Math 818: Problem set 2, due January 28
Instructions: Write up any four of the following problems (but
you should try all of them, and
in the end you should be sure you know how to do all of them).
Your write ups are due Friday, January 28, 2005.
Each problem is worth 10 points, 9 points for correctness
and 1 point for communication. (Your goal is not only
to give correct answers but to communicate
your ideas well. Make sure you use good English,
so proofread your solutions.
Once you finish a solution, you
should restructure awkward sentences, and strike out
anything that is not needed in your approach to the problem.)
- Let V be the space of rxs complex matrices.
Given elements A and B of V,
define < A,B > to be trace(A*B).
- (a) Show that < , > is a hermitian form.
- (b) Find a basis for V which is orthonormal with respect to this form;
conclude that < , > is positive definite.
- (c) Do Problem 6 from the January 2004 Qualifying Exam:
If A is an nxn hermitian matrix with trace(A2)=0, then show A = 0.
- Let A be an nxn complex matrix.
Prove A is hermitian if and only if X*AX is real for
all vectors X in Cn.
- Let A be an nxn complex matrix. Show that A is a positive definite
hermitian matrix if and only if A = P*P for some invertible
- If A is an nxn real symmetric matrix and C is
an nxn real symmetric positive definite matrix,
show that there is a basis B of Rn such that
B is orthogonal with respect to both < , >A and < , >C. (First think about what the Spectral Theorem tells you about
A in the special case that C = In.)
- Let A be a real mxn matrix, let b be an element of Rm,
and consider the system of equations AX=b. Let W be the column space of A,
and let Wperp be the orthogonal complement of W.
- (a) If X is a solution to AtAX = Atb, show that
b - AX is in Wperp; conclude that AX is the orthogonal
projection of b into W. (Hence solving AtAX = Atb
gives you a way to compute orthogonal projections without having to find
an orthogonal basis of W.)
- (b) If b is not in W, then AX = b has no solutions. Show, however,
that AtAX = Atb always has solutions. (A solution
of AtAX = Atb is called a
least squares solution of AX = b. A least squares solution
of AX = b thus need not be an actual solution of AX = b.)
- (c) If AX = b has a solution, show that every least squares
solution is an actual solution.
- Consider data points (0,0), (1,1) and (2,1).
Let A be the matrix whose rows are (0,1), (1,1) and (2,1).
Let b be the vector whose transpose is (0,1,1).
Thus b is the vector whose entries are the y-coordinates
of the given data points, and the first column of A is
the vector whose entries are the x-coordinates of the
data points. The second column of A is just a column of 1's.
The equation A(s,i)t = b has a solution (s,i)t
if and only if the data points lie on the line
y = sx + i (i.e., on the line with slope s and
y-intercept i). In this example, the points are not
collinear so there is no solution (s,i)t.
- (a) Find a least squares solution to A(s,i)t = b.
- (b) Find a reference for least squares lines and determine
the least squares line for the given data points.
Compare it to the line y = sx + i you found in part (a).
Are they the same?