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Let the adjugate of a matrix be defined as the transpose of the cofactor matrix, denoted $A^{*}$. (Also termed the Classical Adjoint)

It can be proven that for any $n\times n$ matrix $A$,

1) if rank($A$)= $n$ then rank(adj($A$)) = $n$

2) if rank($A$) = $n-1$ then rank(adj($A$)) = $1$

2) if rank($A$) < $n-1$ then rank(adj($A$)) = $0$

Suppose we were given an arbitrary $n\times n$ matrix $A$. If we know that rank((adj($A$)) = $1$, does it necessarily imply that the rank of $A$ is $n-1$? For example, suppose $A$ has an unknown entry but it is given that rank(adj($A$)) =$1$, is it a valid approach to conclude that the rank of $A$ is $n-1$ and proceed to determine the unknown value by solving $\det(A)=0$?

Similarly for the other two statements. Are they single direction statements or are they really if and only if statements?

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  • $\begingroup$ I'm not familiar with the statement you posted. What is your definition of adjoint matrix? Either $\operatorname{rank} A = n$, $\operatorname{rank} A = n-1$, or $\operatorname{rank} A < n-1$. Assuming your statement is correct, if $\operatorname{rank}\operatorname{adj} A = 1$ and $\operatorname{rank} A \neq n-1$ you have a contradiction so $\operatorname{rank}(A)$ must be $n-1$. $\endgroup$ – tch Dec 30 '18 at 15:23
  • $\begingroup$ @TylerChen : thank you for the reply, an adjoint matrix is the conjugate transpose of a matrix. I updated the post with some queries. $\endgroup$ – NetUser5y62 Dec 30 '18 at 15:35
  • $\begingroup$ How do you use that definition of $A$ is not invertible? In any case, I don't think that is the definition of conjugate transpose. $\endgroup$ – tch Dec 30 '18 at 15:53
  • $\begingroup$ @TylerChen thank you for correcting that, I didn’t realize that was just a property thats applies when $A$ is invertible. I have removed it from the definition. $\endgroup$ – NetUser5y62 Dec 30 '18 at 16:02
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1)Yes. If $\mathrm{rank}(\mathrm{adj}(A))=n$ then $\mathrm{adj}(A)$ is invertible, so $A=\mathrm{det}(A)\mathrm{adj}(A)^{-1}$ is invertible, because it can not be zero.

2)Yes. If $\mathrm{rank}(\mathrm{adj}(A))=1$ then $A$ is not invertible so $\mathrm{rank}(A)\leq n-1$ but by 3) we can not have $\mathrm{rank}(A)< n-1$, since it would imply $\mathrm{adj}(A)=0$. Thus $\mathrm{rank}(A)= n-1$.

3)Yes. If $\mathrm{adj}(A)=0$ then all $(n−1)×(n−1)$ minors of $A$ are zero, hence $\mathrm{rank}(A)\leq 2$

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If by adjoint of $A$ you mean adjoint operator, i.e. $\operatorname{adj}(A) = A^*$ then your claimed properties are not true. For instance, any Hermetian matrix satisfies $A=A^*$ so the rank of $A$ is the rank of $A^*$.

More generally, for any matrix $A$, the conjugate transpose $A^*$ always has the same rank.

As an explicit example, consider the $3\times 3$ matrix: $$ A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$

This is clearly rank $2$, but the conjugate transpose is also rank 2.

That said, if there exists some definition of "adjoint" satisfying the 3 properties you stated, then it is true that $\operatorname{rank}\operatorname{adj}(A)=1$ implies $\operatorname{rank}(A) = n-1$. To see this, note that if $\operatorname{rank}(A)$ did not equal $n-1$ then $\operatorname{rank}\operatorname{adj}(A)\neq1$

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  • $\begingroup$ $\mathrm{adj}(A)$ is the transpose of the cofactor matrix of $A$ which make it different from $A^*$ the transpose of the conjugate. $\endgroup$ – mouthetics Dec 31 '18 at 15:15

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