Let $A_{m\times n}$ be a matrix.
Row-rank definition of rank$(A)$:-
Row-rank$(A)$: It is the number of linearly independant rows of matrix A (alternatively the dimension of the rowspace of A).
Col-rank$(A)$: It is the number of LI columns of matrix A.
$$ Rank(A) := \text{Rowrank}(A) = \text{Colrank}(A) $$
Determinantal definition of rank(A):- rank$(A)$ is the largest order of a non-zero minor of A.
I want to show the equivalence of the two definitions. For showing $ \text{determinantal-rank}(A) = r \Rightarrow \text{rowrank}(A) = r$: \begin{align} &\exists \text{ a submatrix of A, } N_{r\times r} \text{ s.t } |N| \neq 0. \\ &\Rightarrow \text{rows of } N_{r\times r} \text{ are LI .} \\ &\Rightarrow \text{r rows of } A \text{ are LI} \\ &\Rightarrow \text{rowrank}(A) \geq r \end{align} I am not able to show rowrank$(A) \leq r$ using the fact that all larger minors are vanishing. How to show this?
You have proved that $\text{determinantal-rank}(A) \ge r \Rightarrow \text{rowrank}(A) \ge r$. To prove the opposite: assume that $\text{rowrank}(A) \ge r$. Then there exists a submatrix $r\times n$ with LI rows. Since row rank is equal to column rank, there exist $r$ columns in this submatrix that are LI. You end up with a submatrix $r\times r$ with LI columns/rows, it has a non-zero determinant. Then $\text{determinantal-rank}(A) \ge r$.
Now $$ \text{determinantal-rank}(A) \ge r \iff \text{rowrank}(A) \ge r,\quad \forall r $$ implies that they are equal.