Given a $n \times m$ real matrix $A$ of rank $r$ one can define its SVD as $A = UD V^T$ with $D$ being a $r \times r$ diagonal matrix and $U^TU = V^TV= I$. Here clearly the diagonal entries of $D$ are not guaranteed to be be all positive.
Hence in terms of the $D$ obtained from SVD does one define the trace norm as $\vert \vert A \vert \vert _{Tr} = \sum_{i=1}^r \vert d_i\vert $ ? (where $d_i$ is the $i^{th}$ diagonal element of $D$)
But one could also have defined $\vert \vert A \vert \vert _{Tr} = \sum_{i=1}^n \sigma_i(A)$ where $\sigma_i$ is the $i^{th}$ singular value of $A$ defined as $\sigma_i(A) = \sqrt{\lambda_i (A^T A)}$. (where $\lambda_i$ picks out the $i^{th}$ largest eigenvalue of its matrix argument and the $\sqrt{}$ is taking the positive square-root)
Which is the right way to define it?
For which of the two does one get a good bounding by the Frobenius norm?
Now using the SVD one would get $A^TA = VD^2V^T$ and it doesn't seem to follow from here that $\sqrt{\lambda_i(VD^2V^T)} = \vert d_i \vert $
In the singular value formulation a special case emerges when $A$ is symmetric and then we have $\sigma_i(A) = \vert \lambda_i (A) \vert $. This in the SVD formulation would correspond to $n=m, U =V$ but does it imply anything more in terms of the $d_i$?
First note that in the SVD of $A$, the diagonal entries of $D$ are guaranteed to be nonnegative. So the absolute values are unnecessary in the first definition of the trace norm.
The two definitions of the trace norm are equivalent. Note that $A = UDV^T \implies A^T A = V D^T D V^T \implies A^T A V = V D^T D$, which shows us that the columns of $V$ form a basis of eigenvectors for $A^T A$, and the corresponding eigenvalues are the diagonal entries of $D^T D$. So the singular values of $A$ (the diagonal entries of $D$) are the square roots of eigenvalues of $A^T A$.