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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.
Originally Posted by matrix3599
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
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.
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
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
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