I would like to understand how the following fact : detA⋅I n ⋅adj(adj(A))=(detA) n−1 ⋅A
allows us to determine/assert that if A isn't invertible then (adj(adj(A)) = 0.
If A isn't invertible I can't devide by detA.....
Thanks , Guy
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What does your question refer to? Do you have a link? – Julian Kuelshammer Oct 20 '12 at 16:51
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http://math.stackexchange.com/questions/92837/proof-mathrmadj-mathrmadja-mathrmdetan-2-cdot-a-for-a/92842#92842 – SyndicatorBBB Oct 20 '12 at 16:58
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I could understand all the proof but I don't understand how it helps to assert that if A isn't invertible then adj(adj(A)) = 0. – SyndicatorBBB Oct 20 '12 at 16:59
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As you already noted: This is not possible but it is not claimed in the link you provided.
Julian Kuelshammer
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It is claimed in the comments. Do you mean It's not the right way to prove it? – SyndicatorBBB Oct 20 '12 at 17:02
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I might confuse you. I want to proof that if A isn't invertible then adj(adj(A)) is zero. Isn't the link I provided is the right way to do so>? – SyndicatorBBB Oct 20 '12 at 17:03
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No it is not, and as I said no one claims it there. Just read the comments, can't find it. – Julian Kuelshammer Oct 20 '12 at 17:05
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It is so weird because the following fact was written by my teacher : If A has rank n-1 then adj(adj(A)) = 0. – SyndicatorBBB Oct 20 '12 at 17:06
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The claim is right, but what you thought of as a proof is not. – Julian Kuelshammer Oct 20 '12 at 17:07
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There is one attempt in the components of the link you provided. – Julian Kuelshammer Oct 20 '12 at 17:20
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