CH 8
CH 8
150
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1. Eigenvalues are: 3>2 and 3 and the corresponding eigenvectors are [1, 0]T, and [0, 1]T
respectively.
2. This zero matrix, like any square zero matrix, has the eigenvalue 0. The algebraic
multiplicity and geometric multiplicity are both equal to 2, and we can choose
the basis [1 0]T, [0 1]T.
3. Eigenvalues are 0 and ⫺3, and the corresponding eigenvectors are [2>3, 1]T, and
[1>3, 1]T respectively.
4. Eigenvalues are 5 and 0 and eigenvectors [1 2]T and [⫺2 1]T, respectively. The
matrix is symmetric, and for such a matrix it is typical that the eigenvalues are real
and the eigenvectors orthogonal. Also, make the students aware of the fact that 0 can
very well be an eigenvalue.
5. Eigenvalues are 4i and ⫺4i, and the corresponding eigenvectors are [⫺i, 1]T and
[i, 1]T respectively.
6. Eigenvalues are 1 and 3 and eigenvectors [1 0]T and [1 1]T, respectively. Note
that for such a triangular matrix, the main diagonal entries are still the eigenvalues
(as for a diagonal matrix; cf. Prob. 1), but the eigenvectors are no longer orthogonal.
7. The matrix has a repeated eigenvalue of 0 with eigenvectors [1, 0]T, and [0, 0]T.
8. Eigenvalues: a ⫾ 2⫺k; Eigenvectors: [1> 2⫺k, 1]T and [⫺1> 2⫺k, 1]T,
respectively.
9. Eigenvalues are 0.20 ⫾ 0.40, and the eigenvectors are [i, 1]T and [⫺i, 1]T
respectively.
10. The characteristic equation is
(cos u ⫺ l)2 ⫹ sin2 u ⫽ 0.
Solutions (eigenvalues) are l ⫽ cos u ⫾ i sin u. Eigenvectors are obtained from
(l ⫺ cos u)x 1 ⫹ (sin u)x 2 ⫽ (sin u)(⫾ix 1 ⫹ x 2) ⫽ 0,
say, x 1 ⫽ 1, x 2 ⫽ ⫿i.
Note that this matrix represents a rotation through an angle u, and this linear
transformation preserves no real direction in the x 1x 2-plane, as would be the case if
the eigenvectors were positive real. This explains why these vectors must be complex.
11. Eigenvalues: 4, 1, 7, Eigenvectors are: [⫺1>2, 1, 1]T, [1, ⫺1>2, 1]T, [⫺2, ⫺2, 1]T.
12. 3, [1 0 0]T; 4, [5 1 0]T; 1, [7, ⫺4 2]T
13. Repeated eigenvalue 2. Eigenvectors: [2, ⫺2, 1]T, [0, 0, 0]T, [0, 0, 0]T.
14. Develop the characteristic determinant by the second row, obtaining
(12 ⫺ l)[(2 ⫺ l)(4 ⫺ l) ⫹ 1] ⫽ (12 ⫺ l)(l ⫺ 3)2.
Eigenvectors for the eigenvalues 12 and 3 are [0 1 0]T and [⫺1 0 1]T,
respectively, and we get no basis for R3.
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0 0 1 1 0
x 2 ⫽ x 1) indicates that every point in the plane x 2 ⫽ x 1 is mapped onto itself. The
other eigenvalue 0 with eigenvector [1 ⫺1 0]T indicates that any point on the line
x 2 ⫽ ⫺x 1, x 3 ⫽ 0 (which is perpendicular to the plane x 2 ⫽ x 1) is mapped onto the
origin. The student should perhaps make a sketch to see what is going on geometrically.
24. By Theorem 1 in Sec. 7.8 the inverse exists if and only if det A ⫽ 0. On the other
hand, from the product representation
D(l) ⫽ det (A ⫺ lI) ⫽ (⫺1)n(l ⫺ l1)(l ⫺ l2) Á (l ⫺ ln)
of the characteristic polynomial we obtain
det A ⫽ (⫺1)n(⫺l1)(⫺l2) Á (⫺ln) ⫽ l1l2 Á ln.
Hence Aⴚ1 exists if and only if 0 is not an eigenvalue of A.
Furthermore, let l ⫽ 0 be an eigenvalue of A. Then
Ax ⫽ lx.
Multiply this by Aⴚ1 from the left:
Aⴚ1Ax ⫽ lAⴚ1x.
Now divide by l:
1
x ⫽ Aⴚ1x.
l
Biology (Example 3)
Mechanical vibrations (Example 4)
Short Courses. Of course, this section can be omitted, for reasons of time, or one or two
of the examples can be considered quite briefly.
Comments on Content
The examples in this section have been selected from the viewpoint of modest
prerequisites, so that not too much time will be needed to set the scene.
Example 4 illustrates why real matrices can have complex eigenvalues (as
mentioned before, in Sec. 8.1), and why these eigenvalues are physically meaningful.
(For students familiar with systems of ODEs, one can easily pick further examples
from Chap. 4.)
Comments on Problems
Problems 1–12 are similar to the applications shown in the examples of the text.
Problems 13–15 show an interesting application of eigenvalue problems to production,
typical of various other applications of eigenvalue theory in economics included in various
textbooks in economic theory.
1. Eigenvalue and eigenvectors are ⫺1, [⫺1, 1]T and 2, [1, 1]T. The eigenvectors are
orthogonal.
2. Eigenvalues and eigenvectors are 1.6, [1 ⫺1]T and 2.4, [1 1]T. These vectors are
orthogonal, as is typical of a symmetric matrix. Directions are ⫺45° and 45°,
respectively.
3. Eigenvalues 3, ⫺3 and eigenvectors [22, 1]T and [⫺1> 22, 1]T respectively.
4. Extension factors 9 ⫹ 215 ⫽ 13.47 and 9 ⫺ 215 ⫽ 4.53 in the directions given by
[1 2 ⫹ 15]T and [1 2 ⫺ 15]T (76.7° and ⫺13.3°, respectively).
1
6. 2, [1 1]T; 2, [1 ⫺1]T; directions 45° and ⫺45°, respectively.
7. Eigenvector [2.5, 1]T with eigenvalue 1.
8. [1 1 1]T, as could also be seen without calculation because A has row sums equal
to 1, which would not be the case in general.
9. Eigenvector [⫺0.2, ⫺0.4, 1] with eigenvalue 1.
10. Growth rate 3. The characteristic polynomial is 14(x ⫺ 3)(2x ⫹ 5)(2x ⫹ 1) which gives
the remaining two eigenvalues as 2.5 and 0.5. The sum of all the eigenvalues is the
trace which is zero. Note that the growth rate is not that sensitive to the elements of
the matrix. Working with two decimal digits still retains the intrinsic characteristic
of the problem.
11. Growth rate is 4. Characteristic polynomial is (x ⫺ 4)(x ⫹ 1)(x ⫹ 3).
12. Growth rate 1.3748. The other eigenvalues are complex or negative and are not needed.
The sum of all eigenvalues equals the trace, that is, 0, except for a roundoff error.
This 4 ⫻ 4 Leslie matrix corresponds to a classification of the population into four
classes.
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14. A has the same eigenvalues as AT, and AT has row sums 1, so that it has the eigenvalue
1 with eigenvector x ⫽ [1 Á 1]T.
Leontief is a leader in the development and application of quantitative methods in
empirical economical research, using genuine data from the economy of the United
States to provide, in addition to the “closed model” of Prob. 13 (where the producers
consume the whole production), “open models” of various situations of production
and consumption, including import, export, taxes, capital gains and losses, etc. See
W. W. Leontief, The Structure of the American Economy 1919–1939 (Oxford: Oxford
University Press, 1951). H. B. Cheney and P. G. Clark, Interindustry Economics (New
York: Wiley, 1959).
16. This follows by comparing the coefficient of lnⴚ1 in the development of the
characteristic determinant D (l) with that obtained from the product representation.
18. The first statement follows from
Ax ⫽ lx, (kA)x ⫽ k(Ax) ⫽ k(lx) ⫽ (kl)x,
the second by induction and multiplication of Akxj ⫽ lkj xj by A from the left.
20. det (L ⫺ lI) ⫽ ⫺l3 ⫹ l 12l 21l ⫹ l 13l 21l 32 ⫽ 0. Hence l ⫽ 0. If all three eigenvalues
are real, at least one is positive since trace L ⫽ 0. The only other possibility is
l1 ⫽ a ⫹ ib, l2 ⫽ a ⫺ ib, l3 real (except for the numbering of the eigenvalues). Then
l3 ⬎ 0 because
l1l2l3 ⫽ (a 2 ⫹ b 2)l3 ⫽ det L ⫽ l 13l 21l 32 ⬎ 0.
c d c d c d
0 1 1 1
A⫽ with eigenvectors and
⫺1 0 i ⫺i
corresponding to the eigenvalues i and ⫺i, respectively.
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1. Eigenvalues: 3>5 ⫾ 4>5i, Eigenvectors: [i, 1]T and [⫺i, 1]T respectively. Skew-
symmetric and orthogonal.
2. Eigenvalues a ⫾ ib. Symmetric if b ⫽ 0; then the eigenvalues are real. Skew-symmetric
if a ⫽ 0; then the eigenvalues are pure imaginary (or zero). Orthogonal if a 2 ⫹ b 2 ⫽ 1;
then the eigenvalues have an absolute value of 1.
3. Non-orthogonal; skew-symmetric; Eigenvalues 1 ⫾ 4i with eigenvectors [⫺i, 1]T
and [i, 1]T.
4. The characteristic equation is
cos u ⫺ l ⫺sin u
2 2 ⫽ l2 ⫺ (2 cos u) l ⫹ 1 ⫽ 0.
sin u cos u ⫺ l
Hence the eigenvalues are
l ⫽ cos u ⫾ i sin u.
c d.
cos u sin u
⫺sin u cos u
(c) To a rotation of about 36.87°. No limit. For a student unfamiliar with complex
numbers this may require some thought.
(d) Limit 0, approach along some spiral.
(e) The matrix is obtained by using familiar values of cosine and sine,
c d.
13>2 ⫺12
A⫽
1
2 13>2
16. Let Ax ⫽ lx (x ⫽ 0), Ay ⫽ y (y ⫽ 0). Then
lx T ⫽ (Ax)T ⫽ x TAT ⫽ x TA.
Thus
lx Ty ⫽ x TAy ⫽ x Ty ⫽ x Ty.
Hence if l ⫽ , then x Ty ⫽ 0, which proves orthogonality.
18. det A ⫽ det (AT) ⫽ det (⫺A) ⫽ (⫺1)n det A ⫽ ⫺det A ⫽ 0 if n is odd. Hence the
answer is no. For even n ⫽ 2, 4, Á we have
0 1 0 0
c d,
0 1 ⫺1 0 0 0
E U, etc,
⫺1 0 0 0 0 1
0 0 ⫺1 0
20. Yes, for instance,
1
2 13>2 0
D 13>2 ⫺12 0T .
0 0 1
c d; c d; c d.
⫺11
9
1
18 0 ⫺1
2. Â ⫽ l ⫽ ⫺1, y ⫽ x ⫽ Py ⫽
⫺80
9
11
9 1 ⫺4
Similarly, for the second eigenvalue we obtain
c d; c d.
2>9 ⫺1>9
l ⫽ 1, y ⫽ x ⫽ Py ⫽
1 ⫺40>9
4 ⫺2 4
5. ˆ ⫽ D0
A ⫺2 12T;
0 ⫺2 12
l ⫽ 10, y ⫽ [1>3, 1, 1]T, x ⫽ Py ⫽ [1, 1>3, 1]T
l ⫽ 4, y ⫽ [1, 0, 0]T, x ⫽ Py ⫽ [0, 1, 0]T
l ⫽ 0, y ⫽ [2, 6, 1]T, x ⫽ Py ⫽ [6, 2, 1]T
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6. Project. (a) This follows immediately from the product representation of the
characteristic polynomial of A.
n n
(b) C ⫽ AB, c11 ⫽ a a1lbl1, c22 ⫽ a a2lbl2, etc. Now take the sum of these n sums.
l⫽1 l⫽1
Furthermore, trace BA is the sum of
n n
~
c 11 ⫽ a b1mam1, Á , ~
cnn ⫽ a bnmamn,
m⫽1 m⫽1
2
involving the same n terms as those in the double sum of trace AB.
(c) By multiplications from the right and from the left we readily obtain
~
A ⫽ P 2ÂP ⫺2.
(d) Interchange the corresponding eigenvectors (columns) in the matrix X in (5).
⫺1 1
9. Eigenvalues: ⫺2, 6; Matrix of corresponding eigenvectors: C S
1 1
10. The eigenvalues of A are ⫺1 and 1. A matrix of corresponding eigenvectors is
c d.
0 1>2
X⫽
1 1
c d.
⫺2 1
Xⴚ1 ⫽
0 1
Hence diagonalization gives
c d.
⫺1 0
D ⫽ Xⴚ1AX ⫽
0 1
c d.
7 11
X⫽
13 ⫺1
Its inverse is
c d.
1>150 11>150
Xⴚ1 ⫽
13>150 ⫺7>150
Hence diagonalization gives
c d.
10 0
D ⫽ Xⴚ1AX ⫽
0 ⫺5
⫺18
7 6>7 0
25
13. Eigenvalues: 2, 9, 6; Matrix of corresponding eigenvectors: D 7 ⫺6>7 1T
⫺1 0 0
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2 0 ⫺1
X ⫽ D1 1 1T .
2 1 0
Its inverse is
1 1 ⫺1
Xⴚ1 ⫽ D⫺2 ⫺2 3T .
1 2 ⫺2
Hence diagonalization gives
⫺2 0 0
D ⫽ Xⴚ1AX ⫽ D 0 4 0T .
0 0 1
0 ⫺3 ⫺1>2
1 1 1
16. A has eigenvalues ⫺2, 2, 0. A matrix of corresponding eigenvectors is
1 0 ⫺1
X ⫽ D1 0 1T .
0 1 0
Its inverse is
1>2 1>2 0
Xⴚ1 ⫽ D 0 0 1T .
⫺1>2 1>2 0
Diagonalization thus gives
⫺2 0 0
D ⫽ Xⴚ1Ax ⫽ D 0 2 0T .
0 0 0
18. The symmetric coefficient matrix is
c d.
4 3
C⫽
3 ⫺4
It has eigenvalues 5 and ⫺5. Hence the transformed quadratic form is
5y 21 ⫺ 5y 22 ⫽ 10, or y 21 ⫺ y 22 ⫽ 2.
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c d y.
3>10110 ⫺1>10110
x⫽
1>10110 3>10110
20. The symmetric coefficient matrix is
c d.
9 3
C⫽
3 1
Its eigenvalues are 0 and 10. Hence the transformed form is
10y 22 ⫽ 10.
This represents a pair of parallel straight lines
10y 22 ⫽ 10, thus y2 ⫽ ⫾1.
The matrix X whose columns are normalized eigenvectors of C gives the relation
between y and x in the form
c d y.
1 1 3
x⫽
110 ⫺3 1
c d.
4 16
C⫽
6 13
Its eigenvalues are 1 and 16. Hence the transformed form is
y 21 ⫹ 16y 22 ⫽ 16.
This represents an ellipse. The matrix whose columns are normalized eigenvectors of
C gives the relation between y and x in the form
c d y.
1 ⫺2 1
x⫽
15 1 2
24. Transform Q (x) by (9) to the canonical form (10). Since the inverse transform
y ⫽ X⫺1x of (9) exists, there is a one-to-one correspondence between all x ⫽ 0 and
y ⫽ 0. Hence the values of Q (x) for x ⫽ 0 coincide with the values of (10) on the
right. But the latter are obviously controlled by the signs of the eigenvalues in the
three ways stated in the theorem. This completes the proof.
c d
1⫹i 1 0
A⫽
12 0 1
is unitary and has
det A ⫽ i.
Comments on Problems
Complex matrices appear in quantum mechanics; see Prob. 7, etc.
Problems 13–20 give an impression of calculations for complex matrices.
Normal matrices, defined in Prob.18, play an important role in a more extended theory
of complex matrices.
1 ⫺1
3. Non-Hermitian; Eigenvalues: 14 ⫾ i22, and the matrix of eigenvectors is C S
1 1
4. Skew-Hermitian, as well as unitary, eigenvalues i and ⫺i, eigenvectors [1 1]T and
[1 ⫺1]T, respectively.
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D1 0 0T
0 0 1
6. Hermitian. Eigenvalues and eigenvectors are
⫺4, [i ⫺1 ⫺ i 1]T
0, [1 0 i]T
4, [i 1⫹i 1]T.
8. Eigenvectors are as follows. (Multiplication by a complex constant may change them
drastically!)
For A [1 ⫺ 3i 5]T, [1 ⫺ 3i ⫺2]T
For B [2 ⫹ i i]T, [2 ⫹ i ⫺5i]T
For C [1 1]T, [1 ⫺1]T.
9. Skew-Hermitian; x TAx ⫽ 8 ⫺ 12i.
10. The matrix is non-Hermitian.
x TAx ⫽ [3, ⫺2i][⫺4 ⫹ i, 3 ⫹ 6i]T ⫽ ⫺3i
c d.
1 ⫹ 19i 5 ⫹ 3i
AB ⫽
⫺23 ⫹ 10i ⫺1
c d
0 0
i 0
is not normal. A normal matrix that is not Hermitian, skew-Hermitian, or unitary is
obtained if we take a unitary matrix and multiply it by 2 or some other real factor
different from ⫾1.
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⫺1 1
11. Eigenvalues: 1, 2, and the corresponding matrix of eigenvectors is C S
1 1
12. The eigenvalues are ⫺1 and 1. Corresponding eigenvectors are [2 3]T and [1 2]T,
respectively. Note that this basis is not orthogonal.
2>5 2>3
13. Eigenvalues: 11/2, 13/2, and the matrix of eigenvectors is C S
1 1
14. One of the eigenvalues is 9. Its algebraic and geometric multiplicities are 2.
Corresponding linearly independent eigenvectors are [1 0 ⫺2]T and [0 1 2]T.
The other eigenvalue is 4.5. A corresponding eigenvector is [2 ⫺2 1]T.
15. Eigenvalues: ⫾12i, 0, and the matrix of eigenvectors is
1 1 1
16. The eigenvalues of A are ⫺17 and 9. The similar matrix, having the same
eigenvalues, is
c dc d c d
1>2 1>2 9 17 9 0
 ⫽ P⫺1AP ⫽ ⫽
⫺1>2 1>2 9 ⫺17 0 ⫺17
c 27 d
35
2 ⫺35
2
17. Eigenvalues: ⫺5 and 7. Â ⫽
⫺312 2
18. A has the eigenvalues ⫺2, ⫺1, 2. The inverse of P is
1 ⫺8 31
P ⴚ1 ⫽ D0 1 ⫺3T .
0 0 1
The similar matrix Â, having the same eigenvalues, is
1 ⫺8 31 ⫺4 ⫺26 52 ⫺35 ⫺259 345
 ⫽ D0 1 ⫺3T D 0 2 6T ⫽ D 3 23 ⫺27T .
0 0 1 ⫺1 ⫺7 11 ⫺1 ⫺7 11
c d.
⫺1>17 17
X⫽
1 1
Note that the vectors are orthogonal. Its inverse is
c d.
17 289
⫺290 290
Xⴚ1 ⫽
17 1
290 290
Diagonalization gives
c dc d c d.
17 289
⫺290 290 ⫺316
17 442 316 0
⫽
17 1
290 290 316 26 0 26
22. The symmetric coefficient matrix is
c d.
9 ⫺3
C⫽
⫺3 17
Its eigenvalues are 8 and 18; they are both positive real. The transformed form is
8y 21 ⫹ 18y 22 ⫽ 36.
This is the canonical form; there is no y1y2-term. It represents an ellipse. The matrix
X whose columns are normalized eigenvectors of C gives the relationship between y
and x in the form
c d y.
1 3 1
x⫽
110 1 ⫺3
24. The symmetric coefficient matrix is
c d.
6 8
C⫽
8 ⫺6
Its eigenvalues are 10 and ⫺10. The transformed form is
10y 21 ⫺ 10y 22 ⫽ 10(y1 ⫹ y2)(y1 ⫺ y2) ⫽ 0.
It represents two perpendicular straight lines through the origin. The matrix X whose
columns are normalized eigenvectors of C gives the relationship between x and y in
the form
c d y.
2>525 ⫺1>525
x⫽
1>525 2>525