Re: Linear algebra with pitfall..

On Sat, 12 Apr 2008 01:16:37 -0500, quasi <quasi@xxxxxxxx> wrote:

On Sat, 12 Apr 2008 00:46:30 -0500, quasi <quasi@xxxxxxxx> wrote:

On Sat, 12 Apr 2008 00:42:09 -0500, quasi <quasi@xxxxxxxx> wrote:

On Sat, 12 Apr 2008 00:30:52 -0500, quasi <quasi@xxxxxxxx> wrote:

On Sat, 12 Apr 2008 04:09:39 GMT, rob@xxxxxxxxxxxxxx (Rob Johnson)
wrote:

In article <ftp983$mcr$1@xxxxxxxxxxxxxxxx>,
"mina_world" <mina_world@xxxxxxxxxxx> wrote:
A, B in Matrix_2x2.
E is identity matrix.

True or false.
(A+B)(A-B) = E ==> AB = BA.

True.

Since (A+B)(A-B) = E, A-B is invertible. Thus, there is a C so that
(A-B)C = E. Thus, A+B = (A+B)E = (A+B)(A-B)C = EC = C. Therefore,
we have that (A-B)(A+B) = E. This means that

(A+B)(A-B) = E = (A-B)(A+B)

A^2 + BA - AB - B^2 = A^2 + AB - BA - B^2

2 BA = 2 AB

AB = BA

Essentially the same argument, but perhaps slightly simpler to see,
(and what I should have seen right away) is the following ...

(A + B) (A - B) = E
=> A + B and A - B are inverses of each other
=> (A - B) (A + B) = E
=> (A + B) (A - B) = (A - B) (A + B)
=> AB = BA
(after expanding and simplifying)

Here's a quick followup.

If A, B are n x n matrices with complex coefficients such that

A*(B^2) = I

must A commute with B?

Ok, the answer is yes (I can prove it).

So now, let me ask this ...

Question:

If A,B are n x n matrices with complex coefficients such that

f(A,B)*g(A,B) = 1

for some nonconstant polynomials f,g in C[x,y], must A and B commute?

Remark: I suspect the answer is no.

If the answer happens to be yes (fat chance, but hey, you never know
(until you do)), we could allow f,g to be in C<x,y>, the ring of
polynomials, with coefficients in C, in the non-commuting variables
x,y, and then re-ask the same question.

Well, I see now that the idea is silly, without more restrictions.

For example, let f(x,y) = g(x,y) = x.

Thus, f(A,B)*g(A,B) = 1 => A^2 = I, but that doesn't imply AB = BA.

As a minimum, the functions f,g besides being non-constant, must such
that at least one of them is non-constant in A and one is non-constant
in B.

I don't really know where I'm going with this -- I'm just fooling
around with the idea.

I have in mind a simpler version of the above problem, but I'll start
it in a new thread.

quasi
.

Relevant Pages

  • Re: Linear algebra with pitfall..
    ... E is identity matrix. ... B are n x n matrices with complex coefficients such that ... must A commute with B? ... for some nonconstant polynomials f,g in C, must A and B commute? ...
    (sci.math)
  • Re: math -- f(A)*g(B) = I => AB = BA?
    ... If A,B are n x n matrices with complex coefficients such that ... for some nonconstant polynomials f,g in C, must A,B commute? ... I suspect the answer is no. ...
    (sci.math)
  • math -- f(A)*g(B) = I => AB = BA?
    ... If A,B are n x n matrices with complex coefficients such that ... for some nonconstant polynomials f,g in C, must A,B commute? ... Remarks: ...
    (sci.math)
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