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I've seen different intuitive explanations on math SE, but none of them has really clicked for me. I cobbled together pieces of the explanations I've seen and come up with something that feels like it makes more sense to me. Is the following a correct interpretation of why $rank(A^T) = rank(A)$?

Explanation: Suppose that $A \in \mathbb{R}^{n\times m}$ has rank $k$. This means that the image of $A$'s linear transformation is a $k$ dimensional subspace of $\mathbb{R}^n$. $A^T$ must then take this $k$ dimensional subspace to a subspace of $\mathbb{R}^m$ that is of dimension at most $k$. This means that $$rank(A^T) \leq k = rank(A).$$ We can then apply the same result starting with $A^T$ so that $rank((A^T)^T) \leq rank(A^T)$, which means that $rank(A) \leq rank(A^T)$. Taken together, we have $rank(A) = rank(A^T)$.

Question: Does this make sense? What would I need to do to make this more rigorous?

Edit: With the help of @Paweł Czyż and some more digging, this is my updated argument.

If $A^TAx = 0$, then $x^TA^TAx = ||Ax||^2 = 0$ which means that $Ax = 0$ (by properties of norms). This shows that $N(A) = N(A^TA)$. Because $A^TA$ and $A$ have the same input dimension, the rank-nullity theorem tells us that $rank(A) = rank(A^TA)$. This completes the argument because we then have $rank(A^T) = rank((A^TA)^T) = rank(A)$.

jeg
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  • "$A^T$ must then take this $k$ dimensional subspace to a subspace of $\mathbb{R}^m$ that is of dimension at most $k$. This means that $\operatorname{rank}(A^T) \leq k = \operatorname{rank}(A).$" This does not follow. You've only shown that $A^T$ maps a subspace of $\mathbb{R}^k$ to a dim-at-most-$k$ space. To show $\operatorname{rank}A^T\leq \operatorname{k}$, you have to show that $A^T$ maps all of $\mathbb{R}^k$ to a dim-at-most-$k$ space. – Jair Taylor Jun 08 '20 at 15:15
  • I agree with $\mathrm{rk}, A = \mathrm{rk}, A^TA$ part. From this you can get $\mathrm{rk}, A^T = \mathrm{rk}, AA^T$, but I don't see how you obtained $\mathrm{rk}, AA^T = \mathrm{rk}, A^TA$ from here. (You can't simply take the transpose of both sides of the equation, as we have not proved it). – Paweł Czyż Jun 08 '20 at 18:51
  • Does this answer your question? If $\operatorname{rank}(A)=m$, can we say anything about $\operatorname{rank}(AA^t)$? – user26857 Jan 24 '24 at 18:03

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Consider a map $A\colon \mathbb R^m\to \mathbb R^n$. Its rank is simply $\mathrm{rk}\, A = \dim(\mathrm{im}\, A)$. The transposition map is the unique map $A^T\colon \mathbb R^n\to \mathbb R^m$ such that $$\langle A^Tb, a \rangle_m = \langle b, Aa \rangle_n \quad \text{for all}\; a\in \mathbb R^m,b\in\mathbb R^n.$$

Your argument only proves that $\mathrm{rk}\, A^TA \le \mathrm{rk}\, A$. In fact you can deduce the equality using the rank-nullity theorem and a simple lemma $\ker A^TA = \ker A$. (This follows easily from $\langle A^TA v, v\rangle = |Av|^2$).

With rank-nullity theorem you can also prove that $\mathrm{rk}\, A = \mathrm{rk}\, A^T$ – just apply it to the equality $\ker A^T = (\mathrm{im}\, A)^\perp$, which is trivial from the definition of the transposition given above.

Paweł Czyż
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