On Mon, 16 May 2005, Scott Morrison wrote:
> Hehe... I'd just written some hints for some of Chu-Wee's problems as
> well. I think they haven't been made entirely redundant by Chu-Wee's
> hints, so I'll post them here:
> Apologies about all the strange notation; it's LaTeX, pretty much, and
> not too hard to understand :-)
> Problem 7), second part.
> You might write A = V D V^{-1}, for some diagonal D. Then p(A) = V
> p(D) V^{-1}. Certainly p(D) will still be diagonal, so call its
> diagonal entries x and y. Now expand out p(A), and see what you can
> say about x and y when p(A) is upper triangular.
> Problem 11)
> a) Easy over the complex numbers; just use the Cayley-Hamilton
> theorem. Also easy if you believe in Jordan Canonical Form.
> b) You need the binomial formula here; for commuting quantities A
> and B, (A+B)^k = \Sum_{i=0}^k \binomial{k}{i} A^i B^{k-i}. Say A^n =
> 0, and B^m = 0. Can you choose k large enough so either i >= n or
> k-i >= m, no matter what i is?
> Problem 12)
> Remember that A satisfies its own characteristic equation, so A^2 =
> x A + y I, for some values of x and y. You can use this to write any
> polynomial in A as a linear function of A.
> Problem 16)
> This is a quadratic form; find the associated symmetric bilinear
> form, and diagonalise it. Hopefully you'll find all the diagonal
> entries are positive.
> Problem 17)
> a) If you're willing to use Jordan canonical form, this is easy;
> observe that the JCF of a unipotent matrix is exactly the identity
> plus the JCF of a nilpotent matrix. ***
> b) and c) should be easily answerable from your solutions for
> Problem 11).
> Problem 19)
> Say (I+AB)x = 0. Can you produce an element of the kernel of I+BA?
> (Clue; there's one and only one sensible way of producing an element
> of F^n from an element of F^m, in this context.)
> Problem 22) Suppose AB=BA, and both A and B are diagonalisable.
> Say x is an eigenvector for A with eigenvalue \lambda. Then ABx =
> BAx = \lambda Bx, so Bx is also an eigenvector for A with eigenvalue
> \lambda. Thus E_\lambda is a B-invariant subspace. However we know
> what B-invariant subspaces look like; they are spanned by
> eigenvectors of B. It works the other way round as well; eigenspaces
> for B are A-invariant as well. Now let's define E_{\lambda,\mu} to
> be the set of simultaneous eigenvectors; vectors x so Ax = \lambda
> x, and Bx = \mu x. By the above argument, the entire vector space is
> in fact a direct sum of these E_{\lambda, \mu} subspaces (work this
> step out carefully!). The result follows (explain exactly why!).
> On 16/05/05, Chu-Wee Lim <limch...@math.berkeley.edu> wrote:
> > Hi everyone.
> > Due to some requests, I'm putting up hints to my review problems.
> > Sorry, I really don't have time to type out full solutions to all
> > the problems.
> > Also, take note that Q20 is wrong: no such M exists. This is
> > interesting, but not an easy fact to show.
> > -CW
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