$\newcommand{\span}{\operatorname{span}}\newcommand{\rank}{\operatorname{rank}}$I would like to prove that for every martix,
column rank = row rank
Let $A\in M_{m\times n}$ be some matrix.
fix
$R_A$ = vector space of the rows of $A$ , $C_A$ = vector space of the columns of $A$
$$\rank (R_A) = dim \ (\span \{R_1, R_2, \ldots, R_m\})$$
$$\rank (C_A) = dim \ (\span \{C_1, C_2, \ldots, C_n\})$$
let $x$ be some vector.
$$x\in \operatorname{Null}(A) \Leftrightarrow \forall i \ \ (1 \leq i \leq m) : \langle x,R_i\rangle = 0$$
(to be clear – I'm referring to the inner product of $x$ with each row of $A$)
Using the rank nullity theorem, $\dim \operatorname{Null}(A) = n – \rank(R_A)$
as $n$ = number of columns.
what I would like to do is to say:
$$\dim \operatorname{Null}(A) = n – \rank(R_A)$$
$$\dim \operatorname{Null}(A) = n – \rank(C_A)$$
therefore, $\dim \operatorname{Null}(A) = n – \rank(R_A)= n – \rank(C_A) \Longrightarrow \rank(R_A) = \rank(C_A) $
it that false? is using the rank- nullity theorem this way is cheating? or just doesn't prove what needs to be proven formally?
Best Answer
You slightly abuse the rank-nullity theorem. Since in writing it like that, you already assume that what you want is true.
To get back to your problem. I would transform the problem, however, to asking whether the rank of $A$ equals the rank of $A^\top$ (think about why this is equivalent). A proof of this statement can be found, for example, in https://yutsumura.com/column-rank-row-rank-the-rank-of-a-matrix-is-the-same-as-the-rank-of-its-transpose/