• Mar 19th 2010, 10:14 AM
matrix3599
Hi,

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

Thanks.
• Mar 19th 2010, 05:32 PM
tonio
Quote:

Originally Posted by matrix3599
Hi,

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

Thanks.

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 AM
Opalg
Quote:

Originally Posted by tonio
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.

As an analyst, I would deduce the general case from the invertible one by adding $\varepsilon I$ to A and B (making them invertible), and then letting $\varepsilon\to0$. But of course that argument won't work over the sort of fields that algebraists like to use. (Itwasntme)
• Mar 20th 2010, 05:40 AM
tonio
Quote:

Originally Posted by Opalg
As an analyst, I would deduce the general case from the invertible one by adding $\varepsilon I$ to A and B (making them invertible), and then letting $\varepsilon\to0$. But of course that argument won't work over the sort of fields that algebraists like to use. (Itwasntme)

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 $A$ is singular then $\exists\, \epsilon\in\mathbb{R}\,\,\,s.t.\,\,\,\epsilon I+A$ 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 $\mathbb{R}^{n^2}$ , but this is a little work the OP would have to do.

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

$adj\left((\epsilon I+A)(\epsilon I+B)\right)=adj(\epsilon I+B)adj(\epsilon I+A)$ (we can take an $\epsilon$ that'll work for both matrices), and then another job for the OP would be to

prove that $adj(\epsilon I+A)\xrightarrow [\epsilon\to 0]{} adj A$ , which can prove to be a little messy since the entries of $adj A$ 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 PM
matrix3599
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 PM
tonio
Quote:

Originally Posted by matrix3599
ops, I forgot to said that it's invertible matix.

tonio, thak you.
easy proof , but I am looking for the Linear proof.

Well, as already said, the case for both matrices invertible is easy:

$adj(A)=A^{-1}\,|A|\cdot I\,,\,adj(B)=B^{-1}\,|B|\cdot I\Longrightarrow \,adj(AB)=(AB)^{-1}$ $|AB|\cdot I=B^{-1}A^{-1}\,|A||B|\cdot I$ ...and now use that scalar

matrices commute with ANY matrix and we're done.

Tonio