Physics 116A
Solutions to Homework Set #7
Winter 2012
1. Boas, p. 141, problem 3.9–5. Show that the product AAT is a symmetric
matrix.
Using eq. (9.10) on p. 139 of Boas, (AB)T = B TAT for any two matrices A and
B. Hence,
(AAT )T = (AT )T AT = AAT ,
(1)
where we have used the fact∗ that (AT )T = A for any matrix A. Eq. (1) implies that
AAT is a symmetric matrix, since by definition a symmetric matrix is equal to its
transpose [cf. the table at the top of p. 138 of Boas].
2. Boas, p. 142, problem 3.9–17.
(a) Show that if A and B are symmetric, then AB is not symmetric unless A and
B commute.
If A and B are symmetric, then A = AT and B = B T . We now examine
(AB)T = B T AT = BA ,
after using the fact that A and B are symmetric matrices. We conclude that (AB)T =
AB if and only if AB = BA. That is, AB is not symmetric unless A and B commute.
(b) Show that a product of orthogonal matrices is orthogonal.
Consider orthogonal matrices Q1 and Q2 . By definition [cf. the table at the top
−1
T
T
of p. 138 of Boas], we have Q−1
1 = Q1 and Q2 = Q2 . We now compute
−1
T T
T
(Q1 Q2 )−1 = Q−1
2 Q1 = Q2 Q1 = (Q1 Q2 ) ,
(2)
after using the fact that Q1 and Q2 are orthogonal. In deriving eq. (2), we have used
the following properties of the inverse and the transpose
(AB)−1 = B −1 A−1 ,
and (AB)T = B T AT ,
for any pair of matrices A and B. Thus, we have shown that (Q1 Q2 )−1 = (Q1 Q2 )T ,
which implies that Q1 Q2 is orthogonal.
(c) Show that if A and B are Hermitian, then AB is not Hermitian unless A and
B commute.
∗
The transpose of a matrix interchanges the rows and columns. Thus, if one performs the
transpose operation twice, the original matrix is recovered.
1
If A and B are Hermitian, then A = A† and B = B † . We now examine
(AB)† = B † A† = BA ,
(3)
after using the fact that A and B are Hermitian matrices. In deriving eq. (3), we
have used the fact that:
(AB)† = (AB)T = (A B)T − B T A T = B † A† .
(4)
We conclude that (AB)† = AB if and only if AB = BA. That is, AB is not Hermitian
unless A and B commute.
(d) Show that a product of unitary matrices is unitary.
Consider unitary matrices U1 and U2 . By definition [cf. the table at the top of
p. 138 of Boas], we have U1−1 = U1† and U2−1 = U2† . We now compute
(U1 U2 )−1 = U2−1 U1−1 = U2† U1† = (U1 U2 )† ,
after using the fact that U1 and U2 are unitary and employing the property of the Hermitian conjugation given in eq. (4). Thus, we have shown that (U1 U2 )−1 = (U1 U2 )† ,
which implies that U1 U2 is orthogonal.
3. Boas, p. 142, problem 3.9–19(c). If S is a symmetric matrix and A is an
antisymmetric matrix, show that Tr(SA) = 0.
Given any matrix M = [aij ], its transpose is defined as M T = [aji ]. In particular,
the diagonal elements (corresponding to i = j) of M and M T coincide. Since the
trace of a matrix is the sum of its diagonal elements, it follows that Tr M T = Tr M .
Consider M = SA, where S = S T is a symmetric matrix and A = −AT is an
antisymmetric matrix. Then,
Tr(SA) = Tr(SA)T = Tr(AT S T ) = −Tr(AS) = −Tr(SA) ,
(5)
where we have used eq. (9.10) on p. 139 of Boas, which states that (AB)T = B T AT ,
and eq. (9.13) of Boas, which states that the trace of a product of matrices is not
changed by cyclic permutation [in particular, Tr(AB) = Tr(BA)]. In the penultimate
step in eq. (5), we used S = S T and A = −AT by virtue of the fact that S is symmetric
and A is antisymmetric. That is, eq. (5) indicates that:
Tr(SA) = −Tr(SA) ,
which can only be true if
Tr(SA) = 0 .
2
5. Boas, p. 171, problem 3.12–4. Find the equations of the following conic,
3x2 + 8xy − 3y 2 = 8 ,
(6)
relative to the principal axes.
In matrix form, eq. (6) can be written as:
3
4
x
(x y)
= 8.
4 −3
y
I could work out the eigenvalues by solving the characteristic equation. But, in this
case I can work them out by inspection by noting that for the matrix
3
4
M=
,
4 −3
we have
λ1 λ2 = det M = −25 .
λ1 + λ2 = Tr M = 0 ,
It immediately follows that the two eigenvalues are λ1 = 5 and λ2 = −5. Next, we
compute the eigenvectors.
3
4
x
x
=5
4 −3
y
y
yields one independent relation, x = 2y. Thus, the normalized eigenvector is
1 2
x
.
=√
y λ=5
5 1
Since M is a real symmetric matrix, the two eigenvectors are orthogonal. It follows
that the second normalized eigenvector is:
1 −1
x
.
=√
y λ=−5
2
5
The two eigenvectors form the columns of the diagonalizing matrix,
1 2 −1
.
C=√
2
5 1
(7)
Since the eigenvectors making up the columns of C are real orthonormal vectors, it
follows that C is a real orthogonal matrix, which satisfies C −1 = C T . As a check, we
make sure that C −1 M C is diagonal.
1
1
2 1
3
4
2 −1
2 1
10
5
5
0
−1
C MC =
=
=
.
4 −3
1
2
5 −10
0 −5
5 −1 2
5 −1 2
3
Following eq. (12.3) on p. 162 of Boas,
0
x
x
,
=C
y0
y
where (x0
y 0 ) are the principal axes. Hence, using C T = C −1 , it follows that:
0
0
x
x
x
0
0
T
0
0
−1
(x y) M
= (x y ) C M C
= (x y ) C M C
y
y0
y0
= (x
0
0
5
0
x
= 5(x0 2 − y 0 2 ) ,
y)
y0
0 −5
0
(8)
That is, relative to the principal axes, eq. (6) takes the form
5(x0 2 − y 0 2 ) = 8
6. Boas, p. 172, problem 3.12–16. Find the characteristic frequencies and the
characteristic modes of vibration for the system of masses and springs as in Figure
12.1 on p. 165 of Boas, for the following array: 4k, m, 2k, m, k.
Following the discussion of Example 3 on pp. 165–166 of Boas, the total potential
energy is
V = 21 (4k)x2 + 12 (2k)(x − y)2 + 21 ky 2 = k 3x2 − 2xy + 23 y 2 .
The equations of motion are:
(
mẍ = −∂V /∂x = −6kx + 2ky ,
mÿ = −∂V /∂y = 2kx − 3ky ,
(9)
where ẍ ≡ d2 x/dt2 and ÿ ≡ d2 y/dt2 . Assuming solutions of the form x = x0 eiωt and
y = y0 eiωt , it follows that
ẍ = −ω 2 x
and
ÿ = −ω 2 y .
Substituting these results into eq. (9) yields:
(
−mω 2 x = −∂V /∂x = −6kx + 2ky ,
−mω 2 y = −∂V /∂y = 2kx − 3ky .
In matrix form, these equations are:
6 −2
x
x
=λ
,
−2
3
y
y
4
with
λ≡
mω 2
.
k
(10)
But the coefficients matrix above is precisely the matrix whose eigenvalue problem
was solved in problem 6 on this homework set. Thus, the eigenvalues are λ = 7 and
λ = 2 with corresponding normalized eigenvectors
1 1
1 −2
x
x
and
=√
.
=√
1
y λ=2
y λ=7
5
5 2
Using eq. (10), ω = (λk/m)1/2 . Thus, there are two characteristic frequencies,
r
r
7k
2k
ω1 =
and
ω2 =
.
m
m
The corresponding characteristic modes of vibrations are identified by the unnormalized eigenvectors,
x
−2
x
1
= c1
and
= c2
,
y ω1
1
y ω2
2
where c1 and c2 are non-zero constants.
7. Boas, p. 172, problem 3.15–14. Multiply the following matrices to find the
resulting transformation
From the first matrix we have:
√ 0 1
x
1
x
3
√
=
0
y
y
2 − 3 1
and we also have
x00
y 00
1
=
2
√ 0 −1
3
x
√
y0
− 3 −1
which yields:
x00
y 00
1
=
4
√ √ −1
3
1
3
x
√
√
y
− 3 −1
− 3 1
which gives us equations:
00 1 −4 0
x
x
=
0
−4
y
y 00
4
or the equations x” =-x and y”=-y
8. Boas, p. 172, problem 3.15–17. Multiply the following matrices to find the
resulting transformation
5
We do the same thing as in the above
to larger expressions, skipping a few steps
we have:
 00 

x
2
1 2
 y 00  = 1  −1 −2 2
9
z 00
−2 2 1
problem, but it’s now more annoying due
and writing down just the combined term

 
2 1
2
x
  1 2 −2   y 
2 −2 −1
z
multiplying out these matrices yields:

 
 00 
9 0
0
x
x
 y 00  = 1  0 −9 0   y 
9
0 0 −9
z
z 00
or the expression x” = x, y” = -y, z” = -z.
9. Boas, p. 172, problem 3.15–25. Find the C Matrix which diagonalizes the
following matrix. Then find C−1 and show that C−1 MC = D.
1 0
3 −2
To find the C matrix, we first determine the eigenvalues from the characteristic
equation:
(1 − λ)(−2 − λ) = λ2 + λ − 2 = 0
which yields:
λ = {−2, 1}
first using the eigenvalue λ = -2, we obtain the matrix:
3 0
3 0
which gives us 3x = 0, and thus a solution x=0, y=1, or a vector {0, 1}. From
the eigenvalue λ = 1, we obtain:
0 0
3 −3
which gives us an equation x - y = 0 or the eigenvector
eigenvectors we can assemble the C matrix
1
1 √0
√
2
2 1
6
√1 {1,
2
1}. So from these
the determinant of this matrix is 1 (taking into account the extra
the inverse of this matrix is:
√
2 0
−1 1
√
2 outside), so
We can prove that this diagonalizes the M matrix by calculating:
√
√
1
1
1 √0
1 0
2
0√
2 0
√
=√
3 −2
2
1
0 −2 2
2 −1 1
2
which are the eigenvalues we were looking for.
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