Given a matrix $X$ which has the following form: $\begin{bmatrix} 0 & A \\ 0 & 0 \end{bmatrix}$, where $A$ is again some matrix. How does the nuclear/trace norm of $||X||_{\text{tr}}$ relate to $||A||_{\text{tr}}$?
Can I say that they are equal because there exists some permutation matrix such that A i.e. is the upper left hand block. Furthermore permutation matrices are unitary and the Schatten norms are unitarily invariant which means $||X||_{\text{tr}} = ||A||_{\text{tr}}$?
On
If $A = \sum_{i=1}^{n} \sigma_i \boldsymbol{u}_i\boldsymbol{v}^*_i$, then
\begin{align} X &= \begin{bmatrix} \mathbf O & A\\ \mathbf O & \mathbf O \end{bmatrix}\\ &= \sum_{i=1}^n \sigma_i \begin{bmatrix} \mathbf O & \boldsymbol{u}_i\boldsymbol{v}_i^*\\ \mathbf O & \mathbf O \end{bmatrix}\\ &= \sum_{i=1}^n \sigma_i \begin{bmatrix} \boldsymbol{u}_{i}\\ \boldsymbol{0} \end{bmatrix} \begin{bmatrix} \boldsymbol{0} & \boldsymbol{v}^*_{i} \end{bmatrix} \end{align}
This is the SVD of $X$ and clearly $X$ and $A$ have the same signular values.
Your approach with the permutation matrix works out. An alternative argument would be to just use the definition.
Namely, by definition we have
\begin{align*} \Vert X \Vert_\mathrm{tr} = \mathrm{tr}(\sqrt{X^* X}). \end{align*} However, \begin{align*} X^* X = \begin{pmatrix} 0 & 0 \\ A^* & 0 \end{pmatrix} \begin{pmatrix} 0 & A \\ 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & A^* A \end{pmatrix} \end{align*} Thus, \begin{align*} \sqrt{X^* X} = \begin{pmatrix} 0 & 0 \\ 0 & \sqrt{A^* A} \end{pmatrix} \end{align*} and therefore \begin{align*} \Vert X \Vert_\mathrm{tr} = \mathrm{tr}(\sqrt{X^* X}) = \mathrm{tr}(\sqrt{A^* A}) = \Vert A \Vert_\mathrm{tr}. \end{align*}