UDC 519.214.9

M. V. KOZLOV

ON THE RANK OF MATRICES WITH RANDOM BOOLEAN ELEMENTS

(Presented by Academician A. N. Kolmogorov, 27 XI 1965)

1. Let \(GF(2)\) be the Galois field of 2 elements . Let \(V^m\) be the set of all ordered collections
   \(\mathbf v=(v_1,v_2,\ldots,v_m)\) with components \(v_i\in GF(2)\). \(V^m\) forms an \(m\)-dimensional vector (coordinate) space over \(GF(2)\)*.

Throughout what follows, \(a_i^j,\ i=1,\ldots,n;\ j=1,\ldots,m;\ m\ge n\), are mutually independent random variables taking the values 1 and 0 with probabilities \(\pi_i^j\) and \(1-\pi_i^j\), respectively; \(A_s^m=A_s^m(\|\pi_i^j\|),\ s\le n\), is a random matrix whose \(i\)-th row is \(\mathbf a_i=(a_i^1,a_i^2,\ldots,a_i^m)\), and whose \(j\)-th column is \(\mathbf a^j=(a_1^j,a_2^j,\ldots,a_s^j)\). By \(\alpha_s^m=\alpha_s^m(\|\pi_i^j\|)\) we shall denote the probability that the matrix \(A_s^m\) has rank \(s\). Consider, as \(n\to\infty\), the relation

\[ \left(1-\alpha_n^m(\|\pi_i^j\|)\right) \left(1-\prod_{k=m-n+1}^{m}\left(1-\frac{1}{2^k}\right)\right)^{-1}\to 1. \tag{1} \]

The main result of the present note is

Theorem 1. Let all \(\pi_i^j\in[\delta,1-\delta]\), where \(\delta\le \tfrac12\) is a fixed positive number. Then:

a) if \(1\le m/n\le c<(1+\log(1-\delta))^{-1}\), then (1) holds uniformly in \(\pi_i^j\in[\delta,1-\delta]\);

b) if \(m/n\ge (1+\log(1-\delta))^{-1}(1+\varepsilon/n)\), where \(\varepsilon>0\) is fixed, then there exists a collection of \(\pi_i^j\) lying in \([\delta,1-\delta]\) for which (1) does not hold.

Remark 1. From assertion a) it follows that, uniformly in \(\pi_i^j\in[\delta,1-\delta]\), as \(n\to\infty\),

\[ \alpha_n^m(\|\pi_i^j\|) \left(\prod_{k=m-n+1}^{m}\left(1-\frac{1}{2^k}\right)\right)^{-1}\to 1. \tag{2} \]

But, as analysis of formula (1) shows, \(\alpha_n^m(\|\pi_i^j\|)\to 1\) if \(m-n\to\infty\). Therefore (1) follows from (2) only in the case when \(m-n\) is bounded as \(n\to\infty\).

Remark 2. In (1) it was shown that for \(\pi_i^j=\tfrac12,\ i=1,\ldots,n;\ j=1,\ldots,m;\ m\ge n\), in (2) the equality sign holds instead of \(\to\). In (2) the relation (2) is proved for \(m=n\).

Theorem 1 is also valid when \(\delta\to0\) as \(n\to\infty\); more precisely, the following holds.

Theorem \(1'\). Let all \(\pi_i^j\in[\delta_n^m,1-\delta_n^m]\), where \(\delta_n^m\) is a positive function of \(n\) and \(m\), not exceeding \(\tfrac12\), and \(\varepsilon\) is an arbitrary positive number. Then:

a′) if \(\delta_n^m=(\log n)^{1+\varepsilon}/m,\ m\le n+(\log n)^{1+\varepsilon}-3\log n\), then (1) holds uniformly in \(\pi_i^j\);

* The field elements are 0 and 1; addition and multiplication are defined by the relations
\(0\oplus0=1\oplus1=0,\ 0\oplus1=1\oplus0=1,\ 0\cdot0=0\cdot1=1\cdot0=0,\ 1\cdot1=1.\)

** Operations in \(V^m\) are given by the relations
\(\mathbf v\oplus\mathbf u=(v_1,\ldots,v_m)\oplus(u_1,\ldots,u_m)=(v_1\oplus u_1,\ldots,v_m\oplus u_m)\),
\(w\cdot\mathbf v=(w\cdot v_1,\ldots,w\cdot v_m)\), where \(w,v_i,u_i\in GF(2)\).

*** Here and below, \(\log\) is taken to base 2.

b′) if \(\delta_n^m=(\log n)^{1-\varepsilon}/m\), then there exists a collection
\(\pi_i^j\in[\delta_n^m,1-\delta_n^m]\) for which (1) does not hold for any \(m\).

Below the proof of Theorem 1 is given. First (Lemma 1), assertion a) is proved under the additional condition \(m-n\to\infty\) as \(n\to\infty\). Lemma 2 reduces the case of bounded \(m-n\) to the preceding one.

1. Denote by \(D_\nu\) the event
   \[ \{x_1^\nu\mathbf a_1\oplus x_2^\nu\mathbf a_2\oplus\cdots\oplus x_n^\nu\mathbf a_n=0_m\}, \]
   consisting in the fact that the linear combination of the rows \(\mathbf a_i\) with coefficients \(x_i^\nu\) is the zero row; here \(x_1^\nu x_2^\nu\ldots x_n^\nu\) is the binary notation of the number \(\nu\), \(1\le \nu\le 2^n-1\), and \(0_m\) is the zero vector of the space \(V^m\). The events \(\{\operatorname{rank} A_n^m<n\}\) and \(\bigcup D_\nu\), where the union is taken over all \(\nu\) from 1 to \(2^n-1\), coincide. Therefore we have
   \[ \sum_{1\le \nu\le 2^n-1} P\{D_\nu\}\ge P\left\{\bigcup_{1\le \nu\le 2^n-1}D_\nu\right\}=1-\alpha_n^m\ge \]
   \[ \ge \sum_{1\le \nu\le 2^n-1}P\{D_\nu\}- \sum_{1\le \nu<\mu\le 2^n-1}P\{D_\nu\cap D_\mu\}. \tag{3} \]

Let us estimate \(P\{D_\nu\}\) and \(P\{D_\nu\cap D_\mu\}\). Since
\[ P\{a_1^j\oplus\cdots\oplus a_r^j=0\} = P\{a_1^j\oplus\cdots\oplus a_{r-1}^j=0\}\,P\{a_r^j=0\} + P\{a_1^j\oplus\cdots\oplus a_{r-1}^j=1\}\,P\{a_r^j=1\}, \]
putting \(P\{a_1^j\oplus\cdots\oplus a_r^j=0\}=1/2+\varepsilon_r\), we obtain
\[ \varepsilon_r=\varepsilon_{r-1}(1-2\pi_r^j) \]
and \(\varepsilon_1=1/2-\pi_1^j\). Consequently,
\[ |\varepsilon_r|\le \tfrac12(1-2\delta)^r . \]
In exactly the same way, introducing the notation \(a(r)\) for
\(a_1^j\oplus\cdots\oplus a_r^j\), \(a(s)\) for
\(a_{r+1}^j\oplus\cdots\oplus a_{r+s}^j\), and \(a(t)\) for
\(a_{r+s+1}^j\oplus\cdots\oplus a_{r+s+t}^j\), we obtain
\[ P\{a(r)\oplus a(s)=0,\ a(s)\oplus a(t)=0\}= \]
\[ = P\{a(r)=0\}P\{a(s)=0\}P\{a(t)=0\} + P\{a(r)=1\}P\{a(s)=1\}P\{a(t)=1\}. \]

Using the estimate already obtained above for each of the factors and putting
\(\Delta=1-2\delta\), for the probability (4), as an upper bound, we shall have the expression
\(\tfrac14(1+\Delta^{r+s}+\Delta^{r+t}+\Delta^{s+t})\). In view of the independence of the \(a_i^j\),
\[ P\{D_\nu\}=P\{x_1^\nu a_1^1\oplus\cdots\oplus x_n^\nu a_n^1=0\}\cdots P\{x_1^\nu a_1^m\oplus\cdots\oplus x_n^\nu a_n^m=0\}, \]
and an analogous relation is valid for \(P\{D_\nu\cap D_\mu\}\). Combining the estimates obtained, we arrive at
\[ \sum_{1\le k\le n} C_n^k(1+\Delta^k)^m \le 2^m\sum_{1\le \nu\le 2^n-1}P\{D_\nu\} \le \sum_{1\le k\le n} C_n^k(1+\Delta^k)^m, \tag{5} \]
\[ 2^{2m}\sum_{1\le \nu<\mu\le 2^n-1}P\{D_\nu\cap D_\mu\} \le \sum_{\substack{2\le r+s+t\le n\\ r\le s}} \frac{n!(1+\Delta^{r+s}+\Delta^{r+t}+\Delta^{s+t})^m} {r!\,s!\,t!\,(n-r-s-t)!}. \tag{6} \]

Lemma 1. If \(m-n\to\infty\), \(m/n\le c<(1+\log(1-\delta))^{-1}\), then, uniformly in \(\pi_i^j\in[\delta,1-\delta]\), (1) holds.

Indeed, substituting (5) and (6) into (3), we obtain that, under the conditions of Lemma 1,
\[ \alpha_n^m(\|\pi_i^j\|)=2^{\,n-m}\bigl(1+\theta_{nm}(\|\pi_i^j\|)\bigr), \]
where \(\theta_{nm}\to0\). At the same time, the second factor in (1) is bounded above and below, respectively, by the expressions
\[ ((2^n-1)2^{-m}-\tfrac12\cdot 2^{2n-2m})^{-1} \]
and \(2^{m-n}\).

For the proof of assertion b) of Theorem 1, put \(\pi_n^j=\delta\) for \(j=1,\ldots,m\). In this case
\[ 1-\alpha_n^m=P\{\cup D_\nu\}>P\{D_1\}=P\{\mathbf a_n=0_m\}=(1-\delta)^m, \]
and then the left-hand side of relation (1) is bounded below by the number \(2^\varepsilon\).

1. Let \(A^m\) be the random subspace of \(V^m\) generated by the rows of the matrix \(A_{n-1}^m\). Let \(X^m\) run through all distinct vector subspaces of \(V^m\) of dimension \(m-n+1\). In each \(X^m\) fix some basis, and from the basis vectors, as rows, form a matrix \(X_{m-n+1}^m\). Denote by \(A(X^m)\) the event consisting in the fact that \(A^m\) coincides with the space of solutions of the system of linear homogeneous equa-

with coefficient matrix \(X_{m-n-1}^{m}\):

\[ x_i^1 a^1 \oplus x_i^2 a^2 \oplus \ldots \oplus x_i^m a^m = 0,\quad i=1,\ldots,m-n+1. \tag{7} \]

It is easy to see that \(\bigcup A(X^m)\) over all \(X^m\) represents the event \(\{\operatorname{rank} A_{n-1}^m=n-1\}\). Using this, we obtain
\[ \{\operatorname{rank} A_n^m=n\} = \{(\operatorname{rank} A_{n-1}^m=n-1)\cap(\operatorname{rank} A_n^m=n)\} = \bigcup\{A(X^m)\cap(\operatorname{rank} A_n^m=n)\}. \]
The events under the union sign coincide with
\[ A(X^m)\cap\{x^1 a_n^1\oplus x^2 a_n^2\oplus\ldots\oplus x^m a_n^m\ne 0_{m-n+1}\}, \]
where \(x^j\) is the \(j\)-th column of \(X_{m-n+1}^{m}\). Passing to probabilities and noting that the \(A(X^m)\) are pairwise disjoint, while the events joined in the last relation by the sign \(\cap\) are independent, we obtain

\[ \alpha_n^m=\sum P\{A(X^m)\}\, P\{x^1 a_n^1\oplus\ldots\oplus x^m a_n^m\ne 0_{m-n+1}\}. \tag{8} \]

4. We shall say that a subspace \(X^m\) (of dimension \(m-n+1\)) has property \((t)\) if in \(X^m\) one can choose a basis such that the matrix whose rows are all the vectors of this basis contains at least \(t\) columns of each of the following \(m-n+1\) types:
\[ (1,0,\ldots,0),\quad (0,1,\ldots,0),\ldots,(0,0,\ldots,1). \]
If \(X^m\) has property \((t)\), then it is easy to show that
\[ P\{x^1 a_n^1\oplus\ldots\oplus x^m a_n^m=0_{m-n+1}\} =(1/2)^{m-n+1}(1+\theta\Delta^t)^{m-n+1}, \quad |\theta|\le 1. \tag{9} \]

For clarity of the argument, let us put for the time being \(m-n+1=3\). By a permutation of columns the matrix \(X_3^m\) can be brought to the form

\[ \left\| \begin{array}{cccccccccccccccccccccc} 1&\ldots&1&0&\ldots&0&0&\ldots&0&1&\ldots&1&1&\ldots&1&0&\ldots&0&1&\ldots&1&0&\ldots&0\\ 0&\ldots&0&1&\ldots&1&0&\ldots&0&1&\ldots&1&0&\ldots&0&1&\ldots&1&1&\ldots&1&0&\ldots&0\\ 0&\ldots&0&0&\ldots&0&1&\ldots&1&0&\ldots&0&1&\ldots&1&1&\ldots&1&1&\ldots&1&0&\ldots&0 \end{array} \right\|. \]

Denote by \(l_i\) the number of columns of each of the first 7 types, \(l_i\ge 0\). Consider the following typical cases:
\[ \begin{array}{ll} 1^\circ.\ l_1\ge t,\ l_2\ge t,\ l_3\ge t. & 2^\circ.\ l_1\ge t,\ l_2\ge t,\ l_3<t. \\[2mm] 3^\circ.\ l_1\ge t,\ l_2<t,\ l_3<t. & 4^\circ.\ l_1<t,\ l_2<t,\ l_3<t. \end{array} \]

In case \(1^\circ\) the space \(X^m\) has property \((t)\). Case \(2^\circ\) reduces to \(1^\circ\) if at least one of the numbers \(l_5,l_6\), or \(l_7\) is not less than \(t\). To see this, it is enough to add the 3rd row to the 1st, to the 2nd, or to both, according as \(l_5,l_6\), or \(l_7\) is not less than \(t\). Otherwise \(X^m\) contains a vector of weight* \(<2^2 t\). Cases \(3^\circ\) and \(4^\circ\) are considered analogously. Carrying out these arguments in the general case, we obtain: if \(X^m\) contains no vector of weight \(<2^{m-n}t\), then \(X^m\) has property \((t)\). Consequently, if \(X^m\) does not have property \((t)\), then
\[ A(X^m)\subseteq\{x_1a^1\oplus\ldots\oplus x_m a^m=0_{n-1}\}, \]
where \((x_1,\ldots,x_m)\) is some nonzero vector of weight \(<2^{m-n}t\), depending on the choice of \(X^m\). Combining the last relation over all \(X^m\) not having property \((t)\), and passing to probabilities, we shall have

\[ \sum P\{A(X^m)\} \le P\{U\,x_1a^1\oplus\ldots\oplus x_m a^m=0_{n-1}\} \le \sum_{k=1}^{2^{m-n}t} C_m^k\left(1/2+1/2\,\Delta^k\right)^{n-1}, \tag{10} \]

where \(\sum\) is taken over all \(X^m\) not having property \((t)\), and \(U\) over all nonzero vectors \((x_1,\ldots,x_m)\) of weight \(<2^{m-n}t\).

Lemma 2. If \(0\le m-n\le K\log\log n\), where \(K\) is an arbitrary constant, then
\[ \alpha_n^m(\|\pi_i^j\|) = (1-2^{\,n-m-1})\cdot \alpha_{n-1}^m(\|\pi_i^j\|) + R_{mn}(\|\pi_i^j\|), \]
\[ (1/\sqrt n)\log |R_{mn}|\to-\infty\quad\text{as }n\to\infty \]
uniformly in \(m\) and \(\pi_i^j\in[\delta,1-\delta]\).

* The weight of a vector is the number of coordinates different from zero.

For the proof it is necessary to split the sum (8) into two parts: over \(X^m\) possessing and not possessing property \((t)\), choose \(t=n/(\log n)^{K+1}\), and then use relations (9) and (10).

If \(m-n\) is bounded as \(n\to\infty\), then, repeatedly applying Lemma 2 to \(A_n^m\), and then using Lemma 1, we obtain the proof of assertion a) in this case as well.

The proof of assertion a′) of Theorem 1′ is quite analogous, but it requires a more careful estimate of the right-hand sides of the sums in (5) and (6). As for b′), it is necessary to put \(\pi_i^j=\delta_n^m\) for all \(i\) and \(j\), and then to use the inequality

\[ \alpha_n^m \leq P\left\{\bigcap_{1\leq i\leq n}\mathbf a_i \ne 0_m\right\}=\left(1-(1-\delta_n^m)^m\right)^n. \]

The author takes this opportunity to express his deep gratitude to A. N. Kolmogorov, who posed the present problem (Theorems 1 and 1′) to the author and guided its solution.

Moscow State University
named after M. V. Lomonosov

Received
24 XI 1965

REFERENCES

1. C. Shannon, Collected Works. The Mathematical Theory of Communication, IL, 1957.
2. I. N. Kovalenko, DAN, 161, No. 3, 517 (1965).