Homework #7 for Math H110, fall '02

1. Given M a matrix in Jordan canonical form, how can you quickly compute the JCF of M^2?

A. Squaring a block-diagonal matrix amounts to squaring each block individually. So it's enough to understand what's going on block by block. Call the size of the block N.

If M has lambda down the diagonal, and 1s above, then the square has lambda^2 down the diagonal, and 2.lambda above. (I'm going to ignore the case 2=0 -- you can think about it if you want.) Above that is all 2s.

If lambda isn't zero, then lambda^2 and 2.lambda aren't either. So when we look at M - (lambda^2 . Identity), we get something strictly upper triangular whose Nth power is zero, but whose (N-1)st power isn't. Therefore it has just one Jordan block, with lambda^2 on the diagonal.

If lambda is zero, then we're studying the square of a nilpotent matrix whose suicide diagram has just one line. Therefore M^2 kills two people each time (except at the very end, just one, if N is odd). Therefore _its_ suicide diagram has two lines of length N/2 (round up and down if N is odd).

2a. Find a degree 2 polynomial p(x,y,z,w) in four variables, such that for every matrix
a b
c d
that is not diagonalizable, we have p(a,b,c,d)=0. (Hint: if the matrix has 2 distinct eigenvalues, then it's diagonalizable.)

A. The characteristic polynomial is (x-a)(x-d)-bc = x^2 - (a+d)x + det. Its roots are (a+d) +/- sqrt( (a+d)^2 - 4det) / 2.

For the matrix to not be diagonalizable, its two eigenvalues must be the same, which means that this +/- sqrt must be +/- 0. So one polynomial which works is p(a,b,c,d) = (a+d)^2 - 4(ad-bc) = (a-d)^2 + bc.

(In fact this, and multiples of it, are the only degree 2 that work.)

b. If p(a,b,c,d)=0, does that imply the matrix is not diagonalizable?

A. No, a 2x2 matrix M can have two equal eigenvalues and be diagonalizable. Its JCF would therefore have to be a scalar matrix (the eigenvalue times the identity matrix), i.e. M = X lambda I X^{-1}. But then M = lambda I. So the only counterexamples are the scalar matrices.

3. Let G be a diagonal matrix with s 1's then t -1's down the diagonal. Describe the matrices M such that M G M^T = G. That set is called O(G).

A. Not a very concrete question; the best that one can do simply is to break M up into sxs, txt, sxt, txs blocks and write the equations on them.

4. Say V is the real vector space R^{s+t}, and G and O(G) are as in #3. Say there is a constant B such that for all M in O(G), each matrix entry m_ij of M has | m_ij | < B. What can you say about s, t?

If you're stuck, consider the three cases when V is 2-dimensional.

A. Indeed, let's start with the 2-d case. If s=0 or t=0, then O(G) is just "orthogonal matrices" M, satisfying M M^T = Id. Looking at the diagonal entries of M M^T, we find out that the sum of the squares across any row is 1, so each entry is at most 1. So B=1 works.

Whereas if s=t=1, we're thinking about 2x2 matrices
ab
cd
such that a^2 = b^2 + 1, ac = bd, and d^2 = c^2 + 1. Once one's specified b, that determines a up to sign, therefore a/b, therefore d/c, therefore d and c up to sign. If we think about the case where everybody's positive, we get
sqrt(b^2+1) b
b sqrt(b^2+1)
and now we can take b as big as we like. So there is no bound B.

(The nicest parametrization is b = cosh theta, a = sinh theta, just like one might parametrize orthogonal matrices with cos and sin.)

Okay, now the general case. If s=0 or t=0, we're back in the easy orthogonal case, and can use B=1. Otherwise, s,t > 0. But then we can use the 2x2 matrices we just found, with an (s-1) identity matrix before and a (t-1) identity matrix after, like this,
1 . . . . . .
. 1 . . . . .
. . sqrt(b^2+1) b . . .
. . b sqrt(b^2+1) . . .
. . . . 1 . .
. . . . . 1 .
. . . . . . 1
and show again that there is no B. (Pictured above is s=3,t=4.)