Comments on Efficient Singular
Value Thresholding Computation

We first show that if at least one such $j$ exists, then such a $j$ (and hence $t_{j}$) is unique. Consider the set $J=\{j\mid t_{j}\text{ satisfies \eqref{algebraicEquation_g} and \eqref{constraint_g}}\}.$ Assume $J\neq\emptyset$, let $j^{*}$ be the smallest element in $J$. Now we argue that no $j^{*}+k$, $1\leq k\leq r-j^{*}$, can be in $J$. Consider any $k$ with $1\leq k\leq r-j^{*}$. Suppose for contradiction $j^{*}+k\in J$. That is:

Expanding on the right side of (2.3), we have

where the first inequality follows from the non-increasing values of the singular values, the second inequality follows from the assumption in (2.4) and that $g$ is increasing, the last inequality follows from $t_{j^{*}}\geq\sigma^{j^{*}+1}_{Y}\geq\sigma^{j^{*}+k}_{Y}>t_{j^{*}+k}$ and the last equality follows from the definition of $t_{j^{*}}$.

Hence, it follows that

, leading to $t_{j^{*}+k}>t_{j^{*}}$, hence a contradiction.

Next, we prove that $J$ is indeed not empty.

First, we note that by the property of $g$, a unique solution $t_{j}$ ($>0$) exists for $g(\sum_{i=1}^{j}{\sigma^{i}_{Y}}-jt_{j})=\frac{t_{j}}{\tau}$, for each $j$ satisfying $1\leq j\leq r$. We denote by $t_{j}$ the unique solution corresponding to each $j$. Hence, it suffices to show at least one $t_{j}$ satisfies $\sigma^{j+1}_{Y}\leq t_{j}<\sigma^{j}_{Y}$.

Again by monotonicity of $g$ and $g(0)\leq 1$, it is easily seen that $\tau<\sigma^{1}_{Y}$ implies that $t_{1}<\sigma^{1}_{Y}$. Now suppose it also holds that $\sigma^{2}_{Y}\leq t_{1}$, then we are done. Otherwise, we have $t_{1}<\sigma^{2}_{Y}$. Under this assumption, we claim that $t_{1}<t_{2}$ and $t_{2}<\sigma^{2}_{Y}$. To prove $t_{1}<t_{2}$, assume for the sake of contradiction that $t_{1}\geq t_{2}$, leading to:

where the first inequality follows from $t_{1}<\sigma^{2}_{Y}$. Hence we reach a contradiction, establishing that $t_{1}<t_{2}$.

The desired inequality $t_{2}<\sigma^{2}_{Y}$ then follows since

implies $\sigma^{1}_{Y}-t_{1}<\sigma^{1}_{Y}+\sigma^{2}_{Y}-t_{1}-t_{2}$ by monotonicity of $g$, hence yielding $t_{2}<\sigma^{2}_{Y}$

If $t_{2}\geq\sigma^{3}_{Y}$, then the claim is established. If not, we can repeat this process inductively. More formally, suppose we have just finished the $j$-th iteration (note that the induction basis $j=1$ is verified above) and we have $t_{j}<\sigma^{j}_{Y}$. If it also holds that $t_{j}\geq\sigma^{j+1}_{Y}$, then the claim follows. If not, then we show $t_{j+1}>t_{j}$ and $t_{j+1}<\sigma^{j+1}_{Y}$. First, assume on the contrary, $t_{j+1}\leq t_{j}$

where the first inequality follows from $t_{j}<\sigma^{j}_{Y}$. Hence we reach a contradiction, establishing that $t_{j}<t_{j+1}$.

Next, we note that $t_{j+1}<\sigma^{j+1}_{Y}$ follows since

(where the last inequality follows due to $jt_{j}<jt_{j+1}$,) implying $\sum_{i=1}^{j}{\sigma^{i}_{Y}}-jt_{j}<\sum_{i=1}^{j+1}{\sigma^{i}_{Y}}-jt_{j}-t_{j+1}$, which is equivalent to $t_{j+1}<\sigma^{j+1}_{Y}$.

Thus, we have a strictly increasing sequence $\{t_{j}\}$ with $t_{j}<\sigma^{i}_{Y}$. If it holds that $\sigma^{j+1}_{Y}\leq t_{j}<\sigma^{j}_{Y}$ at some iteration $j$, then such a $j$ certifies that $J$ is not empty. If $\sigma^{j+1}_{Y}\leq t_{j}<\sigma^{j}_{Y}$ never holds for $j$ up to $r-1$, then it must hold for $j=r$, since $-\infty=\sigma^{r+1}_{Y}\leq t_{r}<\sigma^{r}_{Y}$, also certifying that $J$ is not empty. ∎

From the first part of proof for Lemma 2, we know that if $j^{*}$ is the unique $j$ guaranteed by Lemma 2, then for all $k>j^{*}$, we have $t_{k}\geq\sigma^{k}_{Y}$. Thus, if $t_{M}<\sigma^{M}_{Y}$, then we know that $j^{*}$ cannot be less than M. That is, $j^{*}$ must be in the second half of the unsearched space. Conversely, if we hypothetically do a sequential search, then it follows immediately from the second part of proof of Lemma 2 that before $j$ reaches $j^{*}$, $t_{M}<\sigma^{M}_{Y}$ must hold. This establishes that if in the while loop we encounter $t_{M}\geq\sigma^{M}_{Y}$, then it must be the case that $j^{*}\leq M$. That is, $j^{*}$ must lie in the first part of the unsearched space. It then follows that $j^{*}$ always lies between $L$ and $R$, establishing that while loop will eventually halt, returning $t_{j^{*}}$ and $j^{*}$.

∎

To prove this theorem, we build on the techniques introduced in Cai et al., (2010).

The function $h(X)=\tau f(\|X\|_{*})+\frac{1}{2}\|X-Y\|^{2}_{F}$ is strictly convex, since it is the sum of a convex function and a strictly convex function. As a result, the minimizer $\hat{X}$ to $h(X)$ is unique and it suffices to show that $\mathcal{H}_{\tau}(Y)$ is one minimizer.

Per the definition of a subgradient, $S$ is a subgradient of a convex function $f$ at $X_{0}$ if $f(X)\geq f(X_{0})+\langle S,X-X_{0}\rangle$. $\partial f(X_{0})$ is commonly used to denote the set of subgradients of $f$ at $X_{0}$. Recall that the set $\partial\|X_{0}\|_{*}$ of subgradients of the nuclear norm function at $X_{0}$ is: $\partial\|X_{0}\|_{*}=\{U_{X_{0}}V^{*}_{X_{0}}+W\mid W\in\mathbf{R}^{n_{1}\times n_{2}},U^{*}_{X_{0}}W=0,WV_{X_{0}}=0,\|W\|_{2}\leq 1\},$ where the SVD of $X_{0}$ is $U_{X_{0}}\Sigma_{X_{0}}V^{*}_{X_{0}}$ and $\|W\|_{2}$ is the top singular value of $W$.

First, it is easy to check that $f^{\prime}({\|X_{0}\|_{*}})(U_{X_{0}}V^{*}_{X_{0}}+W)\in\partial(f({\|X_{0}\|_{*}}))$ for $W\in\mathbf{R}^{n_{1}\times n_{2}},U^{*}_{X_{0}}W=0,WV_{X_{0}}=0,\|W\|_{2}\leq 1$, by the composition rule for the subgradient. Hence, $\tau g({\|\hat{X}\|_{*}})(U_{\hat{X}}V^{*}_{\hat{X}}+W)+\hat{X}-Y$ is a subgradient for $h$ at $\hat{X}$, for $W$ satisfying $U^{*}_{\hat{X}}W=0,WV_{\hat{X}}=0,\|W\|_{2}\leq 1$, where $g=f^{\prime}$ and $g$ satisfies the assumption given in Lemma 2. Moreover, if there exists such a $W$ and it holds that $0=\tau g({\|\hat{X}\|_{*}})(U_{\hat{X}}V^{*}_{\hat{X}}+W)+\hat{X}-Y$, or equivalently that

then $\hat{X}$ is a minimizer (hence the unique minimizer) to $h(X)$.

We now establish that, with $\hat{X}=\mathcal{H}_{\tau}(Y)$, Eq. (2.5) does hold with $W$ satisfying the given constraints. First, we consider the case that $\tau<\sigma^{1}_{Y}$.

By Lemma 2, since $t_{j^{*}}$ satisfies the equation $\frac{t_{j^{*}}}{\tau}=g({\sum_{i=1}^{j^{*}}\sigma^{i}_{Y}}-j^{*}t_{j^{*}})$, we have $t_{j^{*}}=\tau g({\sum_{i=1}^{j^{*}}{(\sigma^{i}_{Y}}-t_{j^{*}})})$. Since $\sigma^{j^{*}+1}_{Y}\leq t_{j^{*}}<\sigma^{j^{*}}_{Y}$, $Y$’s last $r-j^{*}$ singular values ($\sigma^{k}_{Y}$ with $k\geq j^{*}+1$) have been set to 0, leading to that $\|\hat{X}\|_{*}=\sum_{i=1}^{j^{*}}{(\sigma^{i}_{Y}-t_{j^{*}})}$. Therefore, $t_{j^{*}}=\tau g({\|\hat{X}\|_{*}})$.

Next, we partition $Y$ as follows:

where $U_{a}$’s columns and $V_{a}$’s columns are the first $j^{*}$ left and right singular vectors respectively, associated with the first $j^{*}$ singular values (i.e. singular values larger than $t_{j^{*}}$), while $U_{b}$’s columns and $V_{b}$’s colunms are the remaining $r-j^{*}$ left and right singular vectors respectively, associated with the remaining $r-j^{*}$ singular values (i.e. singular values less than or equal to $t_{j^{*}}$).

Under this partition, it is easily seen that $\hat{X}=U_{a}(\Sigma_{a}-t_{j^{*}}\mathbf{I})V_{a}^{*}.$ We then have

Choose $W=\frac{1}{t_{j^{*}}}U_{b}\Sigma_{b}V_{b}^{*}$. By construction, $U_{a}^{*}U_{b}=0,V_{b}^{*}V_{a}=0$, hence we have $U^{*}_{\hat{X}}W=0,WV_{\hat{X}}=0$. In addition, since $t_{j^{*}}\geq\sigma^{j*+1}_{Y}$, we have $\|W\|_{2}=\frac{\sigma^{j*+1}_{Y}}{t_{j^{*}}}\leq 1$. Hence $W$ thus chosen satisfies the constraints, hence establishing the claim.

Now, if $\tau\geq\sigma^{1}_{Y}$, then $t_{j^{*}}=\sigma^{1}_{Y}$ and $j^{*}=1$ are returned by Algorithm 1. It follows immediately that in this case $\hat{X}=0$. Choosing $W=\tau^{-1}Y$, it is easily seen that $W$ satisfies the constraints. Verification of Eq. (2.5) is instant when $\hat{X}=0$ and $W=\tau^{-1}Y$. ∎