Is it true in general, true for a specific field ($\mathbb R$/$\mathbb C$) or false? could it be that $A=adj(B),A=adj(C)$ but $B\not=C$?
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1What do you call adjoint here ? – Captain Lama May 23 '16 at 09:55
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Edited. I mean Adjugate matrix. – asaf92 May 23 '16 at 09:56
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1The question's body seems to be different from the title's... – DonAntonio May 23 '16 at 09:58
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Indeed, the title seems to ask for existence, and the body for unicity. – Captain Lama May 23 '16 at 10:00
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I ask both for existence and for unicity. – asaf92 May 23 '16 at 10:01
3 Answers
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The fundamental relation is $A\cdot \operatorname{adj}(A) = \operatorname{adj}(A) \cdot A = \det(A)I_n$, so if $A$ is inverible $\operatorname{adj}(A) = \det(A)A^{-1}$.
If you start from some invertible $B$, then you want to solve the equation $B^{-1} = \frac{1}{\det(A)}A$. So the solution, if it exists, must me of the form $A = \lambda B^{-1}$. Then $\det(A) = \lambda^n (\det(B))^{-1}$ and the equation becomes $B^{-1} = \lambda^{-n}\det(B)\cdot \lambda B^{-1}$, so $$\lambda^{n-1} = \det(B).$$
Thus there are as many solutions as there are $(n-1)$th roots of $\det(B)$ is the field : there may be none (if the field if $\mathbb{R}$, $n$ is odd and $\det(B)$ is negative for instance), and there may be several (in $\mathbb{C}$ there are always $n-1$ solutions for instance).
Captain Lama
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No, what I wrote is correct. The equation is $B = \operatorname{adj}(A)$, so $B = \det(A)A^{-1}$, which is equivalent to what is written. – Captain Lama May 23 '16 at 15:54
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We know that for any square $\;n\times n\;$matrix $\;A\;$ we have
$$A\cdot\text{Adj}\,A=|A|I\implies|\text{Adj}\,A|=|\,|A|\cdot A^{-1}|=|A|^{n-1}$$
if $\;A\;$ is invertible . Thus
$$A=\text{Adj}\,B\implies|A|=|\text{Adj}\,B|=|B|^{n-1}$$
Thus, for example, no rational $\;3\times3\;$ matrix $\;A\;$ with $\;|A|=2\;$ can be the adjugate of other rational matrix, since then we'd get the contradiction $\;2=|B|^2\;$ , or no real matrix of the same order with negative determinant.
Perhaps over an algebraically closed field that can be possible, say over $\;\Bbb C\;$ .
DonAntonio
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This only disproves that the claim is not true over $\mathbb R$, but how can I prove that it is true over $\mathbb C$? – asaf92 May 23 '16 at 12:20
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@PanthersFan92 Disclaims over the reals or over any non-algebraically closed field. Over $;\Bbb C;$ I can't say, but Captain's answer seems to give a good start. – DonAntonio May 23 '16 at 12:40
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Suppose $A\in GL_n(F)$ for an arbitrary field $F$. Using the identity $\operatorname{adj}(\operatorname{adj}(M)) = (\det(M))^{n-2} M$, we see that if $A=\operatorname{adj}(B)$, then $\operatorname{adj}(A)=\operatorname{adj}(\operatorname{adj}(B)) = (\det(B))^{n-2} B$. It follows that $B=k\operatorname{adj}(A)$ for some scalar $k$ and $$ \operatorname{adj}(B) =\operatorname{adj}(k\operatorname{adj}(A)) =k^{n-1}\operatorname{adj}(\operatorname{adj}(A)) =k^{n-1}(\det(A))^{n-2} A. $$ Consequently, $B$ exists if and only if the equation $k^{n-1}\det(A)^{n-2}=1$ is solvable over $F$, and $B$ is unique if and only if the solution $k$ is unique.
