Re: Linear algebra with pitfall..

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.

quasi
.

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