Download MA 723: Theory of Matrices with Applications Homework 2

MA 723: Theory of Matrices with Applications
Homework 2
February 11, 2017
Instructions: You are encouraged to collaborate with other students. However you must write
your own solutions and list the names of collaborators. Explain the steps of your solution clearly
and succinctly. Only two problems (randomly chosen) will be graded. Correct answers without
derivation or explanation will be awarded zero points.
Graded
1 ) (HoJ) Let A be nonsingular and let x, y ∈ Cn . Prove the Cauchy determinant identity
det(A + xy ∗ ) = det(A)(1 + y ∗ A−1 x).
See Horn and Johnson Example 1.3.24.
2 ) (HoJ)
1. Using Schur’s theorem show that for every square matrix A
n
X
|λi |2 ≤ trace (AA∗ ),
i=1
with equality iff T is diagonal. For which class of matrices does equality hold?
2. Let c = {max |aij |1 ≤ i, j ≤ n}. Show that
| det(A)| ≤ cn nn/2 .
Hint: Use the formula for determinants in terms of eigenvalues, the arithmetic-geometric
inequality and the previous question.
1. From Schur’s theorem A = U T U ∗ , the diagonals of T contain the eigenvalues. Therefore,
U ∗ AA∗ U = T T ∗ =
n
X
i=1
|λi |2 +
n X
X
|tij |2 .
i=1 j>i
Pn
2
Then trace (AA∗ ) = trace (U ∗ AA∗ U ) ≥
i=1 |λi | . Clearly equality holds only if T is
diagonal; this holds true for normal matrices.
1
2. The determinant is the product of eigenvalues, and using AM-GM inequality
!n
n
X
1
| det(A)|2 = |λ1 . . . λn |2 ≤
|λ2 |
.
n i=1 i
Using the result of part 1,
| det(A)|2 ≤

n
n
n
X
1
1
trace (AA∗ )
=
|aij |2  ≤ (cn2 /n)n .
n
n i,j=1
Taking square roots, we have the desired result.
3 ) (HoJ) Let (λ, x) be an eigenpair of A and let the remaining eigenvalues of A be λ2 , . . . , λn . Using
Schur form of A, show that the eigenvalues of A + xv ∗ are λ + x∗ v, λ2 , . . . , λn .
See Horn and Johnson Theorem 2.4.10.1.
4 ) (HoJ) Suppose A ∈ Cn×n is diagonalizable.
1. Prove that rank (A) is equal to the number of nonzero eigenvalues.
2. Prove that rank (A) = rank (Ak ) for all k = 1, 2, . . .
3. Prove that A is nilpotent iff A = 0.
4. If trace (A) = 0 then rank (A) 6= 1.
Let A = XΛX −1 We prove each item individually.
1. The rank is invariant to similarity. Therefore, rank (A) = rank (Λ). The result follows.
2. From the formula Ak = XΛX −1 , we have
rank (Ak ) = rank (Λk ) = rank (Λ) = rank (A).
3. If A = 0, then clearly it is nilpotent for k = 1. Suppose A 6= 0; but Ak = 0 for some k 6= 1.
We have
λ ∈ σ(A) ⇔ λk ∈ σ(Ak ).
Therefore, if A is nilpotent λk = 0 implies λ = 0. Since all the eigenvalues are zero, the
result follows from part (a).
Pn
4. Since trace (A) = i=1 λi ; either all the eigenvalues are zero, or multiple eigenvalues add
to zero. Having only one nonzero eigenvalue is not possible.
5 ) A spring-mass system with no friction can be modeled as the system of differential equations.
du
0 −1
a
= Au
A=
u(0) =
.
1 0
b
dt
1. Using diagonalization, derive the following formula for eAt .
cos t − sin t
eAt =
.
sin t cos t
2. Show that eAt is an orthogonal matrix.
Page 2
3. Using part (2), or otherwise, show that ku(t)k2 remains constant with time. What is this
constant equal to?
This model has the following interpretation: in the absence of friction, the energy of the
system doesn’t change.
4. Show that if A ∈ Rn×n is Skew-Symmetric, then eA is orthogonal.
1
. Computing the matrix exponential using the
±i
diagonalization formula, gives the desired result.
1. Eigenvalues are ±i with eigenvectors
2. This can be verified by direct multiplication.
3. The energy
cos t − sin t a a p
= a2 + b2 .
=
ku(t)k2 = sin t cos t
b b The second step follows because of the unitary invariance of the 2-norm.
4. Using the infinite series representation, with A = −AT
A A2
A3
+
+
+ . . . )T
1!
2!
3!
A A2
A3
=I − +
−
+ · · · = e−A .
1!
2!
3!
(eA )T = (I +
Therefore, (eA )T eA = e−A eA = I. Therefore, eA is orthogonal.
6 ) (Inspired by Strang, GoT) Suppose there is an epidemic in which every month half of those who
are Night’s watchmen become Free Folk, and a quarter of those who are Free Folk become White
Walkers. Let their populations at month k be denoted by nk , fk and wk . The recurrence can be
expressed as
 



wk
wk+1
1 1/4 0
 fk+1  = 0 3/4 1/2  fk  .
yk+1 = Ayk
0 0 1/2
nk
nk+1
1. Find all the eigenvalues and eigenvectors of A.
2. Suppose, initially, there are a thousand each of Night’s watchmen, Free Folk and White
Walkers, what happens to their population yk as k → ∞.
1. Since A is triangular, it has eigenvalues 1, 3/4, 1/2. Eigenvectors
     
1
−1
1
0 ,  1  , −2 .
0
0
1
2. The solution is yk = Ak x0 , i.e.

 k
1
1 −1 1
yk = 0 1 −2 
0 0
1
0.75k

1
 0
0
0.5k
−1
1
0
−1 

1
1000
−2 1000 .
1
1000
Only the eigenvector corresponding to 1, contributes to the solution in the infinite limit.
T
Finally y∞ = 3000 0 0 .
Page 3
7 ) Meyer 7.2.14.
8 ) Meyer 7.1.20 and 7.2.15.
9 ) Meyer 7.3.12.
Project
For those of you interested in the project option, do the following:
• Indicate your interest when you turn in the homework.
• Setup an appointment (during office hours, preferably) with the instructor to discuss potential
topics.
• Learn LATEX, as a starter, here is a resource http://catalog.lib.ncsu.edu/record/NCSU2512039
Notes and additional questions (no need to turn them in)
1 ) Diagonalizable matrices A and B satisfy AB = BA iff they are simultaneously diagonalizable.
See Meyer Exercise 7.2.16 and HoJ Theorem 1.3.12.
2 ) Toeplitz matrices have constant entries along each diagonal; for tridiagonal Toeplitz matrices, we
can explicitly compute the eigenvalues and eigenvectors. See Meyer Example 7.2.5. A special case
of Toeplitz matrices are circulant matrices, which have very interesting properties. (See Meyer
7.2.20)
Due date: Thursday, February 9, 2017.
Page 4