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Simultaneously diagonalizing matrices exactly

I have a bunch of matrices with integer coefficients that simultaneously commute. I know that there is a basis that simultaneously diagonalizes all of them, and I want to find it exactly so that I can recover all the corresponding eigenvalues as algebraic numbers.

I've tried casting to QQbar and using eigenvectors, but this occasionally tries to divide by zero for no reason I can discern. Any ideas?

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could you write the code?

( 2018-01-22 19:21:07 +0200 )edit

Please give us at least two of the many commuting matrices that can be diagonalized (simultaneously).

( 2018-01-22 21:55:42 +0200 )edit

Let me join the club of asksage junkies : we need your code to understand your problem and (hopefully) provide a solution.

( 2018-01-25 11:26:24 +0200 )edit

Please provide an example of matrices for the computation you are asking about.

Ready-to-copy-paste examples make it way easier to explore a question, thereby increasing chances of an answer.

( 2018-04-03 09:51:29 +0200 )edit

1 Answer

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Getting a simultaneous basis of diagonalization for a list of diagonalizable matrices that pairwise commute is just a matter of getting a diagonalization basis for any one of them!

This can be obtained with jordan_form with the argument transformation=True.

For example, starting from these two matrices a and b:

sage: a = matrix(QQ, [[1, 1, 1], [-2, 3, 2], [0, 1, 2]])
sage: b = matrix(QQ, [[-1, 1, 1], [1, 1, -1], [-3, 1, 3]])


Check that they commute:

sage: a * b
[-3  3  3]
[-1  3  1]
[-5  3  5]
sage: b * a
[-3  3  3]
[-1  3  1]
[-5  3  5]
sage: a * b == b * a
True


If we diagonalize them separately:

sage: da, pa = a.jordan_form(transformation=True)
sage: db, pb = b.jordan_form(transformation=True)


we can notice that the transformation matrices pa and pb have the same columns, permuted:

sage: pa
[ 1  1  0]
[ 1  0  1]
[ 1  1 -1]
sage: pb
[ 0  1  1]
[ 1  1  0]
[-1  1  1]


Obtaining the simultaneous diagonal form is just a matter of reordering the columns of:

sage: da
[3|0|0]
[-+-+-]
[0|2|0]
[-+-+-]
[0|0|1]
sage: db
[2|0|0]
[-+-+-]
[0|1|0]
[-+-+-]
[0|0|0]


For instance, use the transformation matrix pa to get simultaneous diagonalizations:

sage: qa = ~pa
sage: qa * a * pa
[3 0 0]
[0 2 0]
[0 0 1]
sage: qa * b * pa
[1 0 0]
[0 0 0]
[0 0 2]

more

Comments

2

If A has a higher dimensional eigenspace then an eigenvector of A is not necessarily an eigenvector of B. Example: A=[0,1; -1,-1] and B=[1,0; 0,1]. They commute, B is already in diagonal form, and A is not. You need to intersect the eigenspaces of A with those in B and make a basis out of those smaller spaces.

( 2018-04-04 17:36:03 +0200 )edit

Good point.

( 2018-04-04 21:10:21 +0200 )edit

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Asked: 2018-01-22 18:43:40 +0200

Seen: 1,694 times

Last updated: Apr 03 '18