Re: Still ,Matrices, A is square: If AB = Identity, is B the inverse?
- From: "linux" <xx@xxxxxxxxxx>
- Date: Sat, 25 Feb 2006 16:25:10 +0100
"Chip Eastham" <hardmath@xxxxxxxxx> a écrit dans le message de news:
1140876942.794455.279860@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
linux wrote:
"Timothy Murphy" <tim@xxxxxxxxxxxxxxxxxxxxxx> a écrit dans le message de
news: BQXLf.6471$j7.223578@xxxxxxxxxxxxxxxxx
linux wrote:
If In is a matrix the coefficients of which belong to a some ring, the
result is it really still? I think that yes. Can one prove him(it)
without
determinant?
AB = I => BA = I.
One way to see this is to consider the minimal equation of A
(ie the polynomial of minimal degree satisfied by A), say
x^d + a_1x^{d-1} + ... + a_d = 0.
Then a_d <> 0, since otherwise you would get an equation of lower degree
on multiplying on the right by B.
Now it follows that you get a right and left inverse C, say,
as a polynomial in A.
It is easy to show that the inverse is unique, ie B = C.
--
Timothy Murphy
Thanks.
The ring is commutative ?
Confusion.
If A=(a_ij) and the a_ij belong to a ring not commutative , the result is
not true?
The polynomial of minimal degree satisfied by A can is a_0 x^d + a_1x^{d-1}
+ ... + a_d = 0. .
And a_d/a_0 do not exist !
.
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