Simultaneously diagonalizing matrices exactly

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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]