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Assume A is an $m\times n$ matrix with real-valued entries. Is it always true that $\operatorname{rank} A = \dim \operatorname{Col} A = \dim\operatorname{Row} A = \dim \operatorname{Col} A^T = \operatorname{rank} A^T$?

Lozansky
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    See here. Maybe some of the linked questions might be interesting for you, too. – Martin Sleziak Jan 12 '16 at 18:23
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    @MartinSleziak I think this question is not quite the same as the more general question in your first link, because here we have the special assumption that $A$ has real-valued entries, which allows us to use the fact that the range of $A^T$ is the orthogonal complement of the null space of $A$. A similar proof can be given in the general case, but we must substitute annihilators for orthogonal complements (this is explained in Lax's linear algebra book, for example) and I think it's not quite as obvious. – littleO Jan 12 '16 at 18:32
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    @littleO Well, I did not vote to close, I have just provided links. – Martin Sleziak Jan 12 '16 at 18:32

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Yes, it's true, you can prove it using the fact that the range of $A^T$ (denoted here by $\text{Im}(A^T)$ is the orthogonal complement of the null space of $A$ (denoted here by $\text{null}(A))$. This tells us that \begin{equation} \text{dim null}(A) + \text{dim Im}(A^T) = n. \end{equation}

From the rank-nullity theorem, we have: \begin{equation} \text{dim null}(A) + \text{dim Im}(A) = n. \end{equation}

Comparing these two equations, we see that $\text{dim Im}(A) = \text{dim Im}(A^T)$.

littleO
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