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I spoke to someone locally, and we think the real issue is which convention is used to define the singular values of a matrix. If one defines the singular values of a matrix $A$ to be the eigenvalues of the matrix $$\sqrt{A^TA}$$ then if $A$ is $m \times n$ with $m < n$ we have $\sigma_{\min}(A) = 0$ but $\sigma_{\min}(A^T) \neq 0$ in general. This agrees with the identity $\phi(A) = \sigma_{\min}(A)$.

However, if one defines the singular values of $A$ to be the diagonal entries of the matrix $\Sigma$ in the singular value decomposition $$A = U\Sigma V^T$$

then $A$ and $A^T$ have exactly the same singular values, and $\phi(A) \neq \sigma_{\min}(A)$ in general.

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I spoke to someone locally, and we think the real issue is which convention is used to define the singular values of a matrix. If one defines the singular values of a matrix $A$ to be the eigenvalues of the matrix $$\sqrt{A^TA}$$ then if $A$ is $m \times n$ with $m < n$ we have $\sigma_{\min}(A) = 0$ but $\sigma_{\min}(A^T) \neq 0$ in general. This agrees with the identity $\phi(A) = \sigma_{\min}(A)$.

However, if one defines the singular values of $A$ to be the diagonal entries of the matrix $\Sigma$ in the singular value decomposition $$A = U\Sigma V^T$$

then $A$ and $A^T$ have exactly the same singular values, and $\phi(A) \neq \sigma_{\min}(A)$ in general.