product, the transpose, the definitions of symmetric and
orthogonal matrices, all need to be modified for complex
numbers. The new definitions coincide with the old when
the vectors and matrices are r
(the steady states) of the following? h .4 .2.6 .8 ik " 1 0
# , h .4 .2.6 .8 ik " 0 1 # , h .4 .2.6 .8 ik . Problems 2329
are about A = SS 1 and A k = S kS 1 23. Diagonalize A
and compute S kS 1 to p
cannot come in negative amounts. Von Neumann asked
for the maximum rate t at which the economy can
expand and still stay nonnegative, meaning that u1 tu0
0. Thus the problem requires u1 tAu1. It is l
infinitesimal steps, but the two theories stay absolutely in
parallel. It is the same analogy between the discrete and
the continuous that appears over and over in
mathematics. A good illustration is
multiplied by i or i: (K 1I)x1 = " i 1 1 i #"y z # = " 0 0
# and x1 = " 1 i # (K 2I)x2 = " i 1 1 i #"y z # = " 0 0 #
and x2 = " 1 i # . The eigenvalues are distinct, even if
imaginary, and the eigenve
A+iB is a Hermitian matrix (A and B are real), show that
A B B A is symmetric. 48. Prove that the inverse of a
Hermitian matrix is again a Hermitian matrix. 49.
Diagonalize this matrix by constructin
+Wn1 ) = Wn 1 W 1 = 0. In the second case, W = w
ji . Every entry of the original F has absolute value 1. The
factor n shrinks the columns of U into unit vectors. The
fundamental identity of the finit
an additional time t to reach e At(e AT u(0). This solution
at time t + T can also be written as . Conclusion: e At
times e AT equals . 36. Write A = 1 1 0 0 in the form
SS 1 . Find e At from SetS 1 .
certainly stable: A = " 0 4 0 1 2 # has eigenvalues 0 and 1
2 . The s are on the main diagonal because A is
triangular. Starting from any u0, and following the rule
uk+1 = Auk , the solution must even
v1 vn i . The determinant of the left-hand side is the
Wronskian. It never becomes zero, because it is the
product of two nonzero determinants. Both matrices on
the right-hand side are invertible. Rem
a complex pair x iy, the tests still succeed. The trace is
their sum 2x (which is < 0) and the determinant is (x + iy)
(x iy) = x 2 + y 2 > 0. Figure 5.2 shows the one stable
quadrant, trace < 0 and d
trace when A = " a b c d# and B = " q r s t# . Deduce that
ABBA = I is impossible (except in infinite dimensions).
10. Suppose A has eigenvalues 1, 2, 4. What is the trace
of A 2 ? What is the determi
+(y0 2z0)(.7) k " 1 3 1 3 # . Those two terms are c1 k 1
x1 +c2 k 2 x2. The factor k 1 = 1 is hidden in the first
term. In the long run, the other factor (.7) k becomes
extremely small. The solution a
the preceding A, use elimination to solve Ax = 0. (b) Show
that the nullspace you just computed is orthogonal to C(A
H) and not to the usual row space C(A T ). The four
fundamental spaces in the compl
know the main fact about a nonnegative matrix like A:
Not only is the largest eigenvalue 1 positive, but so is
the eigenvector x1. Then (I A) 1 has the same
eigenvector, with eigenvalue 1/(11). If 1 e
Change of Basis = Similarity Transformation The similar
matrix B = M1AM is closely connected to A, if we go
back to linear transformations. Remember the key idea:
Every linear transformation is repres
changed now that A is changed to A? But show that v(t)
grows to infinity from v(0) = 30. 310 Chapter 5
Eigenvalues and Eigenvectors 26. The solution to y 00 = 0
is a straight line y = C +Dt. Convert t
the Americas, Asia, and Europe have assets of $4 trillion.
At the start, $2 trillion are in the Americas and $2 trillion
in Europe. Each year 1 2 the American money stays
home, and 1 4 goes to each of
Heisenbergs uncertainty principle comes from
noncommuting matrices, like position P and momentum
Q. Position is symmetric, momentum is skew-symmetric,
and together they satisfy QPPQ = I. The uncertain
14. What are the eigenvalues and frequencies , and
the general solution, of the following equation? d 2u dt 2
= " 5 4 4 5 # u. 15. Solve the second-order equation d
2u dt2 = " 5 1 1 5 # u with u(0) =
each of the following. 5.5 Complex Matrices 323 (a) A
real symmetric matrix. (b) A stable matrix: all solutions to
du/dt = Au approach zero. (c) An orthogonal matrix. (d) A
Markov matrix. (e) A defect
orthogonal) matrices. Therefore we go directly to the
three properties of U that correspond to the earlier
Properties 13 of A. Remember that U has orthonormal
columns: Unitary matrix U HU = I, UUH = I
Their total number v + w is constant. That comes from
adding the two equations on the right-hand side: the
derivative of v+w is zero. A discrete Markov matrix has its
column sums equal to max = 1. A c
equations Just as for difference equations. the
eigenvalues decide how u(t) behaves as t . As long as
A can be diagonalized, there will be n pure exponential
solutions to the differential equation, an
solutions of the special form cet x, where is an
eigenvalue of A and x is its eigenvector. These pure
solutions satisfy the differential equation, since d/dt(cet
x) = A(cet x). (They were our introduc
equation Fk+2 = Fk+1 +Fk . (2) That is the difference
equation. It turns up in a most fantastic variety of
applications, and deserves a book of its own. Leaves grow
in a spiral pattern, and on the app
Consider all 4 by 4 matrices A that are diagonalized by
the same fixed eigenvector matrix S. Show that the As
form a subspace (cA and A1 + A2 have this same S). What
is this subspace when S = I? What
A = " 0 0 1 0# B = " 0 1 0 0 # (use series for e A and e B ).
5.4 Differential Equations and e At 307 6. The higher
order equation y 00 +y = 0 can be written as a first-order
system by introducing the
x1,.,xk correspond to different eigenvalues 1,.,k ,
then those eigenvectors are linearly independent.
Suppose first that k = 2, and that some combination of x1
and x2 produces zero: c1x1 +c2x2 = 0. Mu
dimension n. In the new definition of length, each x 2 j is
replaced by its modulus |x j | 2 : Length squared kxk 2 = |
x1| 2 +|xn| 2 . (4) Example 2. x = " 1 i # and kxk 2 = 2;
y = " 2+i 24i # and ky