# 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.

• Of course you need $n\ge2$. – egreg May 20 '16 at 17:33

## 1 Answer

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 $\operatorname{det}(A)=0$.

• The terms "adjoint" and "classical adjoint" are indeed used for what you refer to as the adjugate, see the top of the wiki page. – Omnomnomnom May 20 '16 at 17:43
• @Omnomnomnom I've seen the usage and am still inclined to at least discourage it, because I've never seen anyone comment on this linguistic quirk when they introduce the adjoint in the sense of inner product spaces. – Ian May 20 '16 at 18:07
• Discouraging its use is one thing, but it's probably better for the asker if the answerer sticks to terminology from whichever class/textbook the asker is coming from. – Omnomnomnom May 20 '16 at 19:02
• @Omnomnomnom Fair point. I've toned it down a bit in the answer. – Ian May 20 '16 at 19:09