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If $A$ is an $n\times n$ matrix over a field, then adj$(A)$ is an $n\times n$ matrix (obtained from $A$) such that $$\mathrm{adj}(A)\,A=A\,\mathrm{adj}(A)=\mathrm{det}(A)I_n.$$

Question: If $B$ is an $n\times n$ matrix such that $AB=BA=\mathrm{det}(A)I_n$, then is $B=\mathrm{adj}(A)$? If not, then how can we characterize $\mathrm{adj}(A)$ without using minors/cofactors etc.?

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  • $\begingroup$ If $A$ is invertible, then obviously $B=(\det A)A^{-1}$ is unique, hence it is the adjoint matrix. $\endgroup$
    – Crostul
    Jul 29, 2015 at 8:18
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    $\begingroup$ The answer is clearly no. In particular you can take $A=0$. $\endgroup$ Jul 29, 2015 at 8:18
  • $\begingroup$ @Ofir: First part of question is clear now. $\endgroup$
    – Groups
    Jul 29, 2015 at 8:19
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    $\begingroup$ For invertible matrices the answer is yes, for matrices $A$ with rank smaller than $n-1$, adj$(A)=0$. $\endgroup$ Jul 29, 2015 at 8:22

2 Answers 2

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You can use the following theorem the characterize in a sense adj$(A)$.

Let $A$ be an $n\times n$ matrix. Then,

  1. If rank$(A)=n$, then adj$(A)=|A|A^{-1}$.

  2. If rank$(A)<n-1$, then adj$(A)=0$.

  3. If rank$(A)=n-1$, then rank(adj$(A))=1$.

The first part is clear, the second is easy, for the third you need some work.

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I give an answer to your second question, a characterization of the adjugate which does not directly rely on the minors:

Let $p$ be the characteristic polynomial of matrix $A$ of order $n$. Then $$p(x) = xq(x)+(-1)^n\det(A),$$ where $q$ is some other polynomial. Since by Cayley-Hamilton we have $p(A)=0$, we find $$Aq(A) = (-1)^{n+1}\det(A)I.$$

Multiplying by the adjugate $A^*$ on the left we get $$\det(A)q(A)=(-1)^{n+1}\det(A)A^*.$$

If $\det(A)\neq0$ then we find $$A^*=(-1)^{n+1}q(A) \quad (1).$$ If the field $K$ is infinite, this identity is true even if $\det(A)=0$: We can write (1) as $f(A)=0$ for invertible matrices $A$ of order $n$, where $f$ is polynomial in the entries of $A$; since $\det(A)$ is also polynomial in the entries of $A$, if $f$ is not the zero polynomial (in $K[x_{11},\ldots,x_{nn}]$) then $f·\det$ is not the zero polynomial and there must be some matrix $A$ such that $f(A)\det(A)\neq0$, which is not the case.

If the field is finite, consider the field extension $K\subseteq K(t)$ with $t$ transcendental over $K$. Then $A$, $A^*$, $q(A)$ still have coefficients in $K$, but $K(t)$ is infinite, thus the result applies.

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