A j adj A j I
Chap. 6
• ADDITIONAL TOPICS EIGENVALUE PROBLEMS
As explained, (6.4.10) assumes that the eigenvalues are distinct. We shall not give the generalization of (6.4.10) to the multiple eigenvalue case. 11 It is useful to note in passing that in the
special case where a linear transformation has a single eigenvalue , the exponential matrix turns out to be
e e I A I A I A I
A I (6.4.11)
N 1!
Equation (6.3.46) is a special case of (6.4.11).
Exercises
6.4.1 In this exercise, we shall develop the modification of Sylvester’s Theorem, equation (6.4.10),
in the case where N 3 that is appropriate for the case where the characteristic polynomial takes
the form
2 f
det A I 1 2 (6.4.12)
In other words, the first eigenvalue has algebraic multiplicity of two and the other eigenvalue has algebraic multiplicity of one. The first step is to find a generalization of the Lagrange interpolation
formula to replace (6.4.9). If fx is the quadratic
2 fx
0 1 x 2 x (6.4.13)
and we are given two values fx 1 and fx 2 for distinct values of x 1 and x 2 . In order to complete the interpolation based upon (6.4.13), we are also given the slope, f ' x 1 , at x 1 . These
conditions and (6.4.13) yield the following three equations for the unknown coefficients and 1 , 2
1 x 1 0 fx 1
0 1 2 x 1
1 f '
x 1 (6.4.14)
3 x 3 2 fx 3
Equation (6.4.14) is an example of a confluent Vondermonde matrix (transposed). Show that
11 See, Elementary Matrices by Frazier, Duncan and Collar, Cambridge University Press, 1938, Section 3.10.
Sec. 6.4 • More About the Exponential Linear Transformation
x 1 x 2 x 1 x 2 x 2 x 3
As with the derivation of (6.4.9), we are interested in how the polynomial (6.4.13) can be rearranged into the form
fx N 1 xfx 1 M 1 xf x 1 N 2 xfx 2 (6.4.16)
where the quadratics N 1 x , M 1 x and N 2 x need to be determined. Show that these three
quadratics are given by
N 1 x 1 x x x
2 x 1 x x 2 x
M 1 x
2 3 x 1 x 2
2 x 3 x 3 x
Finally, show that for the linear transformation AV : V whose characteristic polynomial is
given by (6.4.12) the exponential linear transformation is given by
A 1 A 1 I 1 A 1 I A 2 I A 1 I
2 e e 2 2 (6.4.18)
6.4.2 Adapt the results of Example 5.3.2 to the notation used in (6.4.18) and show that the exponential matrix is given by
Chap. 6
• ADDITIONAL TOPICS EIGENVALUE PROBLEMS
3 3 3 3 3 3 e 2 e e e e e 13.4070 6.6786 6.6786
A 1 3 3 3 3 3 3
e e e e 2 e e e 6.6786 13.4070 6.6786 (6.4.19)
e e e e e 2 e 6.6786 6.6786 13.4070
This exercise is essentially the same as Example 6.3.6 except that here a basis has been selected to represent the linear transformation that was unspecified in Example 6.3.6. Also, the method of solution was built around the method introduced in Section 6.4.
6.4.3 Repeat Exercise 6.4.1 for the case where all of the eigenvalues are identical. In particular, show that
e e I A 1 I A 1 I (6.4.20)
Equation (6.4.20) confirms (6.4.11) for the special case N 3 .
Sec. 6.5 • Application of the Exponential Linear Transformation