Subgradient of function $f_p(\mathbf{A})$ that has as output the $p^{th}$ largest singular value

Suppose I have a matrix $$\mathbf{A} \in \mathbb{R}^{n \times n}$$. Let $$\{\sigma_1,\sigma_2,.....,\sigma_n \}$$ be its $$n$$ singular values. Calculating a sub-gradient of its operator norm (largest singular value) is a well known result. Are more general results available i.e can we find the sub-gradient of the function ( of the matrix $$A$$) that produces as output the second largest singular value, third largest singular value and generally p-th largest singular value ($$p=1,2,...,n$$). Is it a well posed question at all.

• What do you mean by subgradient of an eigenvalue? Sep 23, 2019 at 9:15
• I meant , if we treat the $p-th$ largest singular value as a function of the matrix $\mathbf{A}$. Once we have a function of the matrix, then we can talk about subgradients of that function. I edited the question to reflect your question Sep 23, 2019 at 9:20

As an application of the Min-Max Theorem You get $$\sigma_{j+1}=\min_{x_1,..,x_j\in\mathbb{R}^n}\quad\max_{x\in\mathbb{R}^n||x||=1\\(x,x_1)=...=(x,x_j)=0}||Ax||.$$ As You see for $$j=0$$ you get $$\sigma_1=|||A|||$$ where $$|||.|||$$ is the operator norm. Of course the singular values are numbered in non-increasing order.Here $$j=0,..,n-1$$. The result can be generalized to ( possibly infinite-dimensional) Hilbert-spaces and compact operators.
• Thank you @Peter Melech. If I understand correctly, the above formulation expresses each of the singular values as an optimization problem. But does this formulation help us in finding the sub-gradient of function $f_p(\mathbf{A})$ Sep 23, 2019 at 10:02
• Good luck! At least $\{x\in\mathbb{R}^n :||x||=1,(x,x_1)=...=(x,x_j)=0\}$ is convex for $j=0,...,n-1$. Sep 23, 2019 at 10:25