Explain why it is typically impossible to compute the upper triangular matrix whose existence
is guaranteed by Schur’s theorem.
Now recall the QR factorization of Theorem 13.2.9 on Page 873. The QR algorithm is a
technique which does compute the upper triangular matrix in Schur’s theorem sometimes.
There is much more to the QR algorithm than will be presented here. In fact, what I am about
to show you is not the way it is done in practice. One first obtains what is called a Hessenburg
matrix for which the algorithm will work better. However, the idea is as follows. Start with A
an n×n matrix having real eigenvalues. Form A = QR where Q is orthogonal and R is upper
triangular. (Right triangular.) This can be done using the technique of Theorem 13.2.9 using
Householder matrices. Next take A1≡ RQ. Show that A = QA1QT. In other words these two
matrices, A,A1 are similar. Explain why they have the same eigenvalues. Continue by letting
A1 play the role of A. Thus the algorithm is of the form An = QRn and An+1 = Rn+1Q.
Explain why A = QnAnQnT for some Qn orthogonal. Thus An is a sequence of matrices each
similar to A. The remarkable thing is that often these matrices converge to an upper triangular
matrix T and A = QTQT for some orthogonal matrix, the limit of the Qn where the limit
means the entries converge. Then the process computes the upper triangular Schur form of the
matrix A. Thus the eigenvalues of A appear on the diagonal of T. You will see approximately
what these are as the process continues.
↑Try the QR algorithm on
( )
− 1 − 2
6 6
which has eigenvalues 3 and 2. I suggest you use a computer algebra system to do the
computations.
↑Now try the QR algorithm on
( 0 − 1 )
2 0
Show that the algorithm cannot converge for this example. Hint: Try a few iterations of the
algorithm. Use a computer algebra system if you like.
↑Show the two matrices A ≡
( )
0 − 1
4 0
and B ≡
( )
0 − 2
2 0
are similar; that is there exists a
matrix S such that A = S−1BS but there is no orthogonal matrix Q such that QTBQ = A.
Show the QR algorithm does converge for the matrix B although it fails to do so for
A.
Let F be an m × n matrix. Show that F∗F has all real eigenvalues and furthermore, they are all
nonnegative.
If A is a real n × n matrix and λ is a complex eigenvalue λ = a + ib,b≠0, of A having eigenvector
z + iw, show that w≠0.
Suppose A = QTDQ where Q is an orthogonal matrix and all the matrices are real. Also D is a
diagonal matrix. Show that A must be symmetric.
Suppose A is an n × n matrix and there exists a unitary matrix U such that
A = U ∗DU
where D is a diagonal matrix. Explain why A must be normal.
If A is Hermitian, show that det
(A )
must be real.
Show that every unitary matrix preserves distance. That is, if U is unitary,
|Ux| = |x|.
Show that if a matrix does preserve distances, then it must be unitary.
↑Show that a complex normal matrix A is unitary if and only if its eigenvalues have magnitude equal
to 1.
Suppose A is an n × n matrix which is diagonally dominant. Recall this means
∑
|aij| < |aii|
j⁄=i
show A−1 must exist.
Give some disks in the complex plane whose union contains all the eigenvalues of the
matrix
( )
1 + 2i 4 2
|( 0 i 3 |)
5 6 7
Show a square matrix is invertible if and only if it has no zero eigenvalues.
Using Schur’s theorem, show the trace of an n×n matrix equals the sum of the eigenvalues and the
determinant of an n × n matrix is the product of the eigenvalues.
Using Schur’s theorem, show that if A is any complex n × n matrix having eigenvalues
{λi}
listed
according to multiplicity, then ∑i,j
|Aij|
2≥∑i=1n
|λi|
2. Show that equality holds if and only if A is
normal. This inequality is called Schur’s inequality. [25]
change the variables so that in the new variables there are no mixed terms, terms involving xy,yz etc.
The eigenvalues of the matrix which you will work with are −
17
2-
,
19
2-
,−1.
Here is a function of three variables.
f (x,y,z) = − x2 + 2xy + 2xz − y2 + 2yz − z2 +x
change the variables so that in the new variables there are no mixed terms, terms involving xy,yz
etc.
and classify them as local minima, local maxima or saddle
points.
Let f
(x,y)
= 3x4− 24x2 + 48 − yx2 + 4y. Find and classify the critical points using the second
derivative test.
Let f
(x,y)
= 3x4− 5x2 + 2 − y2x2 + y2. Find and classify the critical points using the second
derivative test.
Let f
(x,y)
= 5x4− 7x2− 2 − 3y2x2 + 11y2− 4y4. Find and classify the critical points using the
second derivative test.
Let f
(x,y,z)
= −2x4− 3yx2 + 3x2 + 5x2z + 3y2− 6y + 3 − 3zy + 3z + z2. Find and classify the
critical points using the second derivative test.
Let f
(x,y,z)
= 3yx2− 3x2−x2z −y2 + 2y − 1 + 3zy − 3z − 3z2. Find and classify the critical points
using the second derivative test.
Let Q be orthogonal. Find the possible values of det
(Q )
.
Let U be unitary. Find the possible values of det
(U )
.
If a matrix is nonzero can it have only zero for eigenvalues?
A matrix A is called nilpotent if Ak = 0 for some positive integer k. Suppose A is a nilpotent matrix.
Show it has only 0 for an eigenvalue.
If A is a nonzero nilpotent matrix, show it must be defective.
Suppose A is a nondefective n × n matrix and its eigenvalues are all either 0 or 1. Show A2 = A.
Could you say anything interesting if the eigenvalues were all either 0,1,or −1? By DeMoivre’s
theorem, an nth root of unity is of the form
( (2kπ-) ( 2kπ) )
cos n +isin n
Could you generalize the sort of thing just described to get An = A? Hint:Since A is nondefective,
there exists S such that S−1AS = D where D is a diagonal matrix.
This and the following problems will present most of a differential equations course. Most of the
explanations are given. You fill in any details needed. To begin with, consider the scalar initial value
problem
′
y = ay,y(t0) = y0
When a is real, show the unique solution to this problem is y = y0ea
(t− t0)
. Next suppose
y′ = (a+ ib)y,y(t0) = y0 (14.6)
(14.6)
where y
(t)
= u
(t)
+ iv
(t)
. Show there exists a unique solution and it is given by y
is called the Wronskian and this theorem is sometimes called the Wronskian
alternative.)
Hint:Suppose first the general solution is of the form Φ
(t)
c where c is an arbitrary constant vector
in Fn. You need to verify Φ
(t)
−1 exists for some t. In fact, show Φ
(t)
−1 exists for every t. Suppose
then that Φ
(t0)
−1 does not exist. Explain why there exists c ∈ Fn such that there is no solution x to
the equation c = Φ
(t0)
x. By the existence part of Problem 38 there exists a solution
to
x′ = Ax, x(t0) = c
but this cannot be in the form Φ
(t)
c. Thus for every t, Φ
(t)
−1 exists. Next suppose for some
t0,Φ
(t0)
−1 exists. Let z′ = Az and choose c such that
z (t0) = Φ (t0)c
Then both z
(t)
,Φ
(t)
c solve
′
x = Ax, x(t0) = z(t0)
Apply uniqueness to conclude z = Φ
(t)
c. Finally, consider that Φ
(t)
c for c ∈ Fn either is the general
solution or it is not the general solution. If it is, then Φ
(t)
−1 exists for all t. If it is not, then Φ
(t)
−1
cannot exist for any t from what was just shown.
Let Φ′
(t)
= AΦ
(t)
. Then Φ
(t)
is called a fundamental matrix if Φ
(t)
−1 exists for all t. Show there
exists a unique solution to the equation
′
x = Ax + f,x(t0) = x0 (14.11)
(14.11)
and it is given by the formula
∫ t
x (t) = Φ (t)Φ(t0)− 1x0 + Φ (t) Φ (s)−1 f (s)ds
t0
Now these few problems have done virtually everything of significance in an entire undergraduate
differential equations course, illustrating the superiority of linear algebra. The above formula is called
the variation of constants formula.
Hint:Uniquenss is easy. If x1,x2 are two solutions then let u
(t)
= x1
(t)
− x2
(t)
and argue
u′ = Au,u
(t0)
= 0. Then use Problem 38. To verify there exists a solution, you could just
differentiate the above formula using the fundamental theorem of calculus and verify it works.
Another way is to assume the solution in the form
x(t) = Φ (t)c(t)
and find c
(t)
to make it all work out. This is called the method of variation of parameters.
Show there exists a special Φ such that Φ′
(t)
= AΦ
(t)
,Φ
(0)
= I, and suppose Φ
(t)
−1 exists for all t.
Show using uniqueness that
Φ(− t) = Φ(t)−1
and that for all t,s ∈ ℝ
Φ(t+ s) = Φ (t) Φ(s)
Explain why with this special Φ, the solution to 14.11 can be written as
∫
t
x (t) = Φ (t− t0)x0 + t0 Φ(t− s)f (s)ds.
Hint:Let Φ
(t)
be such that the jth column is xj
(t)
where
x′j = Axj,xj (0) = ej.
Use uniqueness as required.
You can see more on this problem and the next one in the latest version of Horn and Johnson,
[19]. Two n × n matrices A,B are said to be congruent if there is an invertible P such
that
B = P AP ∗
Let A be a Hermitian matrix. Thus it has all real eigenvalues. Let n+ be the number of
positive eigenvalues, n−, the number of negative eigenvalues and n0 the number of zero
eigenvalues. For k a positive integer, let Ik denote the k × k identity matrix and Ok the
k × k zero matrix. Then the inertia matrix of A is the following block diagonal n × n
matrix.
( )
| In+ |
( In− )
On0
Show that A is congruent to its inertia matrix. Next show that congruence is an equivalence relation
on the set of Hermitian matrices. Finally, show that if two Hermitian matrices have the same inertia
matrix, then they must be congruent. Hint: First recall that there is a unitary matrix, U such
that
( )
Dn+
U∗AU = |( Dn− |)
O
n0
where the Dn+ is a diagonal matrix having the positive eigenvalues of A, Dn− being defined similarly.
Now let
||D ||
n−
denote the diagonal matrix which replaces each entry of Dn− with its absolute value.
Consider the two diagonal matrices
( )
D −n1+∕2
D = D∗ = |( || ||−1∕2 |)
Dn−
In0
Now consider D∗U∗AUD.
Show that if A,B are two congruent Hermitian matrices, then they have the same inertia matrix.
Hint:Let A = SBS∗ where S is invertible. Show that A,B have the same rank and this implies that
they are each unitarily similar to a diagonal matrix which has the same number of zero entries on the
main diagonal. Therefore, letting VA be the span of the eigenvectors associated with positive
eigenvalues of A and VB being defined similarly, it suffices to show that these have the same
dimensions. Show that
(Ax, x)
> 0 for all x ∈ VA. Next consider S∗VA. For x ∈ VA, explain why
Next explain why this shows that S∗VA is a subspace of VB and so the dimension of VB is at least as
large as the dimension of VA. Hence there are at least as many positive eigenvalues for B as there are
for A. Switching A,B you can turn the inequality around. Thus the two have the same inertia
matrix.
Let A be an m × n matrix. Then if you unraveled it, you could consider it as a vector
in ℂnm. The Frobenius inner product on the vector space of m × n matrices is defined
as
(A,B) ≡ trace(AB ∗)
Show that this really does satisfy the axioms of an inner product space and that it also amounts to
nothing more than considering m × n matrices as vectors in ℂnm.
↑Consider the n × n unitary matrices. Show that whenever U is such a matrix, it follows
that
--
|U |ℂnn = √ n
Next explain why if
{Uk }
is any sequence of unitary matrices, there exists a subsequence
{Uk }
m
m=1∞ such that limm→∞Ukm = U where U is unitary. Here the limit takes place in the sense
that the entries of Ukm converge to the corresponding entries of U.
is small if and only if every eigenvalue of A is close to some
eigenvalue of B. Now prove the following theorem using the above problem and Schur’s theorem.
This theorem says roughly that if A is close to B then the eigenvalues of A are close to
those of B in the sense that every eigenvalue of A is close to an eigenvalue of B. This is
a very important observation when you try to approximate eigenvalues using the QR
algorithm.
Theorem 14.5.2Suppose limk→∞Ak = A. Then
kli→m∞ dist(σ(Ak),σ (A )) = 0
Let A =
( )
a b
c d
be a 2 × 2 matrix which is not a multiple of the identity. Show that A is similar
to a 2 × 2 matrix which has at least one diagonal entry equal to 0. Hint:First note that there exists
a vector a such that Aa is not a multiple of a. Then consider
( )−1 ( )
B = a Aa A a Aa
Show B has a zero on the main diagonal.
↑ Let A be a complex n × n matrix which has trace equal to 0. Show that A is similar to a matrix
which has all zeros on the main diagonal. Hint:Use Problem 39 on Page 257 to argue that you can
say that a given matrix is similar to one which has the diagonal entries permuted in any order
desired. Then use the above problem and block multiplication to show that if the A has k nonzero
entries, then it is similar to a matrix which has k − 1 nonzero entries. Finally, when A is similar to
one which has at most one nonzero entry, this one must also be zero because of the condition on the
trace.
↑An n × n matrix X is a commutator if there are n × n matrices A,B such that X = AB − BA.
Show that the trace of any commutator is 0. Next show that if a complex matrix X
has trace equal to 0, then it is in fact a commutator. Hint:Use the above problem to
show that it suffices to consider X having all zero entries on the main diagonal. Then
define
( )
1 0 {
|| 2 || Xij-if i ⁄= j
A = || .. || , Bij = i−j
( . ) 0 if i = j
0 n