dimension than $r$. Then we provide upper bound to the error of recovered
matrix.

######  Lemma G.1.

Let assume that $\mathbf{M}$ is can be decomposed as rank $r$ matrix
$\mathbf{L}$ with additional small noise in each column that, its $\ell_{2}$
norm is bounded by $\epsilon$. Then, at the end of the termination of the
algorithm LREBN, estimated subspace $\widetilde{\mathbf{U}}^{k}$ has dimension
at most $r$.

######  Proof G.2.

We prove that in the execution of the algorithm, we show if a column
$\mathbf{M}_{:t}$ has been detected as new column that cannot be contained in
pre-selected $\widetilde{\mathbf{U}}^{k}$, then $\mathbf{L}_{:t}$ is indeed
cannot be contained in the $\mathbf{U}^{k}$. To use triangle inequality, we
notice

$\displaystyle\theta(\mathbf{L}_{:t},\mathbf{U}^{k})\geq\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})-\theta(\widetilde{\mathbf{U}}^{k},{\mathbf{U}}^{k})$

Using the lemma G.14 we can notice that following inequalities are get
satisfied:

$\displaystyle\|\mathbf{M}_{\Omega
t}-\mathcal{P}_{\widetilde{\mathbf{U}}^{k}_{\Omega}}\mathbf{M}_{\Omega:t}\|$
$\displaystyle\leq\sqrt{\frac{3d}{2m}}\Big{(}\|\mathbf{M}_{:t}-\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}\mathbf{M}_{:t}\|\Big{)}$
$\displaystyle\leq\sqrt{\frac{3d}{2m}}\Big{(}\|\mathbf{M}_{:t}-\mathbf{L}_{:t}\|+\|\mathbf{L}_{:t}-\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}\mathbf{L}_{:t}\|\Big{)}+\|\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}(\mathbf{L}_{:t}-\mathbf{M}_{:t})\|$
$\displaystyle\leq\sqrt{\frac{3d}{2m}}\Big{(}\epsilon+\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})+\epsilon\Big{)}$

From the design of the algorithm $\|\mathbf{M}_{\Omega
t}-\mathcal{P}_{\widetilde{\mathbf{U}}^{k}_{\Omega}}\mathbf{M}_{\Omega:t}\|>(1+\epsilon)\Big{(}\sqrt{\frac{3d}{2m}}\theta({\mathbf{U}}^{k},\widetilde{\mathbf{U}}^{k})+\sqrt{\frac{3dk\epsilon}{2m}}\Big{)}$
and using this inequality above, we conclude that

$\displaystyle\sqrt{\frac{3d}{2m}}\theta({\mathbf{U}}^{k},\widetilde{\mathbf{U}}^{k})+\sqrt{\frac{3dk\epsilon}{2m}}<\sqrt{\frac{3d}{2m}}\Big{(}\epsilon+\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})+\epsilon\Big{)}$

which follows that

$\displaystyle\theta({\mathbf{U}}^{k},\widetilde{\mathbf{U}}^{k})+\sqrt{k\epsilon}<\Big{(}\epsilon+\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})+\epsilon\Big{)}$

considering the fact that $\epsilon<\frac{1}{4}$ we conclude that

$\displaystyle\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})\geq\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})+2\epsilon-\sqrt{k\epsilon}>\theta({\mathbf{U}}^{k},\widetilde{\mathbf{U}}^{k})$

therefore we conclude that
$\theta({\mathbf{U}}^{k},\widetilde{\mathbf{U}}^{k})<\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})$
and it follows that $\theta(\mathbf{L}_{:t},\mathbf{U}^{k})>0$. Moreover, one
can see that after every time this inequality get satisfied, dimension of
$\mathbf{U}^{k}$ increases by one, and considering the fact that
$\mathbf{U}^{k}$’s are subspace of column space of $\mathbf{L}$, its dimension
cannot increase more than $r$ times.

Then only remaining step to provide an upper bound to recovery error. Note
that, if the algorithm decides completely observe the column, then $\ell_{2}$
norm of the error is upper bounded by $\epsilon$. Then, all we need to do is
to give upper bound to columns those recovered by estimated subspace.

$\displaystyle\|\widetilde{\mathbf{M}}_{:t}-\mathbf{L}_{:t}\|$
$\displaystyle=\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega:}\mathbf{M}_{\Omega
t}-\mathbf{L}_{:t}\|$
$\displaystyle\leq\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega:}\mathbf{M}_{\Omega
t}-\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega:}\mathbf{L}_{\Omega:t}\|+\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega}\mathbf{L}_{\Omega:t}-\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}\mathbf{L}_{:t}\|+\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}\mathbf{L}_{:t}-\mathbf{L}_{:t}\|$
$\displaystyle\leq\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega:}(\mathbf{M}_{\Omega
t}-\mathbf{L}_{\Omega:t})\|+\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega}\mathbf{L}_{\Omega:t}-\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}\mathbf{L}_{:t}\|+\sin{\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})}$
$\displaystyle\leq\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega:}\|\|(\mathbf{M}_{\Omega
t}-\mathbf{L}_{\Omega:t})\|+\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega}\mathbf{L}_{\Omega:t}-\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}\mathbf{L}_{:t}\|+\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})$

Then all we need to do is to give an upper bound to the final term. Lets start
with the second term here: $\mathbf{L}_{:t}=\widetilde{\mathbf{U}}^{k}v+e$
where
$\widetilde{\mathbf{U}}^{k}v=\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}\mathbf{L}_{:t}$
and note
$\|e\|=\sin{\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})}\leq\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})$.
Therefore:

$\displaystyle\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega}\mathbf{L}_{\Omega:t}-\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}\mathbf{L}_{:t}=\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega}(\widetilde{\mathbf{U}}^{k}v+e)-\widetilde{\mathbf{U}}^{k}v=\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega}e$

Hence we conclude that:

$\displaystyle\|\widetilde{\mathbf{M}}_{:t}-\mathbf{L}_{:t}\|$
$\displaystyle\leq\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega:}\|\|(\mathbf{M}_{\Omega
t}-\mathbf{L}_{\Omega:t})\|+\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega}e_{\Omega}\|+\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})$
$\displaystyle\leq\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega:}\|\|(\mathbf{M}_{\Omega
t}-\mathbf{L}_{\Omega:t})\|+\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega}\|\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})+\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k}).$

To give upper bound to this expression, we notice
$\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega:t}\|\leq\frac{\sigma_{1}(\widetilde{\mathbf{U}}^{k})}{\sigma_{k}(\widetilde{\mathbf{U}}^{k}_{\Omega:})}\leq\Theta(\frac{m}{d})$
given the condition that $d\geq
4\mu(\widetilde{\mathbf{U}}^{k})k\log{\frac{k}{\delta}}$ from the lemma G.15.
From lemma G.9 we know that $\mu(\widetilde{\mathbf{U}}^{k})\leq
2\mu(\mathbf{U}^{k})+2\frac{m}{k}\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})^{2}$
and from lemma G.5 we notice that $k\mu(\widetilde{\mathbf{U}}^{k})\leq
r\mu(\mathbf{U})$. Then all together these facts concludes the selected

$d=72\mu(\mathbf{U})r\log^{2}{\frac{1}{\delta}}+8m\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})^{2}\log{\frac{r}{\delta}}\geq
8\mu(\mathbf{U})r\log{\frac{r}{\delta}}+8m\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})^{2}\log{\frac{r}{\delta}}$

satisfies $d\geq 4\mu(\widetilde{\mathbf{U}}^{k})k\log{\frac{k}{\delta}}$ (it
is assumed that $\delta\leq\frac{1}{r^{1/8}}$ ). Therefore, we can bound
$\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega:t}\|$
above by $\Theta(\frac{m}{d})$.
Now, only remaining term in the error bound above is
$\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})$, and we use the following
inequality to compare it with quantities provided as input:

$\displaystyle\|\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}\mathbf{M}_{:t}-\mathbf{L}_{:t}\|\geq\sin\theta(\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}\mathbf{M}_{:t},\mathbf{L}_{:t})\geq\frac{\theta(\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}\mathbf{M}_{:t},\mathbf{L}_{:t})}{2}\geq\frac{\theta(\widetilde{\mathbf{U}}^{k},\mathbf{L}_{:t})}{2}$

and to relate the term
$\|\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}\mathbf{M}_{:t}-\mathbf{L}_{:t}\|$
with observed entries we again use the inequality G.14 and the fact that
$\big{(}1+2\log{\frac{1}{\delta}}\big{)}^{2}\leq 6\log^{2}{\frac{1}{\delta}}$
once $\delta<0.1$, lemma G.5 and lemma G.11

$\displaystyle\|\mathbf{M}_{\Omega:t}-\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}\mathbf{M}_{\Omega:t}\|$
$\displaystyle\geq\sqrt{\frac{1}{m}\Big{(}\frac{d}{2}-\frac{3k\mu(\widetilde{\mathbf{U}}^{k})\beta}{2}\Big{)}}\|\mathbf{M}_{:t}-\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}\mathbf{M}_{:t}\|$
$\displaystyle\geq\sqrt{\frac{1}{m}\Big{(}\frac{d}{2}-\frac{3k\mu(\widetilde{\mathbf{U}}^{k})\beta}{2}\Big{)}}\Big{(}\|\mathbf{L}_{:t}-\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}\mathbf{M}_{\Omega:t}\|-\|\mathbf{L}_{:t}-\mathbf{M}_{:t}\|\Big{)}$
$\displaystyle\geq\sqrt{\frac{1}{m}\Big{(}\frac{d}{2}-\frac{3k\mu(\widetilde{\mathbf{U}}^{k})\beta}{2}\Big{)}}\Big{(}\|\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}\mathbf{M}_{:t}-\mathbf{L}_{:t}\|-\epsilon\Big{)}$
$\displaystyle\geq\sqrt{\frac{1}{m}\Big{(}\frac{d}{2}-\frac{3k\mu(\widetilde{\mathbf{U}}^{k})\beta}{2}\Big{)}}\Big{(}\frac{\theta(\widetilde{\mathbf{U}}^{k},\mathbf{L}_{:t})}{2}-\epsilon\Big{)}$
$\displaystyle\geq\sqrt{\frac{1}{m}\Big{(}\frac{d}{2}-9k\mu(\widetilde{\mathbf{U}}^{k})\log^{2}{\frac{1}{\delta}}\Big{)}}\Big{(}\frac{\theta(\widetilde{\mathbf{U}}^{k},\mathbf{L}_{:t})}{2}-\epsilon\Big{)}$
$\displaystyle\geq\sqrt{\frac{1}{m}\Big{(}\frac{d}{2}-18k\mu(\mathbf{U}^{k})\log^{2}{\frac{1}{\delta}}-18m\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})^{2}\log^{2}{\frac{1}{\delta}}\Big{)}}\Big{(}\frac{\theta(\widetilde{\mathbf{U}}^{k},\mathbf{L}_{:t})}{2}-\epsilon\Big{)}$
$\displaystyle\geq\sqrt{\frac{d}{2m}-\frac{18k}{m}\mu(\mathbf{U}^{k})\log^{2}{\frac{1}{\delta}}-18\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})^{2}\log^{2}{\frac{1}{\delta}}}\Big{(}\frac{\theta(\widetilde{\mathbf{U}}^{k},\mathbf{L}_{:t})}{2}-\epsilon\Big{)}$
$\displaystyle\geq\sqrt{\frac{d}{2m}-\frac{18r}{m}\mu(\mathbf{U})\log^{2}{\frac{1}{\delta}}-18\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})^{2}\log^{2}{\frac{1}{\delta}}}\Big{(}\frac{\theta(\widetilde{\mathbf{U}}^{k},\mathbf{L}_{:t})}{2}-\epsilon\Big{)}$
$\displaystyle\geq\sqrt{\frac{d}{4m}}\Big{(}\frac{\theta(\widetilde{\mathbf{U}}^{k},\mathbf{L}_{:t})}{2}-\epsilon\Big{)}$

From here, we conclude that:

$\displaystyle\theta({\widetilde{\mathbf{U}}^{k},\mathbf{L}_{:t}})\leq
4\sqrt{\frac{m}{d}}\|\mathbf{M}_{\Omega:t}-\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}\mathbf{M}_{\Omega:t}\|+2\epsilon$
$\displaystyle\leq
4\sqrt{\frac{m}{d}}(1+\epsilon)\Big{(}\sqrt{\frac{3d}{2m}}\theta({\mathbf{U}}^{k},\widetilde{\mathbf{U}}^{k})+\sqrt{\frac{3dk\epsilon}{2m}}+2\epsilon\Big{)}$
$\displaystyle\leq(1+\epsilon)\Big{(}\sqrt{24}\theta({\mathbf{U}}^{k},\widetilde{\mathbf{U}}^{k})+\sqrt{8k\epsilon}\Big{)}$

Finally, returning back to the recovery error:

$\displaystyle\|\widetilde{\mathbf{M}}_{:t}-\mathbf{L}_{:t}\|$
$\displaystyle\leq\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega:}\|\|(\mathbf{M}_{\Omega
t}-\mathbf{L}_{\Omega:t})\|+\|\widetilde{\mathbf{U}}^{k}\widetilde{\mathbf{U}}^{k^{+}}_{\Omega}\|\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})+\theta(\mathbf{L}_{:t},\widetilde{\mathbf{U}}^{k})$
$\displaystyle\leq\frac{m}{d}\epsilon+\Big{(}\frac{m}{d}+1\Big{)}\Big{(}\sqrt{24}\theta({\mathbf{U}}^{k},\widetilde{\mathbf{U}}^{k})+\sqrt{8k\epsilon}\Big{)}(1+\epsilon)$

Then all we need to do is to give upper bound to
$\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})$. In the proof below, we
use similar argument to blum. Lets assume
$\mathbf{U}^{k}=\\{u_{1},u_{2},\ldots,u_{k}\\}$ and
$\widetilde{\mathbf{U}}^{k}=\\{\widetilde{u}_{1},\widetilde{u}_{2},\ldots,\widetilde{u}_{k}\\}$
where each of $\|u_{i}-\widetilde{u}_{i}\|\leq\epsilon$ satisfied. Then using
triangle inequality, lemma G.13 and lemma G.3

$\displaystyle\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})$
$\displaystyle\leq\theta(\widetilde{\mathbf{U}}^{k},\widehat{\mathbf{U}})+\theta(\widehat{\mathbf{U}},\mathbf{U}^{k})$
$\displaystyle\leq\frac{\pi}{2}\frac{\theta(u_{k},\widetilde{u}_{k})}{\theta(\widetilde{u}_{k},\mathbf{U}^{k-1})}+\theta(\widetilde{\mathbf{U}}^{k-1},{\mathbf{U}}^{k-1})$
$\displaystyle\leq\frac{\pi}{2}\frac{\theta(u_{k},\widetilde{u}_{k})}{\theta(\widetilde{u}_{k},\widetilde{\mathbf{U}}^{k-1})-\theta(\widetilde{\mathbf{U}}^{k-1},{\mathbf{U}}^{k-1})}+\theta(\widetilde{\mathbf{U}}^{k-1},{\mathbf{U}}^{k-1})$
$\displaystyle\leq\frac{\pi}{2}\frac{\theta(u_{k},\widetilde{u}_{k})}{\sqrt{k\epsilon}+\theta(\widetilde{\mathbf{U}}^{k-1},{\mathbf{U}}^{k-1})-\theta(\widetilde{\mathbf{U}}^{k-1},{\mathbf{U}}^{k-1})}+\theta(\widetilde{\mathbf{U}}^{k-1},{\mathbf{U}}^{k-1})$
$\displaystyle\leq\frac{\pi}{2}\frac{\epsilon}{\sqrt{k\epsilon}}+\theta(\widetilde{\mathbf{U}}^{k-1},{\mathbf{U}}^{k-1})$

and using lemma G.7, we can conclude that
$\theta(\widetilde{\mathbf{U}}^{k},{\mathbf{U}}^{k})\leq\frac{3\pi}{2}\sqrt{k\epsilon}$,
which gives the final bound to
$\|\widetilde{\mathbf{M}}_{:t}-\mathbf{L}_{:t}\|$ to be
$\Theta(\frac{m}{d}\sqrt{k\epsilon})$.

######  Lemma G.3.

Given that $\|\mathbf{M}_{\Omega
t}-\mathcal{P}_{\widetilde{\mathbf{U}}^{k-1}_{\Omega}}\mathbf{M}_{\Omega:t}\|\geq(1+\epsilon)\Big{(}\sqrt{\frac{3d}{2m}}\theta(\widetilde{\mathbf{U}}^{k-1},\mathbf{U}^{k-1})+\sqrt{\frac{3dk\epsilon}{2m}}\Big{)}$
satisfies. Then following also satisfies:

$\displaystyle\theta(\widetilde{u}_{k},\widetilde{\mathbf{U}}^{k-1})\geq\theta(\widetilde{\mathbf{U}}^{k-1},\mathbf{U}^{k-1})+\sqrt{k\epsilon}$

######  Proof G.4.

Note that simple triangle inequality implies $\|\mathbf{M}_{:t}\|\leq
1+\epsilon$

$\displaystyle\theta(\widetilde{u}_{k},\widetilde{\mathbf{U}}^{k-1})=\theta(\mathbf{M}_{:t},\widetilde{\mathbf{U}}^{k-1})$
$\displaystyle\geq\sin\theta(\mathbf{M}_{:t},\widetilde{\mathbf{U}}^{k-1})$
$\displaystyle\geq\frac{\mathbf{M}_{:t}}{1+\epsilon}\sin\theta(\mathbf{M}_{:t},\widetilde{\mathbf{U}}^{k-1})$
$\displaystyle=\frac{1}{1+\epsilon}\|\mathbf{M}_{:t}-\mathcal{P}_{\widetilde{\mathbf{U}}^{k-1}}\mathbf{M}_{:t}\|$
$\displaystyle\geq\frac{1}{1+\epsilon}\sqrt{\frac{2m}{3d}}\|\mathbf{M}_{\Omega
t}-\mathcal{P}_{\widetilde{\mathbf{U}}^{k-1}_{\Omega}}\mathbf{M}_{\Omega:t}\|$

Using, the fact that $\|\mathbf{M}_{\Omega
t}-\mathcal{P}_{\widetilde{\mathbf{U}}^{k-1}_{\Omega}}\mathbf{M}_{\Omega:t}\|\geq(1+\epsilon)\Big{(}\sqrt{\frac{3d}{2m}}\theta(\widetilde{\mathbf{U}}^{k-1},\mathbf{U}^{k-1})+\sqrt{\frac{3dk\epsilon}{2m}}\Big{)}$
we conclude
$\theta(\widetilde{u}_{k},\widetilde{\mathbf{U}}^{k-1})\geq\sqrt{k\epsilon}+\theta(\widetilde{\mathbf{U}}^{k-1},{\mathbf{U}}^{k-1})$.

######  Lemma G.5.

Let $\mathbf{U}^{k}$ be a $k$-dimensional subspace of $\mathbf{U}$ which is
subspace of $\mathbb{R}^{m}$ with dimension $r$. Then following inequality
satisfied:

$\displaystyle k\mu(\mathbf{U}^{k})\leq r\mu(\mathbf{U})$

######  Proof G.6.

$\displaystyle k\mu(\mathbf{U}^{k})=k\frac{m}{k}\underset{1\leq j\leq
m}{\max}\|\mathcal{P}_{\mathbf{U}^{k}}e_{i}\|^{2}$
$\displaystyle=r\frac{m}{r}\underset{1\leq j\leq
m}{\max}\|\mathcal{P}_{\mathbf{U}^{k}}e_{i}\|^{2}$ $\displaystyle\leq
r\frac{m}{r}\underset{1\leq j\leq
m}{\max}\|\mathcal{P}_{\mathbf{U}}e_{i}\|^{2}$
$\displaystyle=r\mu(\mathbf{U})$

and the inequality due to $\mathbf{U}^{k}\subseteq\mathbf{U}$

######  Lemma G.7.

Let assume that $a_{0}=0$ and $a_{k}\leq
a_{k-1}+\frac{\pi}{2}\sqrt{\frac{\epsilon}{k}}$. Then it follows that
$a_{k}\leq\frac{3\pi}{2}\sqrt{k\epsilon}$

######  Proof G.8.

Its trivial to notice that $a_{1}\leq\frac{\pi}{2}\sqrt{\epsilon}\leq
3\frac{\pi}{2}\sqrt{\epsilon}$. Lets assume by induction for a given $k$ any
index $i\leq k$ satisfies $a_{i}\leq 3\frac{\pi}{2}\sqrt{i\epsilon}$ and then
we will prove that $a_{k+1}\leq 3\frac{\pi}{2}\sqrt{(k+1)\epsilon}$. We prove
it by contradiction, by assuming $a_{k+1}>3\frac{\pi}{2}\sqrt{(k+1)\epsilon}$
and conclude to a contradiction.

$\displaystyle a_{k+1}$ $\displaystyle>3\frac{\pi}{2}\sqrt{(k+1)\epsilon}$
$\displaystyle-a_{k}$ $\displaystyle\geq-3\frac{\pi}{2}\sqrt{k\epsilon}$

Therefore, $a_{k+1}-a_{k}\geq
3\frac{\pi}{2}\sqrt{\epsilon}\Big{(}\sqrt{k+1}-\sqrt{k}\Big{)}=3\frac{\pi}{2}\sqrt{\epsilon}\frac{1}{\sqrt{k}+\sqrt{k+1}}\geq
3\frac{\pi}{2}\sqrt{\epsilon}\frac{1}{3\sqrt{k}}=\frac{\pi}{2}\sqrt{\frac{\epsilon}{k}}$
which contradicts to the statement of the lemma. Therefore, assumption cannot
be satisfied which follows $a_{k+1}\leq 3\frac{\pi}{2}\sqrt{(k+1)\epsilon}$

######  Lemma G.9.

Let $\widetilde{\mathbf{U}}^{k}$ and $\mathbf{U}^{k}$ be as defined above
then, coherence number of these spaces satisfies the following inequality:

$\displaystyle\mu(\widetilde{\mathbf{U}}^{k})\leq
2\mu(\mathbf{U}^{k})+2\frac{m}{k}\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})^{2}$

######  Proof G.10.

In order to achieve the goal of comparing $\mu(\widetilde{\mathbf{U}}^{k})$
and $\mu(\mathbf{U}^{k})$, we first need to understand how projection to
standard vectors to $\mathbf{U}^{k}$ differ than projection of them to
$\widetilde{\mathbf{U}}^{k}$:

$\displaystyle\|\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}e_{i}\|\leq\|\mathcal{P}_{\mathbf{U}^{k}}e_{i}\|+\|\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}e_{i}-\mathcal{P}_{\mathbf{U}^{k}}e_{i}\|$
$\displaystyle\leq\|\mathcal{P}_{\mathbf{U}^{k}}e_{i}\|+\|\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}-\mathcal{P}_{\mathbf{U}^{k}}\|\|e_{i}\|$
$\displaystyle=\|\mathcal{P}_{\mathbf{U}^{k}}e_{i}\|+\sin\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})$
$\displaystyle\leq\|\mathcal{P}_{\mathbf{U}^{k}}e_{i}\|+\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})$

Therefore:

$\displaystyle\mu(\widetilde{\mathbf{U}}^{k})=\frac{m}{k}\underset{1\leq j\leq
n}{\max}\|\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}e_{i}\|^{2}$
$\displaystyle\leq\frac{m}{k}\Big{(}2\underset{1\leq j\leq
n}{\max}\|\mathcal{P}_{\mathbf{U}^{k}}e_{i}\|^{2}+2\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})^{2}\Big{)}$
$\displaystyle=2\mu(\mathbf{U}^{k})+2\frac{m}{k}\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})^{2}$

######  Lemma G.11.

Lets assume the setting as discussed in the proof above. Then,

$\displaystyle\frac{d}{4m}>\frac{18r}{m}\mu(\mathbf{U})\log^{2}{\frac{1}{\delta}}+18\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})^{2}\log^{2}{\frac{1}{\delta}}$

######  Proof G.12.

Remind
$d=72\mu(\mathbf{U})r\log^{2}{\frac{1}{\delta}}+8m\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})^{2}\log{\frac{r}{\delta}}$
and it implies
$\frac{d}{4m}=18\mu(\mathbf{U})r\log^{2}{\frac{1}{\delta}}+2m\theta(\widetilde{\mathbf{U}}^{k},\mathbf{U}^{k})^{2}\log{\frac{r}{\delta}}$.
Then all we need to show is
$2m\log{\frac{r}{\delta}}>18\log^{2}{\frac{1}{\delta}}$. However, we always
pick $\delta$ as $9\log{\frac{1}{\delta}}<m$, simply because $m\geq d\geq
9\log{\frac{1}{\delta}}$

######  Lemma G.13 (blum).

Let subspaces $U$, $V$, and $\widetilde{V}$ defined as
$U=span\\{a_{1},\ldots,a_{k-1}\\}$, $V=span\\{a_{1},\ldots,a_{k-1},b\\}$ and
$\widetilde{V}=span\\{a_{1},\ldots,a_{k-1},\tilde{b}\\}$. Then following
inequality satisfied:

$\displaystyle\theta(V,\widetilde{V})\leq\frac{\pi}{2}\frac{\theta(\tilde{b},b)}{\theta(\tilde{b},U)}$

######  Lemma G.14 (Krishnamurthy and Singh (2014)).

Let $\widetilde{\mathbf{U}}^{k}$ be a $k$-dimensional subspace of
$\mathbb{R}^{m}$, and set
$d=\mathrm{max}(\frac{8}{3}k\mu(\widetilde{\mathbf{U}}^{k})\log{\frac{2k}{\delta}},4\mu\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}\log{\frac{1}{\delta}})$.
Given that $\Omega$ stands for uniformly selected subset of $[m]$ then
following inequality get satisfied:

$\frac{d(1-\alpha)-k\mu(\widetilde{\mathbf{U}}^{k})\frac{\beta}{1-\zeta}}{m}\|y-\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}y\|\leq\|y_{\Omega}-\mathcal{P}_{\widetilde{\mathbf{U}}^{k}_{\Omega}}y_{\Omega}\|\leq(1+\alpha)\frac{d}{m}\|y-\mathcal{P}_{\widetilde{\mathbf{U}}^{k}}y\|$

where
$\alpha=\sqrt{2\frac{\mu(\mathcal{P}_{\widetilde{\mathbf{U}}^{k^{\perp}}}y)}{d}\log{\frac{1}{\delta}}}+2\frac{\mu(\mathcal{P}_{\widetilde{\mathbf{U}}^{k^{\perp}}}y)}{3d}\log{\frac{1}{\delta}}$
, $\beta=(1+2\log{1/\delta})^{2}$ and
$\zeta=\sqrt{\frac{8k\mu(\widetilde{\mathbf{U}}^{k^{\perp}})}{3d}\log{\frac{2r}{\delta}}}$

as noted in the paper, this lemma can be used by $\alpha<1/2$ and $\gamma<1/3$

######  Lemma G.15 (eigen).

Consider a finite sequence $\\{\mathbf{X}_{k}\\}\in\mathbb{R}^{n\times n}$
independent random, Hermitian matrices those satisfies:

$\displaystyle
0\leq\lambda_{\mathrm{min}}(\mathbf{X}_{k})\leq\lambda_{\mathrm{max}}(\mathbf{X}_{k})\leq
L.$

Let $\mathbf{Y}=\sum\limits_{k}\mathbf{X}_{k}$ and $\mu_{r}$ be the $r$-th
largest eigenvalue of $\mathbb{E}[\mathbf{Y}]$ (
$\mu_{r}=\lambda_{r}(\mathbb{E}[\mathbf{Y}])$), then for any
$\epsilon\in[0,1)$ following inequality satisfied:

$\displaystyle\mathbf{Pr}(\lambda_{r}(\mathbf{Y})\geq(1-\epsilon)\mu_{r})\geq
1-r\Big{(}\frac{e^{-\epsilon}}{(1-\epsilon)^{1-\epsilon}}\Big{)}^{\frac{\mu_{r}}{L}}\geq
1-re^{\frac{\mu_{r}\epsilon^{2}}{2L}}$

##  Appendix H Proofs for the design of Experiments

###  H.1 Design of coherent row and column space

To design an $m\times n$ sized rank $r$ with space sparsity number equal to
$1$, we generate $m\times{r-1}$ and ${r-1}\times n$ sized matrices
$\mathbf{X,Y}$, where $X_{i,j},Y_{i,j}\sim\mathcal{N}(0,1)$. As we discussed
before, multiplication of these matrices would gives us a rank $r-1$ matrix.
Moreover, as the column space of $\mathbf{M}$ is column space of $\mathbf{Y}$.
Given that $Y_{i,j}\sim\mathcal{N}(0,1)$ implies that coherence of column
space of $\mathbf{Y}$ is small. Therefore, $\mathbf{M}$ has small column space
coherence number. We can use the similar argument to claim the coherence of
column space is also small as row space of $\mathbf{M}$ is the row space of
$\mathbf{X}$.
In order to make column space of this matrix highly coherent, we follow the
following procedure:

  * 1.

generate a random vector $u\in\mathbb{R}^{n}$

  * 2.

randomly select a number $i\in[m]$

  * 3.

replace $\mathbf{M}_{i:}$ with $u$

We can guarantee that the resulting matrix will contain $i$-th standard basis
vector $e_{i}$ in the column space. To observe this phenomenon, lets analyse
the restriction of the matrix $\mathbf{M}$ to the first $r$ columns and all
the rows but row $i$. As this is a submatrix of rank $r-1$ matrix (initial
$\mathbf{M}$) this matrix also has rank at most $r-1$. Therefore, we have non-
trivial coefficients $\alpha_{1},\alpha_{2},\ldots,\alpha_{r}$, that makes
linear combination of columns of submatrix to be equal to zero vector.
Therefore,

$\displaystyle\alpha_{1}\mathbf{M}_{:1}+\alpha_{1}\mathbf{M}_{:2}+\ldots+\alpha_{r}\mathbf{M}_{:r}=we_{i}$

for some $w$. Then all we need to show is $w\neq 0$, however it is
straightforward because
$w=\alpha_{1}u_{1}+\alpha_{2}u_{2}+\ldots+\alpha_{r}u_{r}$ which is nonzero
because $u$ is random. In conclusion, $e_{i}$ is contained in the column space
of $\mathbf{M}$ and therefore from the definition of coherence of the columns
space of the matrix is equal to $\frac{m}{r}$ which is also maximum value.
We note that, we might change the process above and generate several random
vectors and replace them with some rows of the matrix, we would still get high
coherence values as $\Omega(\frac{m}{r})$. Using similar idea we can add
replace columns with random vector to get highly coherent row spaced matrices.
Moreover, we can apply both at the same time, to get highly coherent column
and highly coherent row space.


where the $\lambda^{(i)}$'s are now just coupling constants, independent of the geometry or anything else.[The outcome of our calculation will show that that this step is justified.] In fact, we can just absorb them into the $m_{i}$'s, which gives us

\begin{equation}
\alpha_{x} =\frac{1}{g_{x}^{1/2}}\left(m_{1}(\rho)\left(\frac{\delta\rho}{\delta\chi_{x}}\right)^{2} + m_{2}(\rho)\left(\frac{\delta^{2}\rho}{\delta\chi_{x}^{2}}\right)\right)\label{5alpha x ansatz}
\end{equation}

Before continuing, it is important to note that functional derivatives commute, i.e.
\[\frac{\delta^{2}}{\delta\chi_{x}\delta\chi_{x^{\prime}}} = \frac{\delta^{2}}{\delta\chi_{x^{\prime}}\delta\chi_{x}} \]

and that the functional derivatives acting on the $m_{i}(\rho)$ behave as
\[\frac{\delta m_{i}(\rho)}{\delta\chi_{x}} = \frac{dm_{i}(\rho)}{d\rho[\chi]}\frac{\delta \rho[\chi]}{\delta\chi_{x}}~.\]
We now plug the $\alpha_{x}$ in (<ref>) into the condition eq.(<ref>). Many terms will automatically be symmetric in $x$, $x^{\prime}$, and thus drop out. The remaining terms can only be made symmetric if the following set of equations between the undetermined $m_{i}(\rho)$ are satisfied:[In fact, we have that the LHS of eq.(<ref>) becomes:

\begin{align}
\hat{O}_{x^{\prime}}\alpha_{x}\longrightarrow &\rho\left(\frac{dm_{1}}{d\rho}\frac{\delta^{2}\rho}{\delta\chi_{x^{\prime}}^{2}}\left(\frac{\delta\rho}{\delta\chi_{x}}\right)^{2}+2m_{1}\frac{\delta\rho}{\delta\chi_{x}}\frac{\delta^{3}\rho}{\delta\chi_{x}\delta\chi_{x^{\prime}}^{2}}\right)\label{5Necessary condition local LHS 1}\\
&+\left(\frac{dm_{2}}{d\rho}+ \rho\frac{d^{2}m_{2}}{d\rho^{2}}\right)\left(\frac{\delta\rho}{\delta\chi_{x^{\prime}}}\right)^{2}\frac{\delta^{2}\rho}{\delta\chi_{x}^{2}} \label{5Necessary condition local LHS 2}\\
&+\left(m_{2}+2\rho\frac{dm_{2}}{d\rho}\right)\frac{\delta\rho}{\delta\chi_{x^{\prime}}}\frac{\delta^{3}\rho}{\delta\chi_{x}\delta\chi_{x^{\prime}}^{2}} ,\label{5Necessary condition local LHS 3}
\end{align}

where the “$\longrightarrow$" symbolizes what remains after the symmetric pieces from the LH and RH sides cancel. This expression can itself be made symmetric in $x$, $x^{\prime}$. For instance, the first term in (<ref>) and the expression in (<ref>) could be symmetric if eq.(<ref>) was satisfied — just equate the “coefficients" with $m_{1}$, $m_{2}$, etc. On the other hand, the second term in (<ref>) and (<ref>) could be made symmetric if eq.(<ref>) is satisfied by, again, setting the undetermined coefficients equal.]

\begin{equation}
m_{1} = \frac{1}{2}\left(\frac{1}{\rho}\frac{dm_{2}}{d\rho} - \frac{m_{2}}{\rho^{2}} + 2 \frac{d^{2}m_{2}}{d\rho^{2}}\right)\label{5Necessary condition final 1}
\end{equation}

\begin{equation}
\frac{dm_{1}}{d\rho} = \frac{1}{\rho}\frac{dm_{2}}{d\rho}+\frac{d^{2}m_{2}}{d\rho^{2}}.\label{5Necessary condition final 2}
\end{equation}

Since the functions $m_{1}$ and $m_{2}$ are coupled, it means neither term in eq.(<ref>) is on its own a solution to eq.(<ref>) — they must appear together.

The coupled eqns.(<ref>) and (<ref>) can be combined into the single equation for $m_{2}$ that reads

\begin{equation}
\frac{dm_{2}(\rho)}{d\rho} = - \frac{m_{2}(\rho)}{\rho},
\end{equation}

which has the solution

\begin{equation}
m_{2}(\rho) = \frac{1}{\rho}.\label{5m2}
\end{equation}

This can then be used to determine $m_{1}$, which is most easily achieved through eq.(<ref>), which yields

\begin{equation}
m_{1}(\rho) =- \frac{1}{2}\,\frac{1}{\rho^{2}}.\label{5m1}
\end{equation}

Thus $\alpha_{x}$ has the form

\begin{equation}
\alpha_{x} = \frac{\lambda}{g_{x}^{1/2}}\left(\frac{1}{\rho}\frac{\delta^{2}\rho}{\delta\chi_{x}^{2}}- \frac{1}{2\rho^{2}}\left(\frac{\delta\rho}{\delta\chi_{x}}\right)^{2}\right) = \frac{\tilde{\delta}F_{x}^{0}}{\tilde{\delta}\rho}
\end{equation}

where $\lambda$ is a yet arbitrary coupling constant. It is now possible to integrate this expression, giving us

\begin{equation}
F_{x}^{0} = -\frac{\lambda}{2g_{x}^{1/2}}\int D\chi \frac{1}{\rho}\left(\frac{\delta\rho}{\delta\chi_{x}}\right)^{2}.
\end{equation}

Upon redefinition of the constant $\lambda$ we find that this potential agrees with the “local" quantum potential introduced in (<ref>) and (<ref>), or more explicitly,

\begin{equation}
Q_{x} = \frac{\lambda}{g_{x}^{1/2}}\int D\chi \frac{1}{\rho}\left(\frac{\delta\rho}{\delta\chi_{x}}\right)^{2}.
\end{equation}

Thus the full class of acceptable relativistic potentials can thus be written as

\begin{equation}
F_{x}^{0} = g_{x}^{1/2}\int D\chi \, \rho V_{x}(\chi_{x},X_{x}^{\mu})+Q_{x} + g_{x}^{1/2} \prod_{m}\int D\chi_{m}\rho[\chi_{m}]V_{x}^{(m)}.
\label{5PB Full solution}
\end{equation}

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