An elementary method for the problem of column subset selection in a rectangular matrix

Concerning the Restricted Invertibility problem, Bourgain and Tzafriri [1] obtained the following result for square matrices:

See also [4] for a simpler proof. Vershynin [6] generalized Bourgain and Tzafriri’s result to the case of rectangular matrices and the estimate of $|T|$ was improved as follows.

Recently, Spielman and Srivastava proposed in [3] a deterministic construction of $T$ which allows them to obtain the following result.

The technique of proof relies on new constructions and inequalities which are thoroughly explained in the Bourbaki seminar of Naor [2]. Using these techniques, Youssef [7] improved Vershynin’s result as:

We provide a deterministic algorithm that extracts a submatrix $Y_{r}$ from the matrix $X$ with guaranteed individual lower and upper bounds on each singular value of $Y_{r}$.

Consider the set of vectors $\mathcal{V}_{0}=\{x_{1},\ldots,x_{p}\}$, where the $x_{i}$ are the columns of $X$. At step $r=1$, choose $y_{1}\in\mathcal{V}_{0}$. By induction, let us be given $y_{1},\ldots,y_{r}$ at step $r$. Let $Y_{r}$ denote the matrix whose columns are $y_{1},\ldots,y_{r}$ and let $v_{k}$ be an unit eigenvector of $Y_{r}^{t}Y_{r}$ associated to $\lambda_{k,r}:=\lambda_{k}(Y_{r}^{t}Y_{r})$.

We say that $u(\cdot,\cdot)$ satisfies the hypothesis (H) if $u$ verifies for $r\geq 1$:

(1.1)		$\displaystyle 0\leq u(k,r)$	$\displaystyle\leq$	$\displaystyle u(k+1,r+1),\quad k\in\{0,\cdots,r\};$
(1.2)		$\displaystyle 0\leq u(k+1,r)$	$\displaystyle\leq$	$\displaystyle u(1,r)\ <\ u(0,r)\quad k\in\{1,\cdots,r-1\}.$

We now introduce the "potential" associated to $u(\cdot,\cdot)$ satisfying (H):

$\displaystyle Q_{r}(x)$

$\displaystyle=$

$\displaystyle\sum_{k=1}^{r}\frac{(v_{k}^{t}Y_{r}^{t}x)^{2}}{u(0,r)-u(k,r)},\quad x\in\mathcal{V}_{0}.$

We then choose $y_{r+1}\in\mathcal{V}_{r}:=\{x_{1},\ldots,x_{p}\}\setminus\{y_{1},\ldots,y_{r}\}$ so that

(1.3)

$\displaystyle Q_{r}(y_{r+1})$

$\displaystyle\leq$

$\displaystyle\frac{1}{p-r}\sum_{x\in\mathcal{V}_{r}}Q_{r}(x)=\frac{1}{p-r}\sum_{k=1}^{r}\frac{\sum_{x\in\mathcal{V}_{r}}(v_{k}^{t}Y_{r}^{t}x)^{2}}{u(0,r)-u(k,r)}.$

The following result, for which we propose a short and elementary proof, gives a control on all singular values in the column selection problem.

Consider the set of vectors $\mathcal{V}_{0}=\{x_{1},\ldots,x_{p}\}$. At step $1$, choose $y_{1}\in\mathcal{V}_{0}$. By induction, let us be given $y_{1},\ldots,y_{r}$ at step $r$. Let $Y_{r}$ denote the matrix whose columns are $y_{1},\ldots,y_{r}$ and let $v_{k}$ be an unit eigenvector of $Y_{r}^{t}Y_{r}$ associated to $\lambda_{k,r}:=\lambda_{k}(Y_{r}^{t}Y_{r})$. Let us choose $y_{r+1}\in\mathcal{V}_{r}:=\{x_{1},\ldots,x_{p}\}\setminus\{y_{1},\ldots,y_{r}\}$ so that

(2.6)

$$\sum_{k=1}^{r}\frac{(v_{k}^{t}Y_{r}^{t}y_{r+1})^{2}}{u(0,r)-u(k,r)}\leq\frac{1}{p-r}\sum_{x\in\mathcal{V}_{r}}\sum_{k=1}^{r}\frac{(v_{k}^{t}Y_{r}^{t}x)^{2}}{u(0,r)-u(k,r)}=\frac{1}{p-r}\sum_{k=1}^{r}\frac{\sum_{x\in\mathcal{V}_{r}}(v_{k}^{t}Y_{r}^{t}x)^{2}}{u(0,r)-u(k,r)}.$$

It is clear that (1.4) holds for $r=1$ since then, 1 is the only singular value because the columns are supposed to be normalized.

Assume the induction hypothesis $(H_{r})$: for all $k$ with $1\leq k\leq r<R$, (1.4) holds.

Let us then show that $(H_{r+1})$ holds. By Cauchy interlacing theorem, we have

	$\displaystyle\lambda_{k+1,r+1}$	$\displaystyle\leq$	$\displaystyle\lambda_{k,r},\quad 1\leq k\leq r$
	$\displaystyle\lambda_{k+1,r+1}$	$\displaystyle\geq$	$\displaystyle\lambda_{k+1,r},\quad 0\leq k\leq r-1.$

We then deduce, due to the induction hypothesis $(H_{r})$ and Assumption (H),

(2.7)		$\displaystyle\lambda_{k+1,r+1}$	$\displaystyle\leq$	$\displaystyle 1+\delta_{R}u(k,r)\sqrt{\lambda_{1,r}}\leq 1+\delta_{R}u(k+1,r+1)\sqrt{\lambda_{1,r+1}},\quad 1\leq k\leq r,$
(2.8)		$\displaystyle\lambda_{k+1,r+1}$	$\displaystyle\geq$	$\displaystyle 1-\delta_{R}u(r-k,r)\sqrt{\lambda_{1,r}}$
			$\displaystyle\geq$	$\displaystyle 1-\delta_{R}u(r+1-(k+1)+1,r+1)\sqrt{\lambda_{1,r+1}},\quad 0\leq k\leq r-1.$

It remains to obtain the upper estimate for $\lambda_{1,r+1}$ and the lower one for $\lambda_{r+1,r+1}$. We write

(2.14)

$\displaystyle Y_{r+1}^{t}Y_{r+1}$

$\displaystyle=$

$\displaystyle\left[\begin{array}[]{c}y_{r+1}^{t}\\ Y_{r}^{t}\end{array}\right]\left[\begin{array}[]{cc}y_{r+1}&Y_{r}\end{array}\right]=\left[\begin{array}[]{cc}1&y_{r+1}^{t}Y_{r}\\ Y_{r}^{t}y_{r+1}&Y_{r}^{t}Y_{r}\end{array}\right],$

and it is well known that the eigenvalues of $Y_{r+1}^{t}Y_{r+1}$ are the zeros of the secular equation:

(2.15)

$\displaystyle q(\lambda):=1-\lambda+\sum_{k=1}^{r}\frac{(v_{k}^{t}Y_{r}^{t}y_{r+1})^{2}}{\lambda-\lambda_{k,r}}\ =\ 0.$

We first estimate $\lambda_{1,r+1}$ which is the greatest zero of $q$, and assume for contradiction that

(2.16)

$\displaystyle\lambda_{1,r+1}>1+\delta_{R}u(0,r)\sqrt{\lambda_{1,r}}.$

From $(H_{r})$, we then obtain that for $\lambda\geq 1+\delta_{R}u(0,r)\sqrt{\lambda_{1,r}}$,

$\displaystyle q(\lambda)\leq 1-\lambda+\frac{1}{\delta_{R}\sqrt{\lambda_{1,r}}}\sum_{k=1}^{r}\frac{(v_{k}^{t}Y_{r}^{t}y_{r+1})^{2}}{u(0,r)-u(k,r)}:=g(\lambda).$

Let $\lambda^{0}$ be the zero of $g$. We have $g(\lambda_{1,r+1})\geq q(\lambda_{1,r+1})=0=g(\lambda^{0})$. But $g$ is decreasing, so

$\displaystyle\lambda_{1,r+1}\leq\lambda^{0}=1+\frac{1}{\delta_{R}\sqrt{\lambda_{1,r}}}\sum_{k=1}^{r}\frac{(v_{k}^{t}Y_{r}^{t}y_{r+1})^{2}}{u(0,r)-u(k,r)}.$

Thus, using Lemma 2.1, the equality (1.5) and noting that $r\leq p/2$, we can write:

(2.17)

$\displaystyle\lambda_{1,r+1}\leq 1+\frac{2}{\delta_{R}}\frac{\sqrt{\lambda_{1,r}}\|X\|^{2}}{p}\sum_{k=1}^{r}\frac{1}{u(0,r)-u(k,r)}\ \leq\ 1+\delta_{R}u(0,r)\sqrt{\lambda_{1,r}},$

which yields a contradiction with the inequality (2.16). Thus, we have

(2.18)

$$\lambda_{1,r+1}\leq 1+\delta_{R}u(0,r)\sqrt{\lambda_{1,r}}\leq 1+\delta_{R}u(1,r+1)\sqrt{\lambda_{1,r+1}}.$$

This shows that the upper bound in $(H_{r+1})$ holds.

Finally, to estimate $\lambda_{r+1,r+1}$ which is the smallest zero of $q$, we write

$\displaystyle q(\lambda)\geq 1-\lambda-\frac{1}{\delta_{R}\sqrt{\lambda_{1,r}}}\sum_{k=1}^{r}\frac{(v_{k}^{t}Y_{r}^{t}y_{r+1})^{2}}{u(0,r)-u(k,r)}:=\widetilde{g}(\lambda).$

By means of the same reasonning as above, we show that the lower bound in $(H_{r+1})$ holds.

Set $\mu_{1,r}=\lambda_{1,r}-1\geq 0$.

Since $u(1,r)\leq u(1,R)\leq u(0,R)$, we can write

$\displaystyle\mu_{1,r}$

$\displaystyle\leq$

$\displaystyle\delta_{R}\sqrt{\mu_{1,r}+1}.$

Hence, using that $x\leq A\sqrt{1+x}$ implies $x\leq 2A$, we reach the upper estimate for $\lambda_{1,r}$.

This concludes the proof of Theorem 1.5.