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The title pretty much says it all, I need to prove that $\text{rank}(A) = \text{rank}(A^T)$ using SVD. It seems quite trivial, but I'd like to hear a second opinion. My thoughts are exposed below.

Based on the SVD theorem, for a matrrix $A \in \mathbb{R}^{m \times n}$ there exist othogonal matrices $U \in \mathbb{R}^{m \times m}$ and $V \in \mathbb{R}^{n \times n}$ such that $$ U^TAV = \text{diag}(\sigma_1, \sigma_2, ..., \sigma_p), \quad p = \min\left\{m, n\right\} $$

and $\sigma_i$ are the singular values of $A$. If it holds that $\sigma_1 \geq \sigma_2 \geq ~ ... ~ \geq \sigma_r > \sigma_{r+1} = ~ ... = \sigma_p = 0$, then it is true that $\text{rank}(A) = r$, meaning that the rank of $A$ is the count of non-zero singular values.

Trivially, if we consider that that $$ U^TAV = \text{diag}(\sigma_1, \sigma_2, \dots, \sigma_p) = \text{diag}(\sigma_1, \sigma_2, ..., \sigma_p)^T = V^TA^TU $$

it is obvious that the singular values of $A$ are also singular values of $A^T$, hence the count of non-zero singular values for both coincide. Bottom line is that $\text{rank}(A) = \text{rank}(A^T)$.

I think it would be OK, but I might be missing something, so please correct me if so. Thank you in advance for your time! :)

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    $\begingroup$ Looks fine with me. The fact that $p=\min\{m,n\}$ means that $\Sigma$ is identical in both cases. $\endgroup$ – Michael Grant Oct 28 '14 at 14:40
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The proof you provided has no issues.

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Here is dyadic viewpoint:

Suppose the SVD is given by $A = U \Sigma V^*$ where $\Sigma$ has the same dimensions as $A$. We note that $U,V$ are unitary, hence their columns are linearly independent.

If we let $v_1,...,v_m$ be the columns of $V$, then by looking at $Ax$ we can see that $A = \sum_{k=1}^r \sigma_k u_k v_k^*$.

It is easy to see that $\operatorname{rk} A \le r$ since ${\cal R} A \subset \operatorname{sp} \{ u_1,...,u_r \}$, and since $Av_k = \sigma_k u_k$, we see that $\operatorname{rk} A = r$.

Similarly, since $A^* = \sum_{k=1}^r \sigma_k v_k u_k^*$, we see that $\operatorname{rk} A^* \le r$ since ${\cal R} A^* \subset \operatorname{sp} \{ v_1,...,v_r \}$, and since $A^*u_k = \sigma_k v_k$, we see that $\operatorname{rk} A^* = r$.

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