2

So I'm referring to the paper Exact Matrix Completion via Convex Optimization: https://arxiv.org/pdf/0805.4471.pdf

If you look on page 6 of this paper,

The coherence of a matrix $\mu(U) = \frac{n}{r} \max_{1 \le i \le n} \|P_Ue_i\|_2^2$ where $i$ is the standard basis vector with a $1$ in the $i$-th entry and $U$ is a subspace of $\mathbb{R}^n$ of dimension $r$.

Now we are considering some matrix $M \in \mathbb{R}^{n_1 \times n_2}$ which is expressed as $\sum_ {1 \le k \le r} \sigma_k u_kv_k^*$ (SVD). Now there is a property A0 which states that the coherence matrices obey $\max(\mu(U), \mu(V)) \le \mu_0$.

Now if you look a bit below they show that $\left|\sum_k u_{ik}v_{jk} \right| \le \sqrt{\sum_k|u_{ik}^2|}\sqrt{\sum_k|v_{jk}^2|}$ by cauchy schwarz ( this part is obvious to me)...but then they say that this whole thing is $\le \frac{\mu_0 r}{\sqrt{n_1n_2}}$. Now how they jumped to this makes no sense to me I do notice that $\mu(U) \le \mu_0 \implies \max_i \|P_Ue_i\|_2^2 \le \frac{\mu_0r}{n_1}$. Hence if you multiply the lengths of the maximum projections onto $U$ and $V$ you get the quantity $\frac{\mu_0 r}{\sqrt{n_1n_2}}$. Notice for example $u_{ik}$ is the $i$th entry of the $k$th left singular vector and similarly for $v_{jk}$. If someone could help me prove this result it would be much appreciated.

cashinhand
  • 23
  • I'm not sure I understand which step you are having trouble with. Is it the step showing that $\sqrt{\sum_{k} u_{ik}^2} \sqrt{\sum_{k} v_{jk}^2} \leq \frac{\mu_0 r}{\sqrt{n_1 n_2}}$? – VHarisop Mar 04 '21 at 03:00
  • Yes it's that one thank you! – cashinhand Mar 04 '21 at 04:55

1 Answers1

1

Note that $\max_{i} \| P_U e_i \|_2^2 \leq \frac{\mu_0 r}{n_1}$ implies that $\max_{i} \| u_{i} \|_2^2 \leq \frac{\mu_0 r}{n_1}$, where $u_i$ is the $i^{\text{th}}$ row of the matrix of left singular vectors $U$. This is because $P_U e_i = UU^{\mathsf{T}} e_i$ and thus $\| UU^{\mathsf{T}} e_i \|_2 = \| U^{\mathsf{T}} e_i \|_2 = \|u_i\|_2$ for each $i$. The same can be shown to hold for $P_V$: $\max_{i} \|v_i\|_2^2 \leq \frac{\mu_0 r}{n_2}$.

Now using these identities, observe that

$$ \sqrt{\sum_{k} u_{ik}^2} \leq \max_{i} \sqrt{\sum_{k} u_{ik}^2} = \max_{i} \| u_i\|_2 \leq \sqrt{\frac{\mu_0 r}{n_1}}, \\ \sqrt{\sum_{k} v_{jk}^2} \leq \max_{j} \sqrt{\sum_{k} v_{jk}^2} = \max_{j} \| v_j\|_2 \leq \sqrt{\frac{\mu_0 r}{n_2}}. \\ $$

Multiplying the two gives the claimed result.

VHarisop
  • 4,130
  • Thanks so much. One follow up is this: Since $U$ is unitary, doesn't $|UU^Te_i |_2 = |e_i|_2 = 1$ as well? So based on this couldn't you get another bound which is different than the one you proved above? btw I'm trying to upvote your post but it isn't publicly showing apparently since I'm a new user. – cashinhand Mar 04 '21 at 21:18
  • Here, the columns of $U$ span a subspace of $\mathbb{R}^n$ of dimension $r$ so $U$ cannot be unitary (unless $r = n$). In other words, you have that $U^{\mathsf{T}} U = I$ but not $U U^{\mathsf{T}} = I$. – VHarisop Mar 04 '21 at 21:24
  • Ohhh that makes sense now thank you! – cashinhand Mar 04 '21 at 21:54