Adjoint of an adjoint of a matrix
Can you please help me on this question? $\DeclareMathOperator{\adj}{adj}$
$A$ is a real $n \times n$ matrix; show that:
$\adj(\adj(A)) = (\det A)^{n-2}A$
I don't know which of the expressions below might help
$$ \adj(A)A = \det(A)I\\ (\adj(A))_{ij} = (-1)^{i+j}\det(A(i|j)) $$
Editor's note: adjoint here refers to the classical adjoint.
$\endgroup$ 12 Answers
$\begingroup$I would discourage you from using the word "adjoint" in this context. This is an accepted usage of the word, but there is another concept in linear algebra which is always referred to by the word "adjoint". The two can be easily confused. An unambiguous word that can be used in this context is "adjugate", and I would encourage you to use this word.
Anyway, you know by the first property you stated that when $A$ is invertible, the adjugate of $A$ is a multiple of the inverse of $A$. So the adjugate of the adjugate is a multiple of the inverse of $A^{-1}$, so it is a multiple of $A$. All you need to keep track of is what the constant factors in each of these steps were.
When $A$ is not invertible the situation is quite simple, the result follows from the fact that $\operatorname{det}(A)=0$.
$\endgroup$ 4 $\begingroup$We know the property adj(A).A = |A|I
Now consider
adj(adj(A))*adj(A)=|adj A|I
Post multiply this with A
adj(adj(A))*adj(A)*A=|A|^(n-1).I.A
adj(adj(A))*|A|=|A|^(n-1).A
adj(adj(A))=|A|^(n-2).A
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