- 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(A
^{2})=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**C**^{n}. - 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 matrix P. - 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
**R**^{n}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 = I_{n}.) - Let A be a real mxn matrix, let b be an element of
**R**^{m}, and consider the system of equations AX=b. Let W be the column space of A, and let W^{perp}be the orthogonal complement of W.- (a) If X is a solution to A
^{t}AX = A^{t}b, show that b - AX is in W^{perp}; conclude that AX is the orthogonal projection of b into W. (Hence solving A^{t}AX = A^{t}b 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 A
^{t}AX = A^{t}b always has solutions. (A solution of A^{t}AX = A^{t}b 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.

- (a) If X is a solution to A
- 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?

- (a) Find a least squares solution to A(s,i)