A heads up on Chebfun

December 16, 2011

The applied side of approximation theory is like black magic to me: even if I don’t want to do research in the area, at least I respect it enough to find it interesting. So, I recommend bit-player’s post on Nick Trefethen’s Chebfun package. Especially check out the links (Trefethen’s an engaging author).

I’m looking at products like \(\mat{G}^t \mat{A} \mat{G}\) where the columns of \(\mat{G}\) are (nonisotropic) normal vectors. Specifically, I’d like to know the distribution of the eigen/singular-values of this product. Surprisingly, I was unable to find any results in the literature on this, so I started reading Gupta and Nagar to see if I could work it out using Jacobian magic.

In the preliminary material, they list some basic matrix algebra facts. Your mission, should you choose to accept it, is to prove the following:

  • If \(\mat{A} > \mat{0},\) \(\mat{B} > \mat{0},\) and \(\mat{A} – \mat{B} > \mat{0},\) then \(\mat{B}^{-1} – \mat{A}^{-1} > \mat{0}\)
  • If \(\mat{A}>\mat{0}\) and \(\mat{B} > \mat{0},\) then \(\det(\mat{A}+\mat{B}) > \det(\mat{A}) + \det(\mat{B})\)

The second’s easy, but for the first I found it necessary to use the fact that every positive matrix has a unique positive square root.

Some more, this time on Kronecker products:

  • If \(\mat{A}\) has eigenvalues \(\{\alpha_i\}\) and \(\mat{B}\) has eigenvalues \(\{\beta_j\},\) then \(\mat{A} \otimes \mat{B}\) has eigenvalues \(\{\alpha_i \beta_j\}.\)
  • If \(\mat{A}\) is \(m \times m\) and \(\mat{B}\) is \(n \times n\), then \(\det(\mat{A} \otimes \mat{B}) = \det(\mat{A})^n \det(\mat{B})^m\)

It’s useful here to note that
\[
(\mat{A} \otimes \mat{B}) (\mat{C} \otimes \mat{D}) = \mat{AC} \otimes \mat{BD},
\]
which has some obvious consequences, like that the Kronecker product of two orthogonal matrices is orthogonal. To be clear, the first Kronecker question can be addressed without any tedious calculations.

George Gently

October 9, 2011

Wooo woooo! I just found George Gently‘s coming (came?) back for 2011. I also found out that there was a 2010 season! Too bad Netflix doesn’t have it yet 🙁

Here’s the first Netflix review I (remember having) ever wrote. I was that impressed by the show. Hope I can share that enthusiasm. For some reason Netflix removed all the apostrophes in the review.

George Gently’s an older DI at the Yard who thought he was ready to retire after his wife was murdered because of his crusade against corruption in the force, but a fresh lead on the trail of the man he suspects is behind the murder of his wife brings him back into the game, this time in Northumberland. He’s assisted in his investigation by John Bacchus, a young, ambitious, and brash inspector who, from past bitter experience, Gently sees exhibits all the earmarks of becoming corrupt. So, after the case is resolved, Gently decides to stay on in Northumberland and take the youngster in hand. Thus the first episode sets the stage … after that, each episode deals with a new case.

Yes, the cases are interesting, and the setting is verisimilar, but the appeal of the series for me lies first and foremost in the relationship between Gently and Bacchus. Bacchus is, from the beginning, an unsympathetic character: for no readily apparent reason, he’s unsatisfied with his marriage, and perhaps unconsciously sabotaging it; he’s willing to do whatever it takes — beatings, planting evidence, accepting bribes — to close a case; and his general opinions on race and gender issues reflect those of the times in an unflattering manner. Yet, you can’t help but get the feeling that there’s something salvageable under those youthful excesses. Gently does an excellent job of mentoring Bacchus in a show-and-not-tell manner, but it’s unclear if Bacchus is taking in the messages Gently’s trying to get across or if he’s simply humoring him.

The two episodes of season three are the best of the series. In the first, Bacchus and Gently come to loggerheads over a case involving a child, and in the second, we see concrete hints that Bacchus is becoming a better man. I’d love to see more in this series, but this was a great stopping point. Always leave them wanting more, right?

Looks like I’ll be visiting U. Michigan for a week or so at the end of October. What does one do on such a visit? I’ve been invited to give a talk, but that doesn’t occupy more than an afternoon…

I spent this morning reading through some of the literature on subspace tracking, then wound up visiting Nocedal and Wright for a refresher on the augmented Lagrangian method (Chap 17). Flipping through the book, I came across the following question:

Show that every point on the unit circle is a limit point of the sequence
\[
\vec{x}_k = \left(1 + \frac{1}{2^k}\right) \begin{pmatrix} \cos k \\ \sin k \end{pmatrix}.
\]

Not challenging (esp. if you don’t go into the nasty \(\varepsilon-\delta\) details ), but it’s a cute problem.

A cautionary Mathematica tale

September 26, 2011

Because a picture’s worth a thousand words:

visual proof that mathematica sucks at linear algebra

according to Mathematica, the rank of a matrix and that of a basis for its range can differ

The matrix involved is a projection, not some numerically unstable thing like the Hilbert matrix. You suck at linear algebra, Mathematica.

I still haven’t hammered out the database issues, but I’m tired of my little vacation (i.e. several months!) from posting, so I’m going ahead with the relocation. At some unspecified time in the future (once I figure out the technical details), I’ll migrate the database over from the old location. That site’ll still be up, albeit in its broken state (the math and other escaped code is not being interpreted correctly).

Here’s an inagural problem: Argue that \[\frac{21}{22} \geq \frac{\log 9}{\log 10} \geq \frac{20}{21}\] without using decimal expansions of the involved quantities. Feel free to use arguments that aren’t feasible without a CAS like Mathematica– I sure did.