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?

• 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 $n-1$. – tch Dec 30 '18 at 15:23
• @TylerChen : thank you for the reply, an adjoint matrix is the conjugate transpose of a matrix. I updated the post with some queries. – NetUser5y62 Dec 30 '18 at 15:35
• 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. – tch Dec 30 '18 at 15:53
• @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. – NetUser5y62 Dec 30 '18 at 16:02

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 $$\mathrm{rank}(A)\leq 2$$

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 $$\operatorname{rank}\operatorname{adj}(A)\neq1$$

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