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Math Help - Proof Adj characteristic

  1. #1
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    Proof 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.
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  2. #2
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    Quote Originally Posted by matrix3599 View Post
    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.

    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
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  3. #3
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    Quote Originally Posted by tonio View Post
    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.
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    Quote Originally Posted by Opalg View Post
    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.

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

    tonio, thak you.
    easy proof , but I am looking for the Linear proof.
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    Quote Originally Posted by matrix3599 View Post
    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
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