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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?

Anon
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1 Answers1

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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.

A.Γ.
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  • I believe there might be a flaw in the argument:- "there exists r columns ... that are LI". Since rowrank(A) = colrank(A) , there exists (atleast) r LI columns in A. However if you take the projection of these r LI vectors on a lower dimension space (i.e. only r entries of the r column vectors) then the new r vectors need not be LI. So you can't conclude that the submatrix has r LI columns. – Anon Jun 21 '18 at 08:40
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    @Ashish There exist $r$ LI rows. Take those and build a $r\times n$ submatrix $B$. The row rank of $B$ is $r$, hence, the column rank is $r$ as well and there are $r$ LI columns in $B$. Take those and build a subsubmatrix $C$ of size $r\times r$. It has $r$ LI columns, then $\det C\ne 0$. – A.Γ. Jun 21 '18 at 08:46