Dear professor Paul Marcoux

In the set M_n of nxn square matrices, the Zero matrix commutes with all other matrices, and in particular Zero matrix commutes with any non- diagonalizable matrix. Obviously, commute does not imply diagonalizable.

If we consider the set D_n of the square diagonalizable matrices,

the statement: Commutative matrices are simultaneously diagonalizable

is true over the set D_n. Mathematically, this is what I exactly meant, because the other case is trivial and well-known.

So, your suggestion is to make a general statement:

Diagonalizable commute matrices are simultaneously diagonalizable.

This will exclude all trivial cases. :-)

Best wishes

Dear Liming Yuan ,

If B = P(A) , where P is any polynomial, then

A and B are commuting matrices of the same size.

Example B = 5A3 + 2 A2 - A + 3(I)

where I is the identity.Then A and B are commuting matrices.

Also, if A is invertible, one can introduce negative powers of A.

Example: B = 5A-3 + 2 A2 - A-1 + 7I .

In general, if A and B are commuting matrices, then they are simultaneously diagonalizable.

Best regards

Let P1, P2, ....,Pn be a set of rank 1 oblique projectors, i.e

Pi=ai*bi', where ai and bi are n by 1 matrices, i = 1, 2,....n

and bi'*ai = Identity matrix of order n.

Let these n Projectors satisfy the following condition:

P1+P2+....+Pn= Identity matrix of order n

Then, A = x1P1+x2P2+....+xnPn

and B =y1P1+y2P2+....+ynPn

are simultaneously diagonalizable.

where x1,x2,...xn and y1, y2,...yn are arbitrary field scalars .

Look into the Matrix analysis book of Carl D Meyer for more details in this regard.

in above bi'*ai = 1, for all i =1,2,..,n

my sincere apologies for the typographical error.

And of course Pi*Pj=0 for all i and j unequal.

And of course Pi*Pj=0 for all i and j unequal.

Dear professor Paul Marcoux

Your counterexample is not applicable to the well-known result

" commuting matrices are simultaneously diagonalizable)

Notice that the statement implicitly includes the condition that

both matrices are diagonalizable! If not, then the statement has no meaning.

In your counterexample, the matrices are not diagonalizable, and then can't be considered as a contradiction.:-).

Best regards

Dear professor Paul Marcoux

In the set M_n of nxn square matrices, the Zero matrix commutes with all other matrices, and in particular Zero matrix commutes with any non- diagonalizable matrix. Obviously, commute does not imply diagonalizable.

If we consider the set D_n of the square diagonalizable matrices,

the statement: Commutative matrices are simultaneously diagonalizable

is true over the set D_n. Mathematically, this is what I exactly meant, because the other case is trivial and well-known.

So, your suggestion is to make a general statement:

Diagonalizable commute matrices are simultaneously diagonalizable.

This will exclude all trivial cases. :-)

Best wishes

Dear Prof. Issam Kaddoura

Thank you very much for sharing your knowledge. I appreciate your generosity and kindness.

Best regards,

Liming Yuan

Dear Professor Paul Marcoux

All that you mentioned of facts are correct. I understand your strict use of the linguistic exact meaning of some claims.

Sometimes, when we post short answers at RG platform, one focus on the main and important results related to the question without discussion the special or subsidiary cases that left to the follower. Unless the special cases themselves are the goal of the question. :-)

best wishes

Dear Liming Yuan ,

I would like to add the following remark about commuting matrices.

Why we prefer that A and B are commute matrices?

In the presence of commutative property, we have the advantage to use many of the relations that can be applied on real numbers. As in the example:

(A + B)2 = A2 + 2AB + B2 this relation is true iff A and B are commute.

But this relation is wrong if A and B are not commute.

Another main advantage:

If we know the eigenvalues of A and B, then we can determine the eigenvalues of any combinations of A and B, that is, the set of the eigenvalues of

F(A,B) are F(λ,μ ) where λ and μ are the eigenvalues of A and B respectively

corresponding to the common eigenvectors and F is a polynomial in 2 variables.

This can be extended to polynomials of several variables.

Matrices are nice mathematical objects. :-).

Best regards

The following interesting characterization/condition of commuting matrices is given as follows (see, for example, J. H. M. Wedderburn, Lectures on matrices, American Mathematical Society, Colloquium Publications, Volume 17, 1984, page 106, Theorem 2): “If the matrix B commutes with every matrix that commutes with A, then B is a scalar polynomial of A.”

Dear Dr Peter Breuer,

The proof of a result in my previous answer was generalized in the paper:Paco Lagerstrom, A proof of a theorem on commutative matrices, Bulletin of the American Mathematical Society, Vol. 51, No. 8 (1945), pp. 535-536, given as follows:

THEOREM 1. Let V be a finite-dimensional vector space over a commutative field K and A and B linear endomorphisms of V such that B commutes with every endomorphism that commutes with A. Then there exists a polynomial Q(x) with coefficients in K such that B = Q(A).

When AB = BA and B is non-singular, then A = B*A*B^(-1), i.e., A and B are 'similar'. In particular, when R corresponds to any n-dimensional rotation (i.e., it is orthonormal), then B = R*A*R^(-1) commutes with A, and that is a lot of matrices, described by n(n-1)/2 real parameters (n-dimensional Euler angles).

Also, as noted by Romeo Meštrović, any polynomial in A commutes with A. Actually when matrices A, B, C, .., Z all commute with each other, then any polynomial in all of them commutes with any of them (specifically, if A commutes with B then A commutes with any polynomial in A and B).

By limit transition it follows that when f(z) is a function with convergent Taylor expansion, then f(A) commutes with A. That, again, is a mighty family.

In physics, when H is the quantum Hamiltonian matrix of a system, the group of non-singular matrices that commutes with it defines the symmetry group of the system. Factoring out the (trivial) rotations, this leaves any nontrivial symmetries of the system (an invariant subgroup) ... etc.

It is interesting to notice that if the matrices A and B commutes, then for any positive integer n, (A + B)^n can be expanded by the binomial (Newtoon) formula. Namely, proceeding in the same manner as in the proof of the classical binomial formula (by induction on n or using a combinatorial approach, or by another way), it can be proved that if A and B are square matrices such that AB = BA, then

(A + B)^n = ∑_(k = 0)^nC(n, k)A^(n - k)B^k,

where C(n, k)=n!/(k!(n - k)!) are usual binomial coefficients (k = 0, 1, …,n).

Question. Is it true that the equality (A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3 for square matrices A and B necessarily implies that AB = BA?

If A_1, A_2,...,A_m are square matrices such that all commute with each other, then for any positive integer n, (A_1 +A_2 +∙∙∙ + A_m)^n can be expanded by the multinomial formula (theorem). The proof(s) is (are) the same as those (these) of multinomial formula (theorem) (see,. e.g., https://en.wikipedia.org/wiki/Multinomial_theorem and https://math.stackexchange.com/questions/568973/combinatorial-proof-of-multinomial-theorem-without-induction-or-binomial-theor).

Of course! The multinomial formula (and in particular, the binomial formula) holds for a power of sum of variables (elements) of any commutative ring (not necessarily with unity).

Concerning the question raised by Romeo Meštrović :

Question. Is it true that the equality (A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3 for square matrices A and B necessarily implies that AB = BA?

The answer is No. Counterexample:

Let A2x2 = [-3 -3 ]

[1 1]

B2x2 = [1 1 ]

[0 0 ]

the equality (A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3 holds.

But A and B are non commuting matrices (AB ≠ BA).

Best regards

If we substitue the matrices A = {{-3,-3},{1,1}} ({-3,-3} and {1,1} are the rows of A) and B= {{1,1},{0,0}} ({1,1} and {0,0} are the rows of B) proposed by Issam Kaddoura as a counterexample (a negative answer to my question), in the expression

(A + B)^3 - A^3 - B^3 - 3A^2B - 3AB^2, using Mathematica 9, we obtain as a result the (nonzero!) matrix {{0.,0.},{-3.,-3.}} (whose rows are {0.,0.} and {-3.,-3.}) (see below this expression and its result obtained in Mathematica 9).

({{-3,-3},{1,1}}+{{1,1},{0,0}})^3-({{-3,-3},{1,1}})^3-({{1,1},{0,0}})^3-

3({{-3,-3},{1,1}})^2.{{1,1},{0,0}}-3({{-3,-3},{1,1}}).({{1,1},{0,0}})^2

{{0.,0.},{-3.,-3.}}

Nevertheless, I believe that there exists a counterexample (similar to those proposed by Issam Kaddoura), not only in the space of square matrices of second order, but in the space of square matrices of every order greater than 1.

Dear Romeo Meštrović ,

I have spent more than one hour with calculations to answer your question.

(1) As a tradition, I expect at least a word of thank.

(2) I trust my hand calculations more than any program.

(3) Could you please check your calculations once more.

It seems you made a mistake.

Regards

Dear Issam Kaddoura,

Thank you very much for the effort to finding a desired counterexample.

In order to discover a “source of a mistake“, I present a calculation in Mathematica step by step, given below, which would be useful for a comparison with your calculation:

(A+B)^3:

({{-3,-3},{1,1}}+{{1,1},{0,0}})^3

{{-8,-8},{1,1}}

A^3:

({{-3,-3},{1,1}})^3

{{-27,-27},{1,1}}

B^3:

({{1,1},{0,0}})^3

{{1,1},{0,0}}

3A^2B:

3({{-3,-3},{1,1}})^2.{{1,1},{0,0}}

{{27.,27.},{0.,0.}}

3AB^2:

3({{-3,-3},{1,1}}).({{1,1},{0,0}})^2

{{-9,-9},{3,3}}

Finally,

(A+B)^3-A^3--B^3-3A^2B-3AB^2

{{-8,-8},{1,1}}-{{-27,-27},{1,1}}-{{1,1},{0,0}}-{{27.`,27.`},{0.`,0.`}}-{{-9,-9},{3,3}}

= {{0.,0.},{-3.,-3.}}!?

(If this is not true, this is a mistake of Wolfram Mathematica!)

Best regards

Dear Romeo Meštrović,

Notice that all of your codes are wrong.

The cubic operations are acting on the entries and not the matrices!!

Could you please read the calculations in the attached pdf file.

Best regards

Dear Issam Kaddoura,

You are right, i.e., your counterexample is correct. Namely, using dot code “.”(instead of power code “^”) for the product of matrices, Mathematica 9 confirms your result (a counterexample) as given below (strange!?).

(A+B)^3:

{{-2,-2},{1,1}}.{{-2,-2},{1,1}}.{{-2,-2},{1,1}}

{{-2,-2},{1,1}}

A^3:

{{-3,-3},{1,1}}.{{-3,-3},{1,1}}.{{-3,-3},{1,1}}

{{-12,-12},{4,4}}

B^3:

({{1,1},{0,0}}).({{1,1},{0,0}}).({{1,1},{0,0}})

{{1,1},{0,0}}

3A^2B:

3{{-3,-3},{1,1}}.{{-3,-3},{1,1}}.{{1,1},{0,0}}

{{18,18},{-6,-6}}

3AB^2:

3{{-3,-3},{1,1}}.{{1,1},{0,0}}.{{1,1},{0,0}}

{{-9,-9},{3,3}}

Finally,

(A+B)^3-A^3--B^3-3A^2B-3AB^2

{{-2,-2},{1,1}}-{{-12,-12},{4,4}}-{{1,1},{0,0}}-{{18,18},{-6,-6}}-{{-9,-9},{3,3}}

={{0,0},{0,0}}

I expect from you related counterexamples for some higher dimensions of the spaces of matrices order (which I will check in my Mathematica via dot code “.”).

Best regards

Dear Romeo Meštrović ,

If we find a counterexample for 2x2 matrices, then we can extend this counterexample for any order.

It is a straightforward task. As an example; See the attached file that shows how to extend lower order examples

into 3x3 matrices and we proceed to any order.

Best regards

(A+B)(A+B)(A+B) = AAA + AAB + ABA + BAA + ABB + BAB + BBA + BBB

When things do not commute you MUST maintain their order, on both sides.

In general, AAB+ABA+BAA can NOT be simplifed further to anything like 3A^2B .

The same for ABB + BAB + BBA.

------

As for counterexamples, etc: Keep in mind that "if B commutes with A then it commute with all powers of A". But not vice versa! When B does not commute with A, it may still commute with some power of A (for example with A^2) in which case it then commutes with all powers of that power (for example with all A^(2k)). Knowing this one can create any number of weird looking examples.

Let k be any fixed positive integer. If the square k x k matrices matrices A and B commute, then for any positive integer n ≥ 2 the following equality holds:

(A^n - B^n) = (A-B)(A^(n-1) + A^(n-2)B+ ∙∙∙ + AB^(n-2) + B^(n-1)). (1)

Namely, it is easily to prove (for example, by induction on n) that the equality (1) holds if A and B are elements of any ring R such that AB = BA.

Clearly, for n = 2, the equality (1) is equivalent to AB = BA.

Question. Do there exist positive integers k ≥ 2, n ≥ 3 and two square k x k non-commutative matrices A and B satisfying the equality (1)?

Dear Romeo Meštrović ,

Yes, there exist infinite numbers of noncommuting matrices satisfy the difference equation. Your request is attached for n =3 and k =2, as a pdf file.

Best regards

It is interesting to notice that a calculation shows that the matrices A = {{-3,-3},{1,1}} and B = {{1,1},{0,0}} presented as a counterexample in the previous answer by Issam Kaddoura also satisfy the equality

(A^3 - B^3) = (A-B)(A^2 + BA + B^2),

i.e., replacing AB by BA in the equality (A^3 - B^3) = (A-B)(A^2 + AB + B^2), it remains true. Subtracting the previous two equations, it follows that

(A-B)(AB - BA) = 0, which is sufficient for verifying my observation. Moreover, the equalities A^2B = ABA = {{6,6},{-2,-2}} and B^2A = BAB = {{-2,-2},{0,0}} hold true.

Dear Dr Peter Breuer,

In view of your observation and the fact that B^2= B, we have

ABBABBAAB = A^4B^5= A^4B = {{24,24},{-8,-8}}, while B^5A^4 = BA^4= {{16,16},{0,0}}, and hence, A^4B^5 ≠ B^5A^4.

Dear Dr. Peter Breuer,

By induction on n, it follows immediately that A^n = (-2)^(n - 1)A holds for each positive integer n. Hence, using the fact that B^m =B for each positive integer m, we find that

A^n B^m - B^n A^m = (-2)^(n - 1)(AB -(-2)^(m - n)BA). (1)

Hence, if m – n is even, then

A^n B^m - B^n A^m = (-2)^(n - 1)(AB - 2^(m - n)BA)

=(-2)^(n - 1){{-3 + 2^(1 + m -n), -3 + 2^(1 + m -n)},{1,1}}. (2)

If m – n is odd, then from (1) we have

A^n B^m - B^n A^m = (-2)^(n - 1)(AB + 2^(m - n)BA)

=(-2)^(n - 1) {{-3 - 2^(1 + m -n), -3 - 2^(1 + m -n)},{1,1}}. (3)

From the equalities (2) and (3) we see that A^n B^m ≠ B^n A^m for all pairs of positive integers m and n.

Moreover, since A^n and A are linearly dependent vectors (because of A^n = (-2)^(n - 1)A) and B^m =B for all positive integers n and m, it folllows that the linear space spanned by linear combinations of vectors A^n B^m, where n and m run over all positive integers, is a twodimensional linear space described as a set

{λA + µB: λ, µ ϵ R (or C)}.

Yes! As AB = {{-3,-3},{1,1}}.{{1,1},{0,0}} = {{-3,-3},{1,1}} = A, it follows that AB belongs to the set {λA + µB: λ, µ ϵ R (or C)}. However, my previous assertion can be corrected as follows.

The linear space spanned by linear combinations of vectors of the forms A^n B^m and B^s A^l, where n, m, s and l run over all positive integers, is a twodimensional linear space described as a set S defined by

S = {λA + µBA: λ, µ ϵ R (or C)}.

Proof. Since A^n = (-2)^(n - 1)A, B^m =B and AB = A, the linear space spanned by linear combinations of all vectors of the forms A^n B^m and B^s A^l, where n, m, s and l run over all positive integers, coincides to the set S defined by (1).