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For a matrix $A\in\mathbb R^{n\times m}$, we consider the vector of its singular values $\sigma = [\sigma_1,\dots,\sigma_{\min\{m,n\}}]^T$. We define the $p$-Schatten norm $$ \|A\|_{S,p} := \|\sigma\|_p $$ as the usual $p$-norm of $\sigma$. How do I show that this $p$-Schatten norm satisfies the triangle inequality?

Everywhere, where Schatten norms are used it is implied that it's actually a norm, but I don't have a good idea on how to prove this.

There is this related question on Ky Fan norms, but I actually don't understand the answer.

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  • $\begingroup$ Maybe it helps to know that $\DeclareMathOperator{\tr}{tr}$ $$ \|A\|_{S,p} = \Bigl(\tr\bigl((A^TA)^{p/2}\bigr)\Bigr)^{1/p}. $$ But I still don't know where to go from there. $\endgroup$ – Stefan Hante Jun 14 '18 at 9:00
  • $\begingroup$ Pedersen's "Analysis Now" proves this (for arbitrary trace-class operators on a Hilbert space) with polar decomposition; I imagine something similar can be done with singular value decomposition. Bhatia's "Matrix Analysis" has a proof that handles arbitrary unitarily invariant norms. I've been looking at both but I have failed to fined a quick, punchy justification of subadditivity. $\endgroup$ – Ben Grossmann Jun 14 '18 at 15:45
  • $\begingroup$ If you can access Bhatia's Matrix analysis, Theorem IV.2.1 will do the job $\endgroup$ – Ben Grossmann Jun 14 '18 at 15:50
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If we adapt the proof of IV.2.1 from Bhatia's Matrix Analysis, it suffices to prove the following facts:

  • The Ky-Fan norms satisfy the triangle inequality
  • For vectors $x, y \in \Bbb R^n_+$: if $$ \sum_{j=1}^k x_k^\downarrow \leq \sum_{j=1}^k y_k^\downarrow, \quad k = 1,\dots,n $$ then $\|x\|_p \leq \|y\|_p$

  • For vectors $x, y \in \Bbb R^n_+$: if $x \leq y$ (entrywise), then $\|x\|_p \leq \|y\|_p$


The result regarding the Ky-Fan norms can be proven as follows:

  • If $A$ is a Hermitian matrix and $\lambda^{\downarrow}_j$ denotes the eigenvalues in decreasing order, then $$ \sum_{j=1}^k \lambda_k^\downarrow = \max \sum_{j=1}^k x_j^TAx_j $$ where the maximum is taken over orthonormal $k$-tuples of vectors $\{x_1,\dots,x_k\} \subset \Bbb R^n$ (known as Ky-Fan's max principle).
  • Consequently: for Hermitian matrices $A,B$, we have $$ \sum_{j=1}^k \lambda_k^\downarrow(A + B) \leq \sum_{j=1}^k \lambda_k^\downarrow(A) + \sum_{j=1}^k \lambda_k^\downarrow(B) $$
  • The same applies to the block matrices $$ \tilde A = \pmatrix{0&A\\A^T & 0}, \quad \tilde B = \pmatrix{0 & B\\B^T & 0} $$ and notably, $\lambda_j^\downarrow(\tilde A) = \sigma_j^\downarrow(A)$.
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  • $\begingroup$ That's nice. What do the little downwards arrows mean? $\endgroup$ – Stefan Hante Jun 15 '18 at 11:11
  • $\begingroup$ @Wauzl it means that the relevant quantities are listed from greatest to least. Of course, this is the usual convention for singular values. $\endgroup$ – Ben Grossmann Jun 15 '18 at 13:38
  • $\begingroup$ @BenGrossmann Could you explain why the last equality holds? Can't the eigenvalues of $\tilde A$ even be negative? $\endgroup$ – Florian Ente Dec 13 '20 at 20:19
  • $\begingroup$ @ludoer If $A = U \Sigma V^T$ is a singular value decomposition, then we have $$ \tilde A = \pmatrix{U & 0\\0 & V} \pmatrix{0 & \Sigma\\ \Sigma^T & 0} \pmatrix{U & 0\\ 0 & V}^T. $$ With that in mind, consider the eigenvalues of $$ \pmatrix{0 & \Sigma\\ \Sigma^T & 0}. $$ $\endgroup$ – Ben Grossmann Dec 14 '20 at 1:01
  • $\begingroup$ @ludoer Sorry, I misunderstood your question at first. The last equality holds only for $j = 1,\dots,\min\{m,n\}$. After that, we reach zero-eigenvalues and then the negative eigenvalues $-\sigma_j(A)$. $\endgroup$ – Ben Grossmann Dec 14 '20 at 1:03

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