How to prove a matrix $M$ has incoherence property?

27 Views Asked by At

By incoherence, I am referring to equation 1.18 in the paper The Power of Convex Relaxation: Near-Optimal Matrix Completion

Given a rank 1 matrix $n\times n$ matrix $M = x \mathbf{1}^T$, $x \in \mathbb{R}^n$ where each $x_i \in [1, \sqrt{\mu_0}] $. The paper states that one can easily show the matrix obeys incoherence property with parameter $\mu_0$. However, I am not exactly sure how to prove it.

The incoherence property with parameter $\mu_0$ is satisfied if

\begin{equation} \|P_U e_a\|^2 \le \frac{\mu_0 r}{n}, \quad \|P_V e_b\|^2 \le \frac{\mu_0 r}{n}, \end{equation}

Let $u_i$ and $v_i$ be singular vectors, and let $P_U$ and $P_V$ be defined as follows:

$P_U = \sum_{i \in [r]} u_i u_i^*$

$P_V = \sum_{i \in [r]} v_i v_i^*$

1

There are 1 best solutions below

1
On BEST ANSWER

Hint: The singular value decomposition of the rank $r = 1$ matrix $M = x\vec{1}^T$ is $\sigma_1u_1v_1^T$ where $u_1 = \dfrac{x}{\|x\|}$, $v = \dfrac{\vec{1}}{\|\vec{1}\|} = \dfrac{1}{\sqrt{n}}\vec{1}$, and $\sigma_1 = \|x\| \cdot \|\vec{1}\| = \sqrt{n}\|x\|$.

So, the rank-$1$ projection matrices are $P_U = u_1u_1^T = \dfrac{xx^T}{\|x\|^2}$ and $P_V = v_1v_1^T = \dfrac{1}{n}\vec{1}\vec{1}^T$.

Can you show that these satisfy the bounds $\|P_Ue_a\|^2 \le \dfrac{\mu_0 }{n}$ and $\|P_Ve_b\|^2 \le \dfrac{\mu_0 }{n}$? It may help to note that $\|P_Ue_a\|^2$ is the sum of the squares of the entries in the $a$-th column of $P_U$.