On Polynomial Time Methods for Exact Low-Rank Tensor Completion


In this paper, we investigate the sample size requirement for exact recovery of a high-order tensor of low rank from a subset of its entries. We show that a gradient descent algorithm with initial value obtained from a spectral method can, in particular, reconstruct a \({d\times d\times d}\) tensor of multilinear ranks (rrr) with high probability from as few as \(O(r^{7/2}d^{3/2}\log ^{7/2}d+r^7d\log ^6d)\) entries. In the case when the ranks \(r=O(1)\), our sample size requirement matches those for nuclear norm minimization (Yuan and Zhang in Found Comput Math 1031–1068, 2016), or alternating least squares assuming orthogonal decomposability (Jain and Oh in Advances in Neural Information Processing Systems, pp 1431–1439, 2014). Unlike these earlier approaches, however, our method is efficient to compute, is easy to implement, and does not impose extra structures on the tensor. Numerical results are presented to further demonstrate the merits of the proposed approach.

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Correspondence to Ming Yuan.

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Ming Yuan’s research was supported in part by NSF Grant DMS-1721584.

Communicated by Thomas Strohmer.


Proof of Lemma 1

The first claim is straightforward. It suffices to prove the second claim. Let \(\mathbf {A}=(\mathbf {U},\mathbf {V},\mathbf {W})\cdot \mathbf {C}\) with \(\mathbf {C}\in \mathbb {R}^{r_1(\mathbf {A})\times r_2(\mathbf {A})\times r_3(\mathbf {A})}\) being the core tensor. Clearly, \(\Vert \mathbf {A}\Vert _{\star }=\Vert \mathbf {C}\Vert _{\star }\) and \(\Vert \mathbf {A}\Vert _{\mathrm{F}}=\Vert \mathbf {C}\Vert _{\mathrm{F}}\). Denote by \(\mathbf {C}_1,\ldots , \mathbf {C}_{r_1(\mathbf {A})}\in \mathbb {R}^{r_2(\mathbf {A})\times r_3(\mathbf {A})}\) the mode-1 slices of \(\mathbf {C}\). By convexity of nuclear norm,

$$\begin{aligned} \Vert \mathbf {C}\Vert _{\star }\le \Vert \mathbf {C}_1\Vert _{\star }+\cdots +\Vert \mathbf {C}_{r_1(\mathbf {A})}\Vert _{\star }. \end{aligned}$$

As a result,

$$\begin{aligned} \Vert \mathbf {C}\Vert _{\star }^2\le & {} r_1(\mathbf {A})\big (\Vert \mathbf {C}_1\Vert _{\star }^2+\cdots +\Vert \mathbf {C}_{r_1(\mathbf {A})}\Vert _{\star }^2\big )\\\le & {} r_1(\mathbf {A})\big (r_2(\mathbf {A})\wedge r_3(\mathbf {A})\big )\big (\Vert \mathbf {C}_1\Vert _{\mathrm{F}}^2+\ldots +\Vert \mathbf {C}_{r_1(\mathbf {A})}\Vert _{\mathrm{F}}^2\big )\\= & {} r_1(\mathbf {A})\big (r_2(\mathbf {A})\wedge r_3(\mathbf {A})\big )\Vert \mathbf {C}\Vert _{\mathrm{F}}^2. \end{aligned}$$


$$\begin{aligned} \Vert \mathbf {C}\Vert _{\star }\le \sqrt{r_1(\mathbf {A})\min \{r_2(\mathbf {A}),r_3(\mathbf {A})\}}\Vert \mathbf {C}\Vert _{\mathrm{F}}. \end{aligned}$$

By the same process on mode-2 and mode-3 slices of \(\mathbf {C}\), we obtain

$$\begin{aligned} \Vert \mathbf {C}\Vert _{\star }\le \sqrt{r_2(\mathbf {A})\min \{r_1(\mathbf {A}), r_3(\mathbf {A})\}}\Vert \mathbf {C}\Vert _{\mathrm{F}}, \end{aligned}$$


$$\begin{aligned} \Vert \mathbf {C}\Vert _{\star }\le \sqrt{r_3(\mathbf {A})\min \{r_1(\mathbf {A}), r_2(\mathbf {A})\}}\Vert \mathbf {C}\Vert _{\mathrm{F}}, \end{aligned}$$

which concludes the proof.

Proof of Corollary 1

By Davis–Kahan theorem (see, e.g., Theorem 2 of [32]),

$$\begin{aligned} d_{\mathrm{p}}\big ({\widehat{\mathbf {U}}},\mathbf {U}\big )\le \frac{2\sqrt{r_1}\Vert {\widehat{\mathbf {N}}}-\mathbf {M}\mathbf {M}^\top \Vert }{\sigma _{\min }(\mathbf {M}\mathbf {M}^\top )}. \end{aligned}$$

By choosing \(m_1=d_1, m_2=d_2d_3\) in Theorem 2 and noticing that \(n\ge C_1(\alpha +1)(d_1d_2d_3)^{1/2}\), then

$$\begin{aligned}&\Vert {\widehat{\mathbf {N}}}-\mathbf {M}\mathbf {M}^\top \Vert \le C\alpha ^2 {(d_1d_2d_3)^{3/2}\log d\over n} \\&\quad \times \, \left[ \left( 1+\frac{d_1}{d_2d_3}\right) ^{1/2}+\left( \frac{n}{d_2d_3\log d}\right) ^{1/2}\right] \Vert \mathbf {M}\Vert _{\max }^2 \end{aligned}$$

with probability at least \(1-d^{-\alpha }\). It suffices to control \(\Vert \mathbf {M}\Vert _{\max }\). Recall that \(\mu (\mathbf {T})\le \mu _0\); then,

$$\begin{aligned} \Vert \mathbf {M}\Vert _{\max }=\Vert \mathbf {T}\Vert _{\max }\le \Vert \mathbf {T}\Vert \mu _0^{3/2}\left( r_1r_2r_3\over d_1d_2d_3\right) ^{1/2}. \end{aligned}$$

It is clear by definition that

$$\begin{aligned} {\Vert \mathbf {T}\Vert ^2}/{\sigma _{\min }(\mathbf {M}\mathbf {M}^\top )}\le \kappa ^2(\mathbf {T})\le \kappa _0^2. \end{aligned}$$

As a result, the following bound holds with probability at least \(1-d^{-\alpha }\),

$$\begin{aligned} d_{\mathrm{p}}\big ({\widehat{\mathbf {U}}},\mathbf {U}\big )&\le 2C\alpha ^2\mu _0^3\kappa _0^2r_1^{3/2}r_2r_3{(d_1d_2d_3)^{1/2}\log d\over n} \\&\quad \times \, \left[ \left( 1+\frac{d_1}{d_2d_3}\right) ^{1/2}+\left( \frac{n}{d_2d_3\log d}\right) ^{1/2}\right] \\&\le 2C\alpha ^2\mu _0^3\kappa _0^2r_1^{3/2}r_2r_3\left[ \frac{(d_1d_2d_3)^{1/2}\log d}{n}+\frac{d_1\log d}{n}+\left( \frac{d_1\log d}{n}\right) ^{1/2}\right] . \end{aligned}$$

The claim then follows.

Proof of Lemma 2

For simplicity, define a random tensor \(\mathbf {E}\in \{0,1\}^{d_1\times d_2\times d_3}\) based on \(\omega \in [d_1]\times [d_2]\times [d_3]\) such that \(\mathbf {E}(\omega )=1\) and all the other entries are 0s. Let \(\mathbf {E}_1,\ldots ,\mathbf {E}_n\) be i.i.d. copies of \(\mathbf {E}\). Equivalently, we write

$$\begin{aligned} \beta _n(\gamma _1,\gamma _2)=\underset{\mathbf {A}\in \mathcal{K}(\gamma _1,\gamma _2)}{\sup }\Big |\frac{1}{n}\sum _{i=1}^n\langle \mathbf {A},\mathbf {E}_i\rangle ^2-\mathbb {E}\langle \mathbf {A},\mathbf {E}\rangle ^2\Big | \end{aligned}$$

which is the upper bound of an empirical process indexed by \(\mathcal {K}(\gamma _1,\gamma _2)\). Define \(\delta _{1,j}=2^j\delta _1^-\) for \(j=0,1,2,\ldots ,\lfloor \rfloor {\log \frac{\delta _1^+}{\delta _1^-}}\) and \(\delta _{2,k}=2^k\delta _2^-\) for \(k=0,1,2,\ldots , \lfloor \rfloor {\log \frac{\delta _2^+}{\delta _2^-}}\). For each jk, we derive the upper bound of \(\beta _n(\gamma _1,\gamma _2)\) with \(\gamma _1\in [\delta _{1,j}, \delta _{1,j+1}]\) and \(\gamma _2\in [\delta _{2,k},\delta _{2,k+1}]\). Following the union argument, we can make the bound uniformly true for \(\gamma _1\in [\delta _1^-, \delta _1^+]\) and \(\gamma _2\in [\delta _{2}^-, \delta _2^+]\).

Consider \(\gamma _1\in [\delta _{1,j}, \delta _{1,j+1}]\), \(\gamma _2\in [\delta _{2,k},\delta _{2,k+1}]\), and observe that

$$\begin{aligned} \underset{\mathbf {A}\in \mathcal {K}(\gamma _1,\gamma _2)}{\sup }\big |\langle \mathbf {A},\mathbf {E}\rangle ^2-\mathbb {E}\langle \mathbf {A},\mathbf {E}\rangle ^2\big |\le \gamma _1^2. \end{aligned}$$


$$\begin{aligned} \underset{{\mathbf {A}}\in \mathcal {K}(\gamma _1,\gamma _2)}{\sup }\mathrm{Var}\big (\langle \mathbf {A},\mathbf {E}\rangle ^2\big )\le \underset{{\mathbf {A}}\in \mathcal {K}(\gamma _1,\gamma _2)}{\sup }\mathbb {E}\langle \mathbf {A},\mathbf {E}\rangle ^4\le \frac{\gamma _1^2\Vert \mathbf {A}\Vert _{\mathrm{F}}^2}{d_1d_2d_3}\le \frac{\gamma _1^2}{d_1d_2d_3}. \end{aligned}$$

Applying Bousquet’s version of Talagrand’s concentration inequality [4], with probability at least \(1-e^{-t}\) for all \(t\ge 0\),

$$\begin{aligned} \beta _n(\gamma _1,\gamma _2)\le 2\mathbb {E}\beta _n(\gamma _1,\gamma _2)+2\gamma _1\sqrt{\frac{t}{nd_1d_2d_3}}+2\gamma _1^2\frac{t}{n}. \end{aligned}$$

By the symmetrization inequality,

$$\begin{aligned} \mathbb {E}\beta _n(\gamma _1,\gamma _2)\le 2\mathbb {E}\underset{\mathbf {A}\in \mathcal {K}(\gamma _1,\gamma _2)}{\sup }\Big |\frac{1}{n}\sum _{i=1}^n\varepsilon _{i}\langle \mathbf {A},\mathbf {E}_i\rangle ^2\Big |, \end{aligned}$$

where \(\varepsilon _1,\ldots ,\varepsilon _n\) are i.i.d Rademacher random variables. Since \(|\langle \mathbf {A},\mathbf {E}\rangle |\le \gamma _1\), by the contraction inequality,

$$\begin{aligned} \mathbb {E}\beta _n(\gamma _1,\gamma _2)\le 4\gamma _1\mathbb {E}\underset{\mathbf {A}\in \mathcal {K}(\gamma _1,\gamma _2)}{\sup }\Big |\frac{1}{n}\sum _{i=1}^n\varepsilon _{i}\langle \mathbf {A},\mathbf {E}_i\rangle \Big |. \end{aligned}$$

Denote \({{\varvec{\Gamma }}}=n^{-1}\sum _{i=1}^n\varepsilon _i\mathbf {E}_i\in \mathbb {R}^{d_1\times d_2\times d_3}\). Then,

$$\begin{aligned} \mathbb {E}\underset{\mathbf {A}\in \mathcal {K}(\gamma _1,\gamma _2)}{\sup }\Big |\frac{1}{n}\sum _{i=1}^n\varepsilon _{i}\langle \mathbf {A},\mathbf {E}_i\rangle \Big |\le \mathbb {E}\underset{\mathbf {A}\in \mathcal {K}(\gamma _1,\gamma _2)}{\sup }\Vert {{\varvec{\Gamma }}}\Vert \Vert \mathbf {A}\Vert _{\star }\le \gamma _2\mathbb {E}\Vert {{\varvec{\Gamma }}}\Vert . \end{aligned}$$

It is not difficult to show that

$$\begin{aligned} \mathbb {E}\Vert \varvec{\Gamma }\Vert \le C\Big (\sqrt{\frac{d}{nd_1d_2d_3}}\log d+\frac{\log ^{3/2}d}{n}\Big ). \end{aligned}$$

See, e.g., Lemma 8 of Yuan and Zhang [33]. The above bound holds as long as

$$\begin{aligned} n\ge C\Big \{\mu _0(r_1r_2r_3d_1d_2d_3)^{1/2}\log ^{3/2}d+\mu _0^2r_1r_2r_3d\log ^2d\Big \}. \end{aligned}$$

As a result, with probability at least \(1-e^{-t}\),

$$\begin{aligned} \beta _n(\gamma _1,\gamma _2)\le C\gamma _1\gamma _2\Big (\sqrt{\frac{d}{nd_1d_2d_3}}\log d+\frac{\log ^{3/2}d}{n}\Big )+2\gamma _1\sqrt{\frac{t}{nd_1d_2d_3}}+2\gamma _1^2\frac{t}{n} \end{aligned}$$

for \(\gamma _1\in [\delta _{1,j}, \delta _{1,j+1}]\) and \(\gamma _2\in [\delta _{2,k},\delta _{2,k+1}]\). Now, consider all the combinations of j and k, and we can make the upper bound uniform for all j and k with adjusting t to \({{\bar{t}}}\), and C to 2C.

Proof of lower bound of \(\langle \mathbf {Q}_\mathbf {T}({\widehat{\mathbf {T}}}-\mathbf {T}), \mathbf {H}_1\rangle \)

Recall that

$$\begin{aligned}&\langle \mathbf {Q}_\mathbf {T}({\widehat{\mathbf {T}}}-\mathbf {T}), \mathbf {H}_1\rangle = \Big <(\mathbf {U},\mathbf {V},\mathbf {W})\cdot (\mathbf {C}-\mathbf {G})+(\varvec{\Delta }_\mathbf {X},\mathbf {V},\mathbf {W})\cdot \mathbf {C}+(\mathbf {U},\varvec{\Delta }_\mathbf {Y},\mathbf {W})\cdot \mathbf {C}\\&\quad +(\mathbf {U},\mathbf {V},\varvec{\Delta }_\mathbf {Z})\cdot \mathbf {C}, (\mathbf {D}_\mathbf {X},\mathbf {V},\mathbf {W})\cdot \mathbf {C}+(\mathbf {U},\mathbf {D}_\mathbf {Y},\mathbf {W})\cdot \mathbf {C}+(\mathbf {U},\mathbf {V},\mathbf {D}_\mathbf {Z})\cdot \mathbf {C}\Big >. \end{aligned}$$

Clearly, the right-hand side can be written as \(\zeta _1+\zeta _2+\zeta _3\) where

$$\begin{aligned} \zeta _1&=\Vert (\varvec{\Delta }_\mathbf {X},\mathbf {V},\mathbf {W})\cdot \mathbf {C}+(\mathbf {U},\varvec{\Delta }_\mathbf {Y},\mathbf {W})\cdot \mathbf {C}+(\mathbf {U},\mathbf {V},\varvec{\Delta }_\mathbf {Z})\cdot \mathbf {C}\Vert _{\mathrm{F}}^2\\ \zeta _2&=\Big<(\mathbf {U},\mathbf {V},\mathbf {W})\cdot (\mathbf {C}-\mathbf {G}), (\mathbf {D}_\mathbf {X},\mathbf {V},\mathbf {W})\cdot \mathbf {C}+(\mathbf {U},\mathbf {D}_\mathbf {Y},\mathbf {W})\cdot \mathbf {C}+(\mathbf {U},\mathbf {V},\mathbf {D}_\mathbf {Z})\cdot \mathbf {C}\Big>\\ \zeta _3&=\Big <\varvec{\Delta }_\mathbf {X},\mathbf {V},\mathbf {W})\cdot \mathbf {C}+(\mathbf {U},\varvec{\Delta }_\mathbf {Y},\mathbf {W})\cdot \mathbf {C}+(\mathbf {U},\mathbf {V},\varvec{\Delta }_\mathbf {Z})\cdot \mathbf {C},(\mathbf {D}_\mathbf {X}-\varvec{\Delta }_{\mathbf {X}},\mathbf {V},\mathbf {W})\cdot \mathbf {C}\\&\quad +\,(\mathbf {U},\mathbf {D}_\mathbf {Y}-\varvec{\Delta }_{\mathbf {Y}},\mathbf {W})\cdot \mathbf {C}+(\mathbf {U},\mathbf {V},\mathbf {D}_\mathbf {Z}-\varvec{\Delta }_{\mathbf {Z}})\cdot \mathbf {C}\Big >. \end{aligned}$$


$$\begin{aligned} \zeta _1&\ge \Vert (\varvec{\Delta }_\mathbf {X},\mathbf {V},\mathbf {W})\cdot \mathbf {C}\Vert _{\mathrm{F}}^2+\Vert (\mathbf {U},\varvec{\Delta }_\mathbf {Y},\mathbf {W})\cdot \mathbf {C}\Vert _{\mathrm{F}}^2+\Vert (\mathbf {U},\mathbf {V},\varvec{\Delta }_\mathbf {Z})\cdot \mathbf {C}\Vert _{\mathrm{F}}^2\\&\quad -\,2\Lambda _{\max }^2(\mathbf {C})\Big (\Vert \mathbf {U}^\top \varvec{\Delta }_\mathbf {X}\Vert _{\mathrm{F}}\Vert \mathbf {V}^\top \varvec{\Delta }_\mathbf {Y}\Vert _{\mathrm{F}}+\Vert \mathbf {U}^\top \varvec{\Delta }_\mathbf {X}\Vert _{\mathrm{F}}\Vert \mathbf {W}^\top \varvec{\Delta }_\mathbf {Z}\Vert _{\mathrm{F}}+\Vert \mathbf {V}^\top \varvec{\Delta }_\mathbf {Y}\Vert _{\mathrm{F}}\Vert \mathbf {W}^\top \varvec{\Delta }_\mathbf {Z}\Vert _{\mathrm{F}}\Big )\\&\ge \Lambda _{\min }^2(\mathbf {C})\Big (\Vert \varvec{\Delta }_\mathbf {X}\Vert _{\mathrm{F}}^2+\Vert \varvec{\Delta }_\mathbf {Y}\Vert _{\mathrm{F}}^2+\Vert \varvec{\Delta }_\mathbf {Z}\Vert _{\mathrm{F}}^2\Big )-8\Lambda _{\max }^2(\mathbf {C})d_{\mathrm{p}}^4\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big ) \end{aligned}$$

where we used the fact that

$$\begin{aligned} \Vert \mathbf {U}^\top \varvec{\Delta }_\mathbf {X}\Vert _{\mathrm{F}}\le 2d_{\mathrm{p}}^2(\mathbf {U},\mathbf {X}). \end{aligned}$$

Recall from (23) that on the event \(\mathcal{E}_1\cap \mathcal{E}_2\cap \mathcal{E}_3\), we have

$$\begin{aligned} \frac{\Lambda _{\min }}{2}\le \Lambda _{\min }(\mathbf {C})\le \Lambda _{\max }(\mathbf {C})\le 2\Lambda _{\max }. \end{aligned}$$


$$\begin{aligned} \zeta _1\ge \frac{1}{12}\Lambda _{\min }^2 d_{\mathrm{p}}^2\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big )-32\Lambda _{\max }^2 d_{\mathrm{p}}^4\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big ). \end{aligned}$$

It also implies that on the event \(\mathcal{E}_1\cap \mathcal{E}_2\cap \mathcal{E}_3\),

$$\begin{aligned} \zeta _1\ge \frac{1}{2}\Big (\Vert (\varvec{\Delta }_\mathbf {X},\mathbf {V},\mathbf {W})\cdot \mathbf {C}\Vert _{\mathrm{F}}^2+\Vert (\mathbf {U},\varvec{\Delta }_\mathbf {Y},\mathbf {W})\cdot \mathbf {C}\Vert _{\mathrm{F}}^2+\Vert (\mathbf {U},\mathbf {V},\varvec{\Delta }_\mathbf {Z})\cdot \mathbf {C}\Vert _{\mathrm{F}}^2\Big ). \end{aligned}$$

We can control \(|\zeta _3|\) in the same fashion. Indeed,

$$\begin{aligned} |\zeta _3|^2&\le |\zeta _1| \Lambda _{\max }^2(\mathbf {C})(\Vert \mathbf {D}_\mathbf {X}-\varvec{\Delta }_\mathbf {X}\Vert _{\mathrm{F}}^2+\Vert \mathbf {D}_\mathbf {Y}-\varvec{\Delta }_\mathbf {Y}\Vert _{\mathrm{F}}^2+\Vert \mathbf {D}_\mathbf {Z}-\varvec{\Delta }_\mathbf {Z}\Vert _{\mathrm{F}}^2)\\&\le 4|\zeta _1|\Lambda _{\max }^2 d_{\mathrm{p}}^4\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big ). \\ \end{aligned}$$


$$\begin{aligned} d_{\mathrm{p}}\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big )\le (C\alpha \kappa _0\log d)^{-1} \end{aligned}$$

for large \(C>0\), then under the event \(\mathcal{E}_1\cap \mathcal{E}_2\cap \mathcal{E}_3\),

$$\begin{aligned} \zeta _1\ge \frac{1}{16}\Lambda _{\min }^2d_{\mathrm{p}}^2\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big ) \quad \mathrm{and}\quad |\zeta _3|\le \frac{\zeta _1}{4} \end{aligned}$$

To control \(\zeta _2\), recall that \(\mathbf {X}^\top \mathbf {D}_\mathbf {X}=\mathbf{0}, \mathbf {Y}^\top \mathbf {D}_\mathbf {Y}=\mathbf{0}\) and \(\mathbf {Z}^\top \mathbf {D}_\mathbf {Z}=\mathbf{0}\). Then,

$$\begin{aligned} |\zeta _2|&\le |\langle (\varvec{\Delta }_\mathbf {X},\mathbf {V},\mathbf {W})\cdot (\mathbf {C}-\mathbf {G}), (\mathbf {D}_\mathbf {X},\mathbf {V},\mathbf {W})\cdot \mathbf {C}\rangle |\\&\quad +\, |\langle (\mathbf {U},\varvec{\Delta }_\mathbf {Y},\mathbf {W})\cdot (\mathbf {C}-\mathbf {G}), (\mathbf {U},\mathbf {D}_\mathbf {Y},\mathbf {W})\cdot \mathbf {C}\rangle |\\&\quad +\, |\langle (\mathbf {U},\mathbf {V},\varvec{\Delta }_\mathbf {Z})\cdot (\mathbf {C}-\mathbf {G}), (\mathbf {U},\mathbf {V},\mathbf {D}_\mathbf {Z})\cdot \mathbf {C}\rangle |\\&\le 2\Vert \mathbf {C}-\mathbf {G}\Vert _{\mathrm{F}}\bigg \{\Big (\Vert (\varvec{\Delta }_\mathbf {X},\mathbf {V},\mathbf {W})\cdot \mathbf {C}\Vert _{\mathrm{F}}+\Vert \mathbf {U},\varvec{\Delta }_\mathbf {Y},\mathbf {W})\cdot \mathbf {C}\Vert _{\mathrm{F}}+\Vert (\mathbf {U},\mathbf {V},\varvec{\Delta }_\mathbf {Z})\cdot \mathbf {C}\Vert _{\mathrm{F}}\Big )\\&\quad +\,\Lambda _{\max }(\mathbf {C})\Big (\Vert \mathbf {D}_\mathbf {X}-\varvec{\Delta }_\mathbf {X}\Vert _{\mathrm{F}}+\Vert \mathbf {D}_\mathbf {Y}-\varvec{\Delta }_\mathbf {Y}\Vert _{\mathrm{F}}+\Vert \mathbf {D}_\mathbf {Z}-\varvec{\Delta }_\mathbf {Z}\Vert _{\mathrm{F}}\Big )\bigg \}d_{\mathrm{p}}\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big )\\&\le 2\Vert \mathbf {G}-\mathbf {C}\Vert _{\mathrm{F}}\sqrt{\zeta _1}d_{\mathrm{p}}\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big )\\ {}&\qquad +\,4\Vert \mathbf {C}-\mathbf {G}\Vert _{\mathrm{F}}\Lambda _{\max } d_{\mathrm{p}}^3\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big ). \end{aligned}$$

Recall from (19) that under the event \(\mathcal{E}_1\cap \mathcal{E}_2\cap \mathcal{E}_3\),

$$\begin{aligned} \Vert \mathbf {G}-\mathbf {C}\Vert _{\mathrm{F}}\le C\Lambda _{\max }(\alpha \log d)^{1/2}d_{\mathrm{p}}\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big ). \end{aligned}$$

Therefore, \(|\zeta _2|\le \zeta _1/2\) in view of the lower bound of \(\zeta _1\). In summary, under the event \(\mathcal{E}_1\cap \mathcal{E}_2\cap \mathcal{E}_3\),

$$\begin{aligned} \langle \mathbf {Q}_\mathbf {T}({\widehat{\mathbf {T}}}-\mathbf {T}), \mathbf {H}_1\rangle \ge \frac{1}{4}\zeta _1\ge \frac{1}{64}\Lambda _{\min }^2 d_{\mathrm{p}}^2\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big ). \end{aligned}$$

Upper bound of \(\Vert \mathbf {H}_2\Vert _{\mathrm{F}}\)

It is shown in (28) that if \(d_{\mathrm{p}}\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big )\le (C\alpha \kappa _0\log d)^{-1}\), then

$$\begin{aligned} \zeta _1\ge \frac{1}{2}\Big (\Vert (\varvec{\Delta }_\mathbf {X},\mathbf {V},\mathbf {W})\cdot \mathbf {C}\Vert _{\mathrm{F}}^2+\Vert (\mathbf {U},\varvec{\Delta }_\mathbf {Y},\mathbf {W})\cdot \mathbf {C}\Vert _{\mathrm{F}}^2+\Vert (\mathbf {U},\mathbf {V},\varvec{\Delta }_\mathbf {Z})\cdot \mathbf {C}\Vert _{\mathrm{F}}^2\Big ). \end{aligned}$$

Observe that

$$\begin{aligned} \Vert (\varvec{\Delta }_\mathbf {X},\mathbf {V},\mathbf {W})\cdot \mathbf {C}\Vert _{\mathrm{F}}^2=\Vert \mathcal{M}_2(\mathbf {C})(\varvec{\Delta }_\mathbf {X}\otimes \mathbf {W})\Vert _{\mathrm{F}}=\Vert \mathcal{M}_3(\mathbf {C})(\varvec{\Delta }_\mathbf {X}\otimes \mathbf {V})\Vert _{\mathrm{F}} \end{aligned}$$

which implies that

$$\begin{aligned} \zeta _1\ge \frac{1}{6}\Big (\Vert \mathcal{M}_2(\mathbf {C})(\varvec{\Delta }_\mathbf {X}\otimes \mathbf {W})\Vert _{\mathrm{F}}+\Vert \mathcal{M}_3(\mathbf {C})(\mathbf {U}\otimes \varvec{\Delta }_\mathbf {Y})\Vert _{\mathrm{F}}+\Vert \mathcal{M}_1(\mathbf {C})( \mathbf {V}\otimes \varvec{\Delta }_\mathbf {Z})\Vert _{\mathrm{F}}\Big )^2 \end{aligned}$$

By definition of \(\mathbf {H}_2\), we obtain

$$\begin{aligned} \Vert \mathbf {H}_2\Vert _{\mathrm{F}}&\le \Vert \mathcal{M}_1(\mathbf {C})(\varvec{\Delta }_\mathbf {Y}\otimes \mathbf {W})\Vert _{\mathrm{F}}\Vert \mathbf {D}_\mathbf {X}\Vert _{\mathrm{F}}+\Vert \mathcal{M}_1(\mathbf {C})(\mathbf {V}\otimes \varvec{\Delta }_\mathbf {Z})\Vert _{\mathrm{F}}\Vert \mathbf {D}_\mathbf {X}\Vert _{\mathrm{F}}\\&\quad + \,\Vert \mathcal{M}_2(\mathbf {C})(\varvec{\Delta }_\mathbf {X}\otimes \mathbf {W})\Vert _{\mathrm{F}}\Vert \mathbf {D}_\mathbf {Y}\Vert _{\mathrm{F}}+\Vert \mathcal{M}_2(\mathbf {C})(\mathbf {U}\otimes \varvec{\Delta }_\mathbf {Z})\Vert _{\mathrm{F}}\Vert \mathbf {D}_\mathbf {Y}\Vert _{\mathrm{F}}\\&\quad +\,\Vert \mathcal{M}_3(\mathbf {C})(\varvec{\Delta }_\mathbf {X}\otimes \mathbf {V})\Vert _{\mathrm{F}}\Vert \mathbf {D}_\mathbf {Z}\Vert _{\mathrm{F}}+\Vert \mathcal{M}_3(\mathbf {C})(\mathbf {U}\otimes \varvec{\Delta }_\mathbf {Y})\Vert _{\mathrm{F}}\Vert \mathbf {D}_\mathbf {Z}\Vert _{\mathrm{F}}\\&\quad +\,24\Lambda _{\max }d_{\mathrm{p}}^3\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big ) \end{aligned}$$

where we used the fact \(\Lambda _{\max }(\mathbf {C})\le 2\Lambda _{\max }\) from (23). Clearly,

$$\begin{aligned} \Vert \mathbf {H}_2\Vert _{\mathrm{F}}&\le 2\sqrt{6\zeta _1}\big (\Vert \mathbf {D}_\mathbf {X}\Vert _{\mathrm{F}}+\Vert \mathbf {D}_\mathbf {Y}\Vert _{\mathrm{F}}+\Vert \mathbf {D}_\mathbf {Z}\Vert _{\mathrm{F}}\big )+24\Lambda _{\max }d_{\mathrm{p}}^3\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big )\\&\le 4\sqrt{6\zeta _1}d_{\mathrm{p}}\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big )+24\Lambda _{\max }d_{\mathrm{p}}^3\big ((\mathbf {U},\mathbf {V},\mathbf {W}),(\mathbf {X},\mathbf {Y},\mathbf {Z})\big ). \end{aligned}$$

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Xia, D., Yuan, M. On Polynomial Time Methods for Exact Low-Rank Tensor Completion. Found Comput Math 19, 1265–1313 (2019). https://doi.org/10.1007/s10208-018-09408-6

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  • Concentration inequality
  • Matrix completion
  • Nonconvex optimization
  • Polynomial time complexity
  • Tensor completion
  • Tensor rank
  • U-statistics

Mathematics Subject Classification

  • Primary 90C25
  • Secondary 90C59
  • 15A52