Hi,

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

Thanks.

2. 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

3. 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 $\displaystyle \varepsilon I$ to A and B (making them invertible), and then letting $\displaystyle \varepsilon\to0$. But of course that argument won't work over the sort of fields that algebraists like to use.

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

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

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

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

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

5. ops, I forgot to said that it's invertible matix.

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

6. 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:

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

matrices commute with ANY matrix and we're done.

Tonio