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 real. We have listed these
changes in a table at the end
(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 prove this formula for A k : A = "
3 2 2 3# has A k = 1
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 like the
Perron-Frobenius theorem, with A on the other s
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 compound interest,
when the time step gets shorter. Sup
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 eigenvectors are independent. They
go into the columns of S: S
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 constructing its eigenvalue
matrix and its eigenvector matrix S: A
+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 finite Fourier transform is U
HU = I. Thus U is a unitary ma
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 . 37. If A 2 = A, show that
the infinite series produces
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 eventually approach zero:
u0 = " 0 1 # , u1 = " 4 1 2 # , u
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. Remark. Not all
differential equations come to us as a fir
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 determinant > 0. It also shows
the parabolic boundary li
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 determinant of (A 1 ) T ? 11. If the
eigenvalues of A are 1, 1
+(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 approaches a limiting state
u = (y,z): Steady state " y
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 complex case are N(A) and
C(A) as before, and then N(A H) an
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 exceeds 1,
that last number is negative. The matrix (I A
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 represented by a matrix.
The matrix depends on the choice of
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 to a matrix equation: d
dt " y y 0 # = " 0 1 0 0#"y y 0
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 Asia and Europe. For Asia
and Europe, 1 2 stays home a
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 uncertainty
principle follows directly from the Schwarz inequali
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) = " 1 0 # and u 0 (0) =
" 0 0 # . 16. In most application
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 defective matrix
(nondiagonalizable). (f) A singular matrix.
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, and U H = U
1 . This leads directly to Property 10 ,
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 continuous Markov
matrix, for differential equations, ha
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, and any specific
solution u(t) is some combination u(t) =
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 introduction to eigenvalues
at the start of the chapter.) In th
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 apple or oak you find five
growths for every two turns aro
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 is its dimension? 38.
Suppose A 2 = A. On the left side
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 velocity y 0 as another
unknown: d dt " y y 0 # = " y
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. Multiplying by A,
we find c11x1 +c22x2 = 0. Subtracting 2
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 kyk 2 = 25. For real vectors there was
a close connection