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.?

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

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.

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.