# Why is minimizing the nuclear norm of a matrix a good surrogate for minimizing the rank?

A method called "Robust PCA" solves the matrix decomposition problem

$$L^*, S^* = \arg \min_{L, S} \|L\|_* + \|S\|_1 \quad \text{s.t. } L + S = X$$

as a surrogate for the actual problem

$$L^*, S^* = \arg \min_{L, S} rank(L) + \|S\|_0 \quad \text{s.t. } L + S = X,$$ i.e. the actual goal is to decompose the data matrix $X$ into a low-rank signal matrix $L$ and a sparse noise matrix $S$. In this context: why is the nuclear norm a good approximation for the rank of a matrix? I can think of matrices with low nuclear norm but high rank and vice-versa. Is there any intuition one can appeal to?

-
Consider this question: Why is the $\ell_1$ norm a surrogate for $\ell_0$? – p.s. Jun 28 '12 at 16:33
You write "low-rank" but the formula has $\arg \max$... – user31373 Jun 28 '12 at 23:01
@LeonidKovalev: Oops. Of course I meant $\min$. – blubb Jun 29 '12 at 7:26

Why does compressed sensing work? Because the $\ell_1$ ball in high dimensions is extremely "pointy" -- the extreme values of a linear function on this ball are very likely to be attained on the faces of low dimensions, those that consist of sparse vectors. When applied to matrices, the sparseness of the set of eigenvalues means low rank, as @mrig wrote before me.

-
Thank you, the reduction to sparse coding of eigenvalues was indeed the link I was looking for! – blubb Jun 29 '12 at 7:39

The nuclear norm can be thought of as a convex relaxation of the number of non-zero eigenvalues (i.e. the rank).

-

A nuclear norm of a matrix is equivalent to the L1-norm of the vector of its eigenvalues. Thus, you are injecting sparsity to the vector of eigenvalues. Essentially, this sparsity means you are reducing the rank of the original matrix.

-

Thanks for asking this question. I have problems understanding the same thing. Particularly as follows.

Although the nuclear norm is a relaxation for the rank, I don't understand how the nuclear norm can work as surrogate for minimising rank unless it is an upper bound for the rank.

As was pointed out in the original post, the nuclear norm does not appear to provide a bound (lower or upper) to the matrix rank. For e.g. consider the following.

1)
A: #<BASIC-MATRIX (2 2)      2.00000      0.00000 \\
0.00000      2.00000 >
(RANK A): 2
(NUCLEAR-NORM A): 4.0d0

2)
A: #<BASIC-MATRIX (2 2)      0.10000      0.00000 \\
0.00000      0.10000 >
(RANK A): 2
(NUCLEAR-NORM A): 0.20000000298023224d0


In 1) the nuclear norm is larger than the rank while in 2) it is lower than the rank.

-
It's not that you want something that's actually a lower or upper bound for your quantity of interest, it's that you want something that grows bigger/smaller together with it. For instance, sqrt(x) can be either smaller or bigger than x, but if you want to minimize sqrt(x), you can perhaps more easily minimize x. – Ken Williams Aug 13 '15 at 19:06
Welcome to Math.SE. I think you should ask a separate question and link to this question. That way, you'll get more attention and people are not forced to answer in a comment. :) – blubb Feb 19 at 6:17