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Topic: Can these rational matrices commute? (Read 1068 times) |
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ecoist
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Can these rational matrices commute?
« on: Mar 2nd, 2006, 8:58pm » |
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Let A and B be 2 by 2 matrices with rational entries satisfying A2=2I and B2=3I, where I is the 2 by 2 identity matrix. Can AB=BA?
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Barukh
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Re: Can these rational matrices commute?
« Reply #1 on: Mar 3rd, 2006, 7:49am » |
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The answer is No. Here is why.
Direct explorations of the given conditions show that the elements of the matrices should satisfy the following conditions:
a112 + a12a21 = 2 (1a)
b112 + b12b21 = 3 (1b)
a11b12 = a12b11 (2)
a11b21 = a21b11 (3)
Now, multiply equations (2) and (3); multiply (1a) by b112, and (1b) by a112. After simplification, we are left with the equation 3a112 = 2b112, which doesn’t have solutions in rationals.
Even if this solution is correct, I am not fully satisfied with it, and would like to know if something more elegant exists.
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ecoist
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Re: Can these rational matrices commute?
« Reply #2 on: Mar 3rd, 2006, 8:30am » |
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I don't see where you got equations (2) and (3).
I know two solutions, one elegant (to my mind) and one not so elegant.
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Barukh
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Re: Can these rational matrices commute?
« Reply #3 on: Mar 3rd, 2006, 9:13am » |
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on Mar 3rd, 2006, 8:30am, ecoist wrote:| I don't see where you got equations (2) and (3). |
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You get them easily if you show that a11 + a22 = b11 + b22 = 0.
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JocK
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Re: Can these rational matrices commute?
« Reply #4 on: Mar 3rd, 2006, 9:47am » |
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on Mar 2nd, 2006, 8:58pm, ecoist wrote:| Let A and B be 2 by 2 matrices with rational entries satisfying A2=2I and B2=3I, where I is the 2 by 2 identity matrix. |
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I doubt whether such A and B exist.
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Obob
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Re: Can these rational matrices commute?
« Reply #5 on: Mar 3rd, 2006, 11:24am » |
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Sure they do. Mathematica tells me that
{{1,1},{1,-1}}^2 = 2I
{{1,2},{1,-1}}^2 = 3I
More generally, the matrices A={{a,b},{c,d}} with A^2=2I are exactly those which satisfy
a=-d
b=(2-d^2)/c
and the matrices with A^2=3I are exactly those which satisfy
a=-d
b=(3-d^2)/c.
Here d and c are arbitrary rational numbers (c not 0) which parameterize the corresponding a and b.
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JocK
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Re: Can these rational matrices commute?
« Reply #6 on: Mar 3rd, 2006, 12:30pm » |
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on Mar 3rd, 2006, 11:24am, Obob wrote:Sure they do. Mathematica tells me that
{{1,1},{1,-1}}^2 = 2I
{{1,2},{1,-1}}^2 = 3I
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Ah yes, of course.. silly me.
I was thinking of matrices A and B with eigenvalues [sqrt]2 and [sqrt]3, respectively. Complete nonsense: if the square of a matrix M has eigenvalues m, that doesn't mean the matrix M needs to have eigenvalues [sqrt]m.
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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JocK
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Re: Can these rational matrices commute?
« Reply #7 on: Mar 3rd, 2006, 1:17pm » |
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on Mar 2nd, 2006, 8:58pm, ecoist wrote:| Let A and B be 2 by 2 matrices with rational entries satisfying A2=2I and B2=3I, where I is the 2 by 2 identity matrix. Can AB=BA? |
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2nd take:
The matrices A and B commute iff a matrix S can be found such that
A = S {{[sqrt]2, 0},{0,-[sqrt]2}} S-1
and
B = S {{[sqrt]3, 0},{0,-[sqrt]3}} S-1
both have rational elements only. Such a matrix S does not exist because the ratio between [sqrt]2 and [sqrt]3 is irrational.
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| « Last Edit: Mar 3rd, 2006, 1:22pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Barukh
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Re: Can these rational matrices commute?
« Reply #8 on: Mar 4th, 2006, 12:01am » |
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Ah, that's nice, JocK! I always forget to use this eigen decomposion (probably, didn't study it too much).
Now, the question arises how to show in the simplest way that the eigenvalues of A, B have the specified values?
Ome way (probably, not the simplest), is to consider:
A- I = 0 (definition)
(A-I)(A+I) = 0
A2-2I = 2I -2I = 0
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ecoist
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Re: Can these rational matrices commute?
« Reply #9 on: Mar 4th, 2006, 5:26am » |
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The "elegant" solution I have is an eigenvalue argument, but it cannot beat Barukh's solution for brevity and simplicity. Is it time to post this so-called elegant solution, perhaps hidden?
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Barukh
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Re: Can these rational matrices commute?
« Reply #10 on: Mar 4th, 2006, 6:49am » |
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on Mar 4th, 2006, 5:26am, ecoist wrote:| The "elegant" solution I have is an eigenvalue argument, but it cannot beat Barukh's solution for brevity and simplicity. |
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You probably had in mind JocK's solution? 
Quote:| Is it time to post this so-called elegant solution, perhaps hidden? |
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It's OK with me...
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ecoist
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Re: Can these rational matrices commute?
« Reply #11 on: Mar 4th, 2006, 9:58am » |
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Ok. Here it is.
| hidden: | Assume that AB=BA. We derive a contradiction. Let u and v be eigenvectors of A such that Au= 2u and Av=-2v. Thus u and v are linearly independent and A has exactly two eigenspaces, each of dimension 1. Then
ABu=BAu=B(2u=2Bu.
Hence Bu is an eigenvector of A belonging to the eigenvalue 2 (Bu is not 0 because B is nonsingular and v is not 0). Hence Bu=su, for some scalar s, because the eigenspaces of A have dimension 1. Since the eigenvalues of B are 3 and -3, s is one of these two numbers. So s2 is plus or minus 6 and ABu=s2u.
Similarly, ABv=s2v. But then AB=s2I, a scalar matrix with s2 an irrational square root of 6. This contradicts that AB is a rational matrix, being the product of rational matrices A and B. |
(advice, please, on inserting math symbols?)
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| « Last Edit: Mar 5th, 2006, 11:38am by Icarus » |
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Icarus
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Re: Can these rational matrices commute?
« Reply #12 on: Mar 5th, 2006, 11:48am » |
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To see the current state of math symbolry, see the [ FAQ : (updated) how to write math symbols? ] thread in the "Suggestions, help, and FAQ" forum, starting at Reply #41 (where I have linked).
We used to have a fairly nice symbolry on hand, that William custom-added to the site as a set of smileys. Unfortunately, he later had to upgrade the YaBB software he used, and in the process, the symbolry was lost. Being a busy grad student, he has never gotten around to re-adding the symbolry. However the gifs are still accessible, but you have to add them in as images, not smileys.
I have updated your post to use the surd image, though you will note that the math symbols are not compatible with hiding.
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Barukh
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Re: Can these rational matrices commute?
« Reply #13 on: Mar 12th, 2006, 12:09am » |
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After looking closer into JocK's solution, I feel I don't understand one step: JocK claims that if matrices A, B commute, then they have the same eigenvectors (since the matrix S consists of them).
Is it true? 
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ecoist
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Re: Can these rational matrices commute?
« Reply #14 on: Mar 12th, 2006, 10:35am » |
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Yes, Barukh. Commuting matrices must fix eigenspaces (If XY=YX and Xv=[lambda]v, then X(Yv)=Y(Xv)=Y[lambda]v=[lambda](Yv); so Yv lies in the [lambda] eigenspace of X.). Since the eigenspaces of A and B are 1-dimensional, they have the same eigenvectors, assuming A and B commute.
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