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$\newcommand{\span}{\operatorname{span}}\newcommand{\rank}{\operatorname{rank}}$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?

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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/

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  • $\begingroup$ I'm looking for a simple proof to this theorem, as all the proofs i saw were too long and complicated and i think that could be proven in a more concise form $\endgroup$ – Jneven Jul 18 '18 at 12:56
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    $\begingroup$ @Jneven The theorem you want to prove is quite significant. It's one of the deepest in beginning linear algebra. I suspect that there is no proof as "simple" as what you are hoping for $\endgroup$ – Ethan Bolker Jul 19 '18 at 10:04
  • $\begingroup$ @EthanBolker I mean I would like to love to have the most "elegant" proof, which is sometimes not so possible, but that is why I'm asking :) $\endgroup$ – Jneven Jul 19 '18 at 10:28
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    $\begingroup$ I don't think there is a really elegant solution here. It depends on what you are allowed to use but as @EthanBolker states, the theorem is mostly used in the beginning of linear algebra. You might be able to use deeper theorems but you should be aware that you don't use any circular reasoning $\endgroup$ – Stan Tendijck Jul 19 '18 at 11:36

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