Wikipedia:Reference desk/Archives/Mathematics/2007 February 21

= February 21 =

division
6x^7/2x^2=

Having a little hw trouble there? Looks like basic algebra. Divide the coefficients and subtract the exponents. Rya Min 03:04, 21 February 2007 (UTC)


 * (6 / 2) x ^ (7 - 2). Whenever you multiply exponents, they're added. When you divide, it's subtracted. So x^2 * x^3 is x^5, and x^3 / x^2 is x^1 (or x). --Wirbelwind ヴィルヴェルヴィント (talk) 03:06, 21 February 2007 (UTC)


 * Let A=x^7
 * Let B=x^2
 * 6x^7/2x^2 = (6*A)/(2*B) = 3 * (A/B)
 * And 3 * (A/B) = 3 * (x^7/x^2)
 * The rest is pretty easy. 211.28.127.77 07:03, 21 February 2007 (UTC)
 * And 3 * (A/B) = 3 * (x^7/x^2)
 * The rest is pretty easy. 211.28.127.77 07:03, 21 February 2007 (UTC)
 * The rest is pretty easy. 211.28.127.77 07:03, 21 February 2007 (UTC)

When two diagonizable matrices commute, are they simultaneously diagonizable?
Hello,

my question may not be that hard, but I'm really confused.

There is a theorem saying that commuting hermitian matrices are simultaneously diagonizable. I was wondering about the relevance of the matrices being hermitian.

I found a simple example... but one of the two was not diagonizable at all!

So this is my question : let$$ A$$ and $$B$$ be diagonizable matrices over some field, and let $$A B = BA$$. Can an invertible matrix $$Q$$ be found such that both $$Q^{-1}A Q$$ and $$Q^{-1}B Q$$ are diagonal matrices?

My guess is yes... :| But is this true? Is it true that the matrices being hermitian isn't that important after all?!

Thank you very much, Evilbu 15:59, 21 February 2007 (UTC)
 * Example: $$A= QD_1Q^{-1}$$ and $$B=QD_2Q^{-1}$$ commute, simultaneously diagonalizable, and are not necessary hermitian.(Igny 18:06, 21 February 2007 (UTC))
 * Thank you. But the main question still remains: what about the other way around, what if they are just both diagonizable and commuting?


 * Yes, you are right. In fact, you can do it with more than just two; any commuting family of diagonalizable matrices is simultaneously diagonalizable; i.e., there exists a matrix Q such that for all A in the family, QAQ-1 is diagonal. –King Bee (T • C) 18:51, 21 February 2007 (UTC)
 * Thanks, that's great! Do you think you can give me a hint how to prove this? I mean, I have already figured out that I can write the vectorspace as a direct sum of eigenspaces of A, and that (because of the commutating property) all of these spaces are invariant with respect to B. Any basis of eigenvectors for the restriction of B to one of these eigenspaces, would be a basis of eigenvectors of that eigenspace of A as well.  But there's my problem : WHY is this restriction of B diagonizable? Thank you,Evilbu 19:35, 21 February 2007 (UTC)
 * Hmm...I actually cannot remember how to do it. I think you're on the right track though; I remember showing that two commuting matrices are simultaneously unitarily upper triangularizable, and I the proof of that has a lot to do with eigenspaces, as you mentioned. –King Bee (T • C) 15:23, 22 February 2007 (UTC)
 * Sorry about this, but I can't remember a totally elementary proof, though I'm sure there is one. Going slightly high-tech, recall (or learn ;-)) that a matrix is diagonalisable iff its minimal polynomial splits into distinct linear factors. Since the restriction of B to an eigenspace of A is a zero of the minimal polynomial of B, it has minimal polynomial dividing that of B and so it is diagonalisable. Thus your proof goes through. Algebraist 15:28, 22 February 2007 (UTC)
 * Eigenspaces are, indeed, key. Maybe it would help to begin with two real symmetric matrices, each with distinct eigenvalues. Then each eigenvalue has a distinct one-dimensional eigenspace, and that space is (as always) orthogonal to all the others. If A1 and A2 commute then their eigenspaces must align. (Discussed further below.) The general case is slightly more interesting. Of course we can have eigenvalues with multiplicity; but consider a pair of 2×2 rotation matrices. These diagonalize over the complex numbers, and always commute.
 * Formally, suppose A1 maps the linear subspace E into itself; in particular, let E be the eigenspace corresponding to eigenvalue &lambda;. Then I claim that A2 also maps E into itself. For, let v be a vector in E; then
 * $$\begin{align}

A_1 (A_2 \mathbf{v}) &{}= A_2 A_1 \mathbf{v} = A_2 \lambda \mathbf{v} \\ &{}= \lambda (A_2 \mathbf{v}). \end{align}$$
 * That is, A2v is an eigenvector with eigenvalue &lambda;, thus in E. We can leverage this to prove simultaneous diagonalizability. --KSmrqT 20:46, 22 February 2007 (UTC)

A riddle...
"If A is 2, B is 3 and F is 8, how high must you count to get to Heaven?"

This question was asked on this desk a few months ago, but since there was no useful answer I'm asking again. This is what I know about the riddle:


 * It's NOT a trick question. There is a numerical answer that corresponds to the word "heaven".
 * F is NOT supposed to be 7. F is definitely 8. It is nothing but 8 and never will be anything but 8. DON'T respond with "if F is 7" because that will just make me angry.
 * The answer is not 70

This puzzle comes from an online webcomic called Jack and the correct answer appended to the site's URL should lead to a secret comic. Thanks for any help. Lovablebeautyme 21:38, 21 February 2007 (UTC)

I gave the answer before it's "70" - see below for another explanation - here is the secret page http://pholph.com/seventy/ - see it works>87.102.9.28 15:58, 22 February 2007 (UTC)


 * Why is it not 70? What is so special about 70? Are you absolutely positive it is not 70? (Igny 22:16, 21 February 2007 (UTC))


 * It could be anything. That's how riddles are. Some context might help, could you link to the actual question? One obvious solution is that each letter in Heaven is a digit in the answer. If each letter corresponds to its position in the alphabet, Heaven would be (8)(5)(1)(22)(5)(14). It doesn't, though, A=2 and B=3 and F=8, which means their positions would have to be seriously scrabbled about to make it work. If you can think of values for them, just translate from base 26 (or whatever it winds up as) to base 10 and you'll probably have the answer. Or you could just cut the knot and webcrawl through his URLs, until you find one of the form http://www.pholph.com/(somenumber). Odds are you'll find it doesn't exist. Black Carrot 22:42, 21 February 2007 (UTC)

I gave the answer to this one last time - here it is again. Count 1 through the alphabet - except on a vowel when you add 2 - thus:

A B C D E F G H I  J  K  L  M  N  O  P  Q  R  S  T  U  V  W  X  Y  Z  2 3 4 5 7 8 9 10 12 13 14 15 16 17 19 20 21 22 23 24 26 27 28 29 30 31

So heaven = 10+7+2+27+7+17=19+34+17=19+51=70.

Now 70 = "Three score and ten" -- See:

Giving the rhyme

"If A is two, B is three and F is eight, how high must you count to get to heaven's gate?" (See now it rhymes)

Three score and ten have a man = 70 = life expectancy

This is the answer - what was wrong with my original answer?87.102.9.28 15:54, 22 February 2007 (UTC)