# Proof verification: 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?

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/