Advanced Linear Algebra Homework for Spring 2002
[these problems are being modified, so check frequently]
1/29 Resolve the parallelogram conjecture, and do the other exercises
suggested in class.
1/31 Do the exercises suggested in class about abelian groups, rings and
fields (division rings), especially: assuming the real numbers R
is a field, verify that the complex numbers C = R + i R is also
a field, and that the quaternions H = R + i R + j R + k R is
a (noncommutative) field, where i^2 = j^2 = k^2 = ijk = 1.
Also try this: for what n is the ring Z/nZ a field? Can you
describe any other finite fields?
2/5 Do the exercises suggested in class: 0v = 0, (1)v = v, k0 = 0;
the {moon} is a vector space over any field; etc.
Also try these:
Let S be any subset of V. Check that span(S) is a subspace.
Show that S={1, x, x^2, ... , x^n, ...} is a linearly independent
set in the vector space K[x] of all polynomial functions of x,
and thus a basis for K[x], since clearly span(S)=K[x].
Show that the sets S={e_1, e_2, ... , e_n} and
T={e_1, e_1+e_2, ... e_1+...+e_n} are bases for K^n.
2/7 Curtis 2.4#3cefg,4cdeg,7,8,10
2/12 Curtis 2.5#1,5
2/14 Curtis 2.7#1,3,4[Hint: use dimension theorem and previous problem]
2/19 [No class because of Washington and Lincoln]
2/21 Let L = [.]_B be the linear map V > F^n which assigns a vector
in V its coordinates with respect to B. Let S be the standard
basis of F^n. What is the matrix [L]_SB?
Show that Hom(V,W) = {linear maps V > W} is a vector space of
dimension dim(Hom(V,W)) = dim(V) dim(W).
2/26 Curtis 3.11#3,4,5,7,9,10
2/28 Curtis 3.12#4,5,8
3/5 Curtis 3.13#1,3,7,12
3/7 Let V L> W M> X be linear maps. Suppose B, C, and D are bases
for V, W and X, respectively. Verify that the matrix for the
composition ML (w.r.t. bases B and D) is the matrix product of
the matrices for M and L (w.r.t....). (Try the case where V, W and
X are each 2dimensional first!)
Try to work through the details of rank+nullity theorem proof
of the dimension theorem: use the linear map from the direct
sum of subspace V and W onto the sum V+W defined by (v,w) > vw,
and show that the kernel is isomorphic to the intersection of
V and W.
3/12 Let V = F^n where F = Z/2Z is the field of two elements. How
many ordered bases are there for V? (This is a tricky counting
problem: try n=2 case first! How many unordered bases are there?
Hamming code problems [I'll distribute these in class]
3/14 Think of the following problem as takehome midterm:
Let L: V > W be a linear map between finite dimensional
vector spaces. Let k = dim(im(L)) = rank(L), m = dim(W) and
n = dim(V). Then there are bases B for V and C for W such
that the m x n matrix for L w.r.t these bases is
 I_k 0 
[L]_CB =  
 0 0 
where I_k is the k x k identity matrix and where 0 denotes
a matrix of all 0's of the appropriate size.
In other words, any m x n matrix A can be put in the above
form by multiplying it on the left and right by invertible
m x m and n x n matrices X and Y:
 I_k 0 
XAY =   .
 0 0 
This means that the only invariant of a linear map when we
are free to choose bases for range and domain independently
is rank(L), which is simply an integer between 0 and m.
Try carrying this out explicitly for the matrix
 1 2 3 
A =  4 5 6  .
 7 8 9 
[You should also read Chapters 5, 6 and 7 of Curtis over the
break (we'll return to ideas from 4 later).]
3/26 Curtis 7.22#6,7,8,9,10
3/28 Curtis 7.22#2(cf. hw from 2/21 above),4,5,11,12
4/2 Show that the space of nforms on F^n is a vector space over
F. Try to compute the dimension of this vector space. This
useful for the following:
Verify that my definition of det (an nform \omega with
\omega(e_1,...,e_n) = 1) agrees with Curtis'. Also verify this
gives an algorithm for computing det (Theorem 16.9 in Curtis 5.16).
Curtis 5.16#4 and 5.18#4
4/4 Curtis 5.19#7,10
4/9 Curtis 7.23#3,4,6
4/11 Curtis 7.24#4,10
Also show that if N:W > W is nilpotent of order c (N^c = 0, but
N^{c1} \= 0) and w is a nonzero vector in W which is not in
ker(N^{c1}), then {w, N(w), N^2(w),...,N^{c1}(w)} is linearly
independent in W.
4/16 Curtis 7.24#1,2[skip part g in each],3
4/18 In class we defined the exponential of an n x n matrix (over a
subfield of the complex numbers for the series to converge)
exp(A) = I + A + ... + A^k/k! + ....
Use the triangular form theorem A = D + N, where D is
diagonalizable and N is nilpotent of some order c \le n,
to get a simpler formula for exp(A). [Hint: in general,
exp(A+B) = exp(A)exp(B) *only* when A and B commute.]
Carry this out to compute exp(A) for the 2 x 2 matrix
14 25
A =   ,
 9 16
which is similar (conjugate) to the matrix
1 1
B =   .
0 1
4/23 Curtis 7.25#2,3,4,8[read/think through this one, at least]
4/25 From my email summarizing our streamlined method, avoiding
some of the ring theory in Curtis, and using instead "Fitting's Lemma":
Over the past few classes we developed a somewhat alternative approach
to the canonical form theorems.
First we discussed finding a basis in which a nilpotent linear map
N:V > V has d x d matrix (d = dim V) in the canonical form
0 0 0 ... 0 0
1 0 0 ... 0 0
0 1 0 ... 0 0 0
0 0 1 ... 0 0
: : : : :
0 0 0 ... 1 0


0  0

where the p x p block corresponds to order of nilpotency p of N. This
is what I call Fitting's Lemma. We proved that if v is a nonzero
vector in V which is not in ker(N^{p1}), then
{v, N(v), N^2(v),..., N^{p1}(v)}
is linearly independent in V, and that it can be extended to a basis
with the properties desired above. (This is essentially the problem
that I suggested for 4/11, with slightly different notation.)
Conseqently, p is at most d.
Note that this works over any field F, not necessarily algebraically
closed. I emphasized this result because it is also at the heart of
rational canonical form, also valid over any F.
Next (a bit last time, the rest today) we applied this to an arbitrary
linear map L: V > V under the assumption that its minimal polynomial
factors into powers of linear factors
m(x) = (x  a_1)^{p_1} ... (x  a_k)^{p_k}
(which will be true automatically when F is closed field).
A key point is that the map N_i:V_i > V_i, defined as (L  a_i)
restricted to V_i = ker((L  a_i)^{p_i}), is nilpotent of order p_i
(by minimality of m). This L restricted to V_i has the block form
indicated above, except with a_i along the main diagonal.
A lemma which I started to state in general, but which I decided to
give only the special case in class, verifies that V_i is in fact an
Linvariant subspace (try as an exercise):
Suppose L, M are commuting linear maps V > V, that is LM = ML.
Then ker(M) and im(M) are Linvariant subspace (similarly for
the roles of M and L reversed).
We applied this with M=(La_i)^{p_i}, which clearly commutes with L.
Finally, we deduced the HamiltonCayley theorem from this:
1) m(x) divides the characteristic polynomial p(x) = det(xL)
2) L satisfies p(x), i.e. p(L)=0
at least when F is closed (since then clearly the factors of m divide those
of p); but a little thought shows that this must really be true for
arbitrary F as well (exercise)!
One final note: problem 23#6 is not correct if the characteristic of F
is 2, since then the matrix
1 0
 
1 1
satisfies the hypotheses, but not the conclusion  you might think
about this in terms of canonical form theorem above (the square of
this matrix is I, so it has minimal polynomial x^2  1 = (x1)^2
in characteristic 2!
4/30 Curtis 4.15#2,9[this relies on the exercise suggested in class,
i.e. V = W \directsum W^\perp for any subspace W of V],12,13
Carry out the KAN factorization of your favorite 2 x 2 matrices,
including my favorite:
1 2
C =   .
3 4
Determine the symmetries of the capital letters of the Roman
alphabet, assuming these are written in the maximally symmetric
form (in particular, L and Q have nontrivial symmetries, and O
has an infinite family), and express as matrices with respect
to the standard basis of R^2.
Find the QR factorization of the (singular) 3 x 3 matrix
1 2 3
 
A = 4 5 6 .
 
7 8 9
5/2 Back to KAN factorization. For small matrices, this isn't very
practical, but the factorization C = KAN gives a nice way to
compute the inverse of C: in fact, the inverses of K and A are very
easy, being K* and 1/A (meaning the diagonal matrix with reciprocal
entries); the inverse of N is also easy: since N = I  P where P
is nilpotent, its inverse is I + P + P^2 + ... + P^k where k+1 is
the order of nilpotency (verify this). Carry this out for the
2 x 2 example from 4/30. [Here and forever more, A* denotes the
(conjugate, if complex entries) transpose of a matrix A.]
5/7 Let W be the subpace of R^4 perpendicular to [1 1 1 1]*. Compute
matrices for orthogonal projections onto W^\perp and onto W. Find
the closest points on W and W^\perp to the vector [1 2 3 4]*.
Curtis 9.30#1,2,3
5/9 Curtis 9.31#1,3,58[really one extended problem]
Suppose we have an endomorphism L: V > V of a real vector space V,
and we hope to find eigenspace and eigenvalues, or at least invariant
subspaces. Today we discussed how to do this using complex
eigenspaces and eigenvalues for L, by regarding V as a real subspace
of a complex vector space V_C. To make sense of this in general
requires the notion of tensor product, but we can be rather concrete:
if V has basis B (over R) then V_C also has basis B (over C); in other
words, we simply extend the coefficient field from R to C. (Note that
the same trick works for any field extension F to K: a vector space
V_F over F can be viewed as an Fsubspace of V_K). Equivalently (and
even more concretely) we know that we may simply regard V as R^n and
V_C as C^n, by choosing some basis B as above.
Note that complex conjugation defines a real (but not complex) linear
mapping #:C^n > C^n which fixes R^n, that is, R^n is the 1eigenspace
of # (in class I used a "bar" instead of #, but can't here without
TeX). In fact, if U is any subspace of C^n which is fixed by # then
the vectors in U have all entries real. Consider any complex subspace
W of C^n and its image #W under complex conjugation. Inside W+#W is
the (real) subspace U of vectors of the form u=w+#w for some w in W;
again, these are just the vectors fixed by #.
Return now to our Rlinear map L: R^n > R^n, which we may extend to a
Clinear map C^n > C^n. Suppose w is an eigenvector of L with
eigenvalue m (both may be complex), and let W be the 1dimensional
complex subspace spanned by w. Then W+#W is an Linvariant subpace,
and because L is real, its real subspace U is also Linvariant. There
are two possibilities:
1) If the eigenvalue m is real, then L(w+#w)=mw+#(mw)=m(w+#w) so
u=w+#w is a real eigenvector, and W=#W. Conversely, if W=#W, then U
is 1dimensional over R, spanned by a real eigenvector u, and in this
case the eigenvalue m must be real.
2) When m is not real, W and #W are distinct, U is 2dimensional over
R, and L restricted to U can be put in the form
(m+#m)/2 (m#m)/2i
 
(m#m)/2i (m+#m)/2 
in a suitable basis for U (and conversely).
Note that in the special case m is a unit complex number (as in
class), 2) is just the corresponding rotation matrix with cosines and
sines of the argument (angle) of m; and 1) occurs for a reflection
with m=1 or 1 (cf. Curtis Theorems 30.2 and 30.3).
5/14 The following is part of your final exam:
Let V be a finite dimensional vector space over C, with hermitian
inner product (think of C^n with (x,x)=x*x if you wish).
The key lemma for a linear map A: V > V which lets us show that
it can be diagonalized with respect to an orthonormal
(unitary) basis when selfadjoint (symmetric or hermitian) is
this: if a subspace W is Ainvariant, then its orthogonal
complement W^\perp is A*invariant. Prove this lemma, and apply
it to show we can diagonalize skewadjoint, unitary, or, more
generally, normal (A*A=AA*) maps in the same way.
To do this you should prove and use the lemma (cf. Curtis 32.13):
Suppose L, M are commuting linear maps V > V, that is LM = ML.
Then L and M have a common 1dimensional invariant subspace W.
[Compare to the lemma from 4/25, and contrast with the following:
commuting linear maps need not have the same invariant subspaces,
since obviously a scalar multiple of the identity (for which any
subspace of V is invariant) commutes with any L (which may have only
particular invariant subspaces).]
To show that a normal linear map can be diagonalized, let L=A and
M=A*, and observe that A**=A. Thus the n1 dimensional subspace
W^\perp is also invariant under under both A and A*, so we can use
the dimension reduction argument as before (induction on n if you like).
In applications, one of the most useful ways to compute eigenvectors
and eignvalues of a symmetric matrix A is to find the maximum of the
associated quadratic form Q(x) = x*Ax among unit vectors x. Show that
the maximum point v is an eigenvector of A belonging to the largest
eigenvalue, the maximum value Q(v). More generally, the critical
points and values of Q correspond to the eigenvectors and eigenvalues
of A.
To find such maximum value, please justify the following algorithm:
pick a random unit vector (say, v_0 = e_1) to begin, then apply A to
get v_1 = Av_0, then rescale to get unit vector u_1; we can iterate
this to get a sequence of unit vectors u_n (in practice, simply use
v_n = Av_{n1} and rescale only at the end to get u_n) which converge
to the eigenvector v. (In general, the eigenvectors of A are fixed
points [up to "+/" sign] of this rescaled action of A on the set
of unit vectors.)
Carry out this algorithm for the 2 x 2 symmetric matrix
 1 2 
A =   .
 2 3 
Fibonacci numbers will appear, and the u_n should converge to a
multiple of [1 T]* where T is the "golden mean" (1 + \sqrt5)/2.
By this method, you should find the largest eigenvalue of A is T^3.
Check algebraicly that this is a root of the characteristic polynomial.
What happens when you begin the iteration at v_0 = e_2?
Actually, a better algorithm for the largest eigenvalue/vector
of A is to iterate
v > (1h)v + hAv ,
rescaling to the sphere of unit vectors as appropriate. For h
small this is a discrete approximation to the gradient flow of Q,
which will generically converge to the largest eigenvalue/vector.
HAVE A MATHEMATICALLY WONDERFUL SUMMER!!!!!!!!!