Hi,

How can I proof that adj(AB)=adj(B)*adj(A)?

I tried to use with the characteristic of A*adj(A)=|A|*I

Thanks.

Printable View

- Mar 19th 2010, 10:14 AMmatrix3599Proof Adj characteristic
Hi,

How can I proof that adj(AB)=adj(B)*adj(A)?

I tried to use with the characteristic of A*adj(A)=|A|*I

Thanks. - Mar 19th 2010, 05:32 PMtonio

If A,B are invertible matrices the proof is very easy and short, but in the general case it is, imo, a very involved, long proof which uses, if I remember correctly, exterior products and devilish stuff like that. Even worse, the only proof I have is veryyyyy long...and in spanish.

Tonio - Mar 20th 2010, 02:01 AMOpalg
- Mar 20th 2010, 05:40 AMtonio

I suppose that's easier than the algebraic stuff, but then we'd need quite a few things to clear out:

1) Is it true that if is singular then is invertible? Of course it is since the set of singular nxn matrices is an open set in the set of all nxn matrices,

with the euclidean topology inherited from , but this is a little work the OP would have to do.

2) Thus, assuming the case for invertible matrices, we'd have that

(we can take an that'll work for both matrices), and then another job for the OP would be to

prove that , which can prove to be a little messy since the entries of are (n-1)x(n-1) determinants (the minors of A), so it'd probably

have to rely on the limit of functions composition (though this should not be a big problem, I think, since the determinant function is continuous on the set of all

square matrices)...

So it still looks like a not-too-short way to prove it...but still it looks easier than the algebraic one.

Tonio - Mar 20th 2010, 12:42 PMmatrix3599
ops, I forgot to said that it's invertible matix.

tonio, thak you.

easy proof , but I am looking for the Linear proof. - Mar 20th 2010, 01:52 PMtonio