Triangle inequality for Schatten norms

Question

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.

Answer 1

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, $\lambda_j^\downarrow(\tilde A) = \sigma_j^\downarrow(A)$.