Mixing
“Biggest” coefficients of a linear combination between vectors of zeros and ones
$begingroup$
Let $n$ be a positive integer. Denote by $B_n$ the set of $ntimes(n+1)$-matrices of rank $n$ and with coefficients in ${0,1}$. I would like to measure how "complex" the coefficients of a linear combination of the columns of a matrix of $B_n$ can be. More precisely, I'd like to compute (or estimate the asymptotic behaviour of)
$$
P_n := max left{
prod_{i=1}^{n+1}|lambda_i|,;
Min B_n,;
sum_{i=1}^{n+1}lambda_i m_{star,i} = 0,;
lambda_1,dots,lambda_{n+1}in mathbb{Z},; gcd(lambda_1,dots,lambda_{n+1})=1
right}
$$
where $m_{star,i}$ stands for the $i$-th column vector of the matrix $M$.
$P_2=1$ and $P_3 = 2$, and for every $1le kle n-2$ with $gcd(k,n-1)=1$, $P_n$ is bounded below by $(n-1)(n-(k+1))^{k}k^{n-k}$ (indeed, setting $v$ the vector whose $k$ first coefficients are 1 and $n-k$ last coefficients are 0 and $hat e_i$ the vector whose only 0 coefficient is at line i and whose other coefficients are 1, we have the following combination: $
(n-1) v + (n-(k+1))(hat e_1 +dots + hat e_k) = k(hat e_{k+1} +dots + hat e_n).)
$
linear-algebra combinatorics
asked Jan 12 at 13:59
AlexandreVAlexandreV
84
$endgroup$
add a comment |
$begingroup$
Let $n$ be a positive integer. Denote by $B_n$ the set of $ntimes(n+1)$-matrices of rank $n$ and with coefficients in ${0,1}$. I would like to measure how "complex" the coefficients of a linear combination of the columns of a matrix of $B_n$ can be. More precisely, I'd like to compute (or estimate the asymptotic behaviour of)
$$
P_n := max left{
prod_{i=1}^{n+1}|lambda_i|,;
Min B_n,;
sum_{i=1}^{n+1}lambda_i m_{star,i} = 0,;
lambda_1,dots,lambda_{n+1}in mathbb{Z},; gcd(lambda_1,dots,lambda_{n+1})=1
right}
$$
where $m_{star,i}$ stands for the $i$-th column vector of the matrix $M$.
$P_2=1$ and $P_3 = 2$, and for every $1le kle n-2$ with $gcd(k,n-1)=1$, $P_n$ is bounded below by $(n-1)(n-(k+1))^{k}k^{n-k}$ (indeed, setting $v$ the vector whose $k$ first coefficients are 1 and $n-k$ last coefficients are 0 and $hat e_i$ the vector whose only 0 coefficient is at line i and whose other coefficients are 1, we have the following combination: $
(n-1) v + (n-(k+1))(hat e_1 +dots + hat e_k) = k(hat e_{k+1} +dots + hat e_n).)
$
linear-algebra combinatorics
asked Jan 12 at 13:59
AlexandreVAlexandreV
84
$endgroup$
add a comment |
$begingroup$
Let $n$ be a positive integer. Denote by $B_n$ the set of $ntimes(n+1)$-matrices of rank $n$ and with coefficients in ${0,1}$. I would like to measure how "complex" the coefficients of a linear combination of the columns of a matrix of $B_n$ can be. More precisely, I'd like to compute (or estimate the asymptotic behaviour of)
$$
P_n := max left{
prod_{i=1}^{n+1}|lambda_i|,;
Min B_n,;
sum_{i=1}^{n+1}lambda_i m_{star,i} = 0,;
lambda_1,dots,lambda_{n+1}in mathbb{Z},; gcd(lambda_1,dots,lambda_{n+1})=1
right}
$$
where $m_{star,i}$ stands for the $i$-th column vector of the matrix $M$.
$P_2=1$ and $P_3 = 2$, and for every $1le kle n-2$ with $gcd(k,n-1)=1$, $P_n$ is bounded below by $(n-1)(n-(k+1))^{k}k^{n-k}$ (indeed, setting $v$ the vector whose $k$ first coefficients are 1 and $n-k$ last coefficients are 0 and $hat e_i$ the vector whose only 0 coefficient is at line i and whose other coefficients are 1, we have the following combination: $
(n-1) v + (n-(k+1))(hat e_1 +dots + hat e_k) = k(hat e_{k+1} +dots + hat e_n).)
$
linear-algebra combinatorics
asked Jan 12 at 13:59
AlexandreVAlexandreV
84
$endgroup$
Let $n$ be a positive integer. Denote by $B_n$ the set of $ntimes(n+1)$-matrices of rank $n$ and with coefficients in ${0,1}$. I would like to measure how "complex" the coefficients of a linear combination of the columns of a matrix of $B_n$ can be. More precisely, I'd like to compute (or estimate the asymptotic behaviour of)
$$
P_n := max left{
prod_{i=1}^{n+1}|lambda_i|,;
Min B_n,;
sum_{i=1}^{n+1}lambda_i m_{star,i} = 0,;
lambda_1,dots,lambda_{n+1}in mathbb{Z},; gcd(lambda_1,dots,lambda_{n+1})=1
right}
$$
where $m_{star,i}$ stands for the $i$-th column vector of the matrix $M$.
$P_2=1$ and $P_3 = 2$, and for every $1le kle n-2$ with $gcd(k,n-1)=1$, $P_n$ is bounded below by $(n-1)(n-(k+1))^{k}k^{n-k}$ (indeed, setting $v$ the vector whose $k$ first coefficients are 1 and $n-k$ last coefficients are 0 and $hat e_i$ the vector whose only 0 coefficient is at line i and whose other coefficients are 1, we have the following combination: $
(n-1) v + (n-(k+1))(hat e_1 +dots + hat e_k) = k(hat e_{k+1} +dots + hat e_n).)
$
linear-algebra combinatorics
linear-algebra combinatorics
asked Jan 12 at 13:59
AlexandreVAlexandreV
84
asked Jan 12 at 13:59
AlexandreVAlexandreV
84
edited Jan 13 at 7:07
AlexandreV
asked Jan 12 at 13:59
AlexandreVAlexandreV
84
asked Jan 12 at 13:59
AlexandreVAlexandreV
84
asked Jan 12 at 13:59
AlexandreVAlexandreV
84
84
add a comment |
add a comment |
1 Answer
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votes
$begingroup$
We can show that $P_nle (n+1)^{n(n+1)}$ for each $nge 2$, becuase a few months ago I proved a following lemma.
For a natural number $n$ let $[n]$ denotes a set ${1,dots, n}$. Given a subset $Y$ of a vector space $X$ over $mathbb R$ by $langle Yrangle$ we denote the linear hull of $Y$ in $X$, that is a set of all finite sums $f_1y_1+dots+f_ky_k$, where $f_iinmathbb R$ and $y_iin Y$ for each $i$.
Lemma. Let $K$ and $N$ be positive integers,
$V={v_1,dots, v_k}subset [0,K]^N$ be a linearly dependent over $mathbb R$ system of vectors with
integer entries. There exist integers
$f_1,dots, f_k$ which are not all zeroes such that $|f_i|le (kK)^{k-1}$ for each $i$ and
$f_1v_1+dots+f_kv_k=0$.
Proof. Let $W$ be a maximal linearly independent subset of a set $V$. Since the set $V$ is linearly
dependent, $|W|le k-1$. For each $iin [N]$ let
$e^i=(e^i_1,dots,e^i_N)inmathbb R^N$ be $i$-th standard orth, that is $e^i_i=1$ and $e^i_j=0$ for each $jne i$. Let
$B_0={e^1,dots,e^n}$ be the standard basis of the linear space $mathbb R^N$.
By [Lan, Ch. III, Theorem 2], there exists a basis $B$ of the space $mathbb R^N$
such that $Wsubset Bsubset Wcup B_0$. Let $C=B_0setminus (Bsetminus W)$ and
$p_{C}:mathbb R^Nto langle Crangle$ be the orthogonal
projection, that is $p_{C}(x)=sum{x_ie^i:x_iinmathbb R$, $e^iin C}$
for each vector $x=(x_1,dots,x_N)in mathbb R^N$. Thus
$ker p_{C}={xin mathbb R^N:p_{C}(x)=0}=langle B_0setminus Crangle=
langle Bsetminus Wrangle$. We have $ker p_{C}cap langle Wrangle=langle Bsetminus
Wranglecaplangle Wrangle=0$, because otherwise the set $B$ is linearly dependent.
Thus the restriction $p_{C}|langle Wrangle$ of the map $p_{C}$ on the set $langle Wrangle$
is injective.
Put $K'=(kK)^{k-1}$. Define a map $f$ from the subset $D^k$ of points of the set $[0, K']^k$ with all
integer coordinates to $langle Wranglecap mathbb Z^Nsubset mathbb R^N$ as follows.
Let $d=(d_1,dots,d_k)in D^k$. Put $f(d)=p_C(dv)$, where $dv=d_1v_1+dots d_kv_k$.
Since $d_iin [0, K']$ and $v_iin [0,K]^N$ for each $iin [k]$, each coordinate of a vector
$dv$ (and, hence, of the vector $f(d)=p_C(dv)$ too) is at most $kK'K$. Since
$$|C|=|B_0setminus (Bsetminus W)|=|B_0|-|Bsetminus W|=|B_0|-(|B|-|W|)=
N-(N-|W|)=|W|le k-1,$$
$|f(Q)|le (kK'K+1)^{k-1}$. We have $|D^k|>|f(Q)|$, because $(1+(kK)^{k-1})^{frac 1{k-1}}>(1+(kK)^k)^{frac 1{k}}$,
because when $a>1$ is a constant and $x>0$ a function $(1+a^x)^{frac 1x}$ decreases.
Therefore the function $f$ is not injective. So there exist distinct elements $d=(d_1,dots,d_k)$
and $d'=(d'_1,dots,d'_k)$ of $D^k$ such that $p_C(dv)=f(d)=f(d')=p_C(d'v)$. Since $dv$ and $dv'$
belong to $langle Wrangle$ and the restriction $p_{C}|langle Wrangle$ is injective,
$dv=d'v$. It remains to put $f_i=d_i-d'_i$ for each $iin [k]$.$square$
Remark that for each $B_n$, the sequence $(lambda_1,dots,lambda_{n+1})$ is determined up to a multiplication by $(-1)$. The lemma implies that $|lambda_i|le (n+1)^n$ for each $i$, so $P_nle (n+1)^{n(n+1)}$.
References
[L] Serge Lange, Algebra, Addison-Wesley, 1965 (Russian translation, Moskow, Mir, 1968).
answered Jan 24 at 3:59
Alex RavskyAlex Ravsky
41.1k32282
$endgroup$
add a comment |
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$begingroup$
We can show that $P_nle (n+1)^{n(n+1)}$ for each $nge 2$, becuase a few months ago I proved a following lemma.
For a natural number $n$ let $[n]$ denotes a set ${1,dots, n}$. Given a subset $Y$ of a vector space $X$ over $mathbb R$ by $langle Yrangle$ we denote the linear hull of $Y$ in $X$, that is a set of all finite sums $f_1y_1+dots+f_ky_k$, where $f_iinmathbb R$ and $y_iin Y$ for each $i$.
Lemma. Let $K$ and $N$ be positive integers,
$V={v_1,dots, v_k}subset [0,K]^N$ be a linearly dependent over $mathbb R$ system of vectors with
integer entries. There exist integers
$f_1,dots, f_k$ which are not all zeroes such that $|f_i|le (kK)^{k-1}$ for each $i$ and
$f_1v_1+dots+f_kv_k=0$.
Proof. Let $W$ be a maximal linearly independent subset of a set $V$. Since the set $V$ is linearly
dependent, $|W|le k-1$. For each $iin [N]$ let
$e^i=(e^i_1,dots,e^i_N)inmathbb R^N$ be $i$-th standard orth, that is $e^i_i=1$ and $e^i_j=0$ for each $jne i$. Let
$B_0={e^1,dots,e^n}$ be the standard basis of the linear space $mathbb R^N$.
By [Lan, Ch. III, Theorem 2], there exists a basis $B$ of the space $mathbb R^N$
such that $Wsubset Bsubset Wcup B_0$. Let $C=B_0setminus (Bsetminus W)$ and
$p_{C}:mathbb R^Nto langle Crangle$ be the orthogonal
projection, that is $p_{C}(x)=sum{x_ie^i:x_iinmathbb R$, $e^iin C}$
for each vector $x=(x_1,dots,x_N)in mathbb R^N$. Thus
$ker p_{C}={xin mathbb R^N:p_{C}(x)=0}=langle B_0setminus Crangle=
langle Bsetminus Wrangle$. We have $ker p_{C}cap langle Wrangle=langle Bsetminus
Wranglecaplangle Wrangle=0$, because otherwise the set $B$ is linearly dependent.
Thus the restriction $p_{C}|langle Wrangle$ of the map $p_{C}$ on the set $langle Wrangle$
is injective.
Put $K'=(kK)^{k-1}$. Define a map $f$ from the subset $D^k$ of points of the set $[0, K']^k$ with all
integer coordinates to $langle Wranglecap mathbb Z^Nsubset mathbb R^N$ as follows.
Let $d=(d_1,dots,d_k)in D^k$. Put $f(d)=p_C(dv)$, where $dv=d_1v_1+dots d_kv_k$.
Since $d_iin [0, K']$ and $v_iin [0,K]^N$ for each $iin [k]$, each coordinate of a vector
$dv$ (and, hence, of the vector $f(d)=p_C(dv)$ too) is at most $kK'K$. Since
$$|C|=|B_0setminus (Bsetminus W)|=|B_0|-|Bsetminus W|=|B_0|-(|B|-|W|)=
N-(N-|W|)=|W|le k-1,$$
$|f(Q)|le (kK'K+1)^{k-1}$. We have $|D^k|>|f(Q)|$, because $(1+(kK)^{k-1})^{frac 1{k-1}}>(1+(kK)^k)^{frac 1{k}}$,
because when $a>1$ is a constant and $x>0$ a function $(1+a^x)^{frac 1x}$ decreases.
Therefore the function $f$ is not injective. So there exist distinct elements $d=(d_1,dots,d_k)$
and $d'=(d'_1,dots,d'_k)$ of $D^k$ such that $p_C(dv)=f(d)=f(d')=p_C(d'v)$. Since $dv$ and $dv'$
belong to $langle Wrangle$ and the restriction $p_{C}|langle Wrangle$ is injective,
$dv=d'v$. It remains to put $f_i=d_i-d'_i$ for each $iin [k]$.$square$
Remark that for each $B_n$, the sequence $(lambda_1,dots,lambda_{n+1})$ is determined up to a multiplication by $(-1)$. The lemma implies that $|lambda_i|le (n+1)^n$ for each $i$, so $P_nle (n+1)^{n(n+1)}$.
References
[L] Serge Lange, Algebra, Addison-Wesley, 1965 (Russian translation, Moskow, Mir, 1968).
answered Jan 24 at 3:59
Alex RavskyAlex Ravsky
41.1k32282
$endgroup$
add a comment |
$begingroup$
We can show that $P_nle (n+1)^{n(n+1)}$ for each $nge 2$, becuase a few months ago I proved a following lemma.
For a natural number $n$ let $[n]$ denotes a set ${1,dots, n}$. Given a subset $Y$ of a vector space $X$ over $mathbb R$ by $langle Yrangle$ we denote the linear hull of $Y$ in $X$, that is a set of all finite sums $f_1y_1+dots+f_ky_k$, where $f_iinmathbb R$ and $y_iin Y$ for each $i$.
Lemma. Let $K$ and $N$ be positive integers,
$V={v_1,dots, v_k}subset [0,K]^N$ be a linearly dependent over $mathbb R$ system of vectors with
integer entries. There exist integers
$f_1,dots, f_k$ which are not all zeroes such that $|f_i|le (kK)^{k-1}$ for each $i$ and
$f_1v_1+dots+f_kv_k=0$.
Proof. Let $W$ be a maximal linearly independent subset of a set $V$. Since the set $V$ is linearly
dependent, $|W|le k-1$. For each $iin [N]$ let
$e^i=(e^i_1,dots,e^i_N)inmathbb R^N$ be $i$-th standard orth, that is $e^i_i=1$ and $e^i_j=0$ for each $jne i$. Let
$B_0={e^1,dots,e^n}$ be the standard basis of the linear space $mathbb R^N$.
By [Lan, Ch. III, Theorem 2], there exists a basis $B$ of the space $mathbb R^N$
such that $Wsubset Bsubset Wcup B_0$. Let $C=B_0setminus (Bsetminus W)$ and
$p_{C}:mathbb R^Nto langle Crangle$ be the orthogonal
projection, that is $p_{C}(x)=sum{x_ie^i:x_iinmathbb R$, $e^iin C}$
for each vector $x=(x_1,dots,x_N)in mathbb R^N$. Thus
$ker p_{C}={xin mathbb R^N:p_{C}(x)=0}=langle B_0setminus Crangle=
langle Bsetminus Wrangle$. We have $ker p_{C}cap langle Wrangle=langle Bsetminus
Wranglecaplangle Wrangle=0$, because otherwise the set $B$ is linearly dependent.
Thus the restriction $p_{C}|langle Wrangle$ of the map $p_{C}$ on the set $langle Wrangle$
is injective.
Put $K'=(kK)^{k-1}$. Define a map $f$ from the subset $D^k$ of points of the set $[0, K']^k$ with all
integer coordinates to $langle Wranglecap mathbb Z^Nsubset mathbb R^N$ as follows.
Let $d=(d_1,dots,d_k)in D^k$. Put $f(d)=p_C(dv)$, where $dv=d_1v_1+dots d_kv_k$.
Since $d_iin [0, K']$ and $v_iin [0,K]^N$ for each $iin [k]$, each coordinate of a vector
$dv$ (and, hence, of the vector $f(d)=p_C(dv)$ too) is at most $kK'K$. Since
$$|C|=|B_0setminus (Bsetminus W)|=|B_0|-|Bsetminus W|=|B_0|-(|B|-|W|)=
N-(N-|W|)=|W|le k-1,$$
$|f(Q)|le (kK'K+1)^{k-1}$. We have $|D^k|>|f(Q)|$, because $(1+(kK)^{k-1})^{frac 1{k-1}}>(1+(kK)^k)^{frac 1{k}}$,
because when $a>1$ is a constant and $x>0$ a function $(1+a^x)^{frac 1x}$ decreases.
Therefore the function $f$ is not injective. So there exist distinct elements $d=(d_1,dots,d_k)$
and $d'=(d'_1,dots,d'_k)$ of $D^k$ such that $p_C(dv)=f(d)=f(d')=p_C(d'v)$. Since $dv$ and $dv'$
belong to $langle Wrangle$ and the restriction $p_{C}|langle Wrangle$ is injective,
$dv=d'v$. It remains to put $f_i=d_i-d'_i$ for each $iin [k]$.$square$
Remark that for each $B_n$, the sequence $(lambda_1,dots,lambda_{n+1})$ is determined up to a multiplication by $(-1)$. The lemma implies that $|lambda_i|le (n+1)^n$ for each $i$, so $P_nle (n+1)^{n(n+1)}$.
References
[L] Serge Lange, Algebra, Addison-Wesley, 1965 (Russian translation, Moskow, Mir, 1968).
answered Jan 24 at 3:59
Alex RavskyAlex Ravsky
41.1k32282
$endgroup$
add a comment |
$begingroup$
We can show that $P_nle (n+1)^{n(n+1)}$ for each $nge 2$, becuase a few months ago I proved a following lemma.
For a natural number $n$ let $[n]$ denotes a set ${1,dots, n}$. Given a subset $Y$ of a vector space $X$ over $mathbb R$ by $langle Yrangle$ we denote the linear hull of $Y$ in $X$, that is a set of all finite sums $f_1y_1+dots+f_ky_k$, where $f_iinmathbb R$ and $y_iin Y$ for each $i$.
Lemma. Let $K$ and $N$ be positive integers,
$V={v_1,dots, v_k}subset [0,K]^N$ be a linearly dependent over $mathbb R$ system of vectors with
integer entries. There exist integers
$f_1,dots, f_k$ which are not all zeroes such that $|f_i|le (kK)^{k-1}$ for each $i$ and
$f_1v_1+dots+f_kv_k=0$.
Proof. Let $W$ be a maximal linearly independent subset of a set $V$. Since the set $V$ is linearly
dependent, $|W|le k-1$. For each $iin [N]$ let
$e^i=(e^i_1,dots,e^i_N)inmathbb R^N$ be $i$-th standard orth, that is $e^i_i=1$ and $e^i_j=0$ for each $jne i$. Let
$B_0={e^1,dots,e^n}$ be the standard basis of the linear space $mathbb R^N$.
By [Lan, Ch. III, Theorem 2], there exists a basis $B$ of the space $mathbb R^N$
such that $Wsubset Bsubset Wcup B_0$. Let $C=B_0setminus (Bsetminus W)$ and
$p_{C}:mathbb R^Nto langle Crangle$ be the orthogonal
projection, that is $p_{C}(x)=sum{x_ie^i:x_iinmathbb R$, $e^iin C}$
for each vector $x=(x_1,dots,x_N)in mathbb R^N$. Thus
$ker p_{C}={xin mathbb R^N:p_{C}(x)=0}=langle B_0setminus Crangle=
langle Bsetminus Wrangle$. We have $ker p_{C}cap langle Wrangle=langle Bsetminus
Wranglecaplangle Wrangle=0$, because otherwise the set $B$ is linearly dependent.
Thus the restriction $p_{C}|langle Wrangle$ of the map $p_{C}$ on the set $langle Wrangle$
is injective.
Put $K'=(kK)^{k-1}$. Define a map $f$ from the subset $D^k$ of points of the set $[0, K']^k$ with all
integer coordinates to $langle Wranglecap mathbb Z^Nsubset mathbb R^N$ as follows.
Let $d=(d_1,dots,d_k)in D^k$. Put $f(d)=p_C(dv)$, where $dv=d_1v_1+dots d_kv_k$.
Since $d_iin [0, K']$ and $v_iin [0,K]^N$ for each $iin [k]$, each coordinate of a vector
$dv$ (and, hence, of the vector $f(d)=p_C(dv)$ too) is at most $kK'K$. Since
$$|C|=|B_0setminus (Bsetminus W)|=|B_0|-|Bsetminus W|=|B_0|-(|B|-|W|)=
N-(N-|W|)=|W|le k-1,$$
$|f(Q)|le (kK'K+1)^{k-1}$. We have $|D^k|>|f(Q)|$, because $(1+(kK)^{k-1})^{frac 1{k-1}}>(1+(kK)^k)^{frac 1{k}}$,
because when $a>1$ is a constant and $x>0$ a function $(1+a^x)^{frac 1x}$ decreases.
Therefore the function $f$ is not injective. So there exist distinct elements $d=(d_1,dots,d_k)$
and $d'=(d'_1,dots,d'_k)$ of $D^k$ such that $p_C(dv)=f(d)=f(d')=p_C(d'v)$. Since $dv$ and $dv'$
belong to $langle Wrangle$ and the restriction $p_{C}|langle Wrangle$ is injective,
$dv=d'v$. It remains to put $f_i=d_i-d'_i$ for each $iin [k]$.$square$
Remark that for each $B_n$, the sequence $(lambda_1,dots,lambda_{n+1})$ is determined up to a multiplication by $(-1)$. The lemma implies that $|lambda_i|le (n+1)^n$ for each $i$, so $P_nle (n+1)^{n(n+1)}$.
References
[L] Serge Lange, Algebra, Addison-Wesley, 1965 (Russian translation, Moskow, Mir, 1968).
answered Jan 24 at 3:59
Alex RavskyAlex Ravsky
41.1k32282
$endgroup$
We can show that $P_nle (n+1)^{n(n+1)}$ for each $nge 2$, becuase a few months ago I proved a following lemma.
For a natural number $n$ let $[n]$ denotes a set ${1,dots, n}$. Given a subset $Y$ of a vector space $X$ over $mathbb R$ by $langle Yrangle$ we denote the linear hull of $Y$ in $X$, that is a set of all finite sums $f_1y_1+dots+f_ky_k$, where $f_iinmathbb R$ and $y_iin Y$ for each $i$.
Lemma. Let $K$ and $N$ be positive integers,
$V={v_1,dots, v_k}subset [0,K]^N$ be a linearly dependent over $mathbb R$ system of vectors with
integer entries. There exist integers
$f_1,dots, f_k$ which are not all zeroes such that $|f_i|le (kK)^{k-1}$ for each $i$ and
$f_1v_1+dots+f_kv_k=0$.
Proof. Let $W$ be a maximal linearly independent subset of a set $V$. Since the set $V$ is linearly
dependent, $|W|le k-1$. For each $iin [N]$ let
$e^i=(e^i_1,dots,e^i_N)inmathbb R^N$ be $i$-th standard orth, that is $e^i_i=1$ and $e^i_j=0$ for each $jne i$. Let
$B_0={e^1,dots,e^n}$ be the standard basis of the linear space $mathbb R^N$.
By [Lan, Ch. III, Theorem 2], there exists a basis $B$ of the space $mathbb R^N$
such that $Wsubset Bsubset Wcup B_0$. Let $C=B_0setminus (Bsetminus W)$ and
$p_{C}:mathbb R^Nto langle Crangle$ be the orthogonal
projection, that is $p_{C}(x)=sum{x_ie^i:x_iinmathbb R$, $e^iin C}$
for each vector $x=(x_1,dots,x_N)in mathbb R^N$. Thus
$ker p_{C}={xin mathbb R^N:p_{C}(x)=0}=langle B_0setminus Crangle=
langle Bsetminus Wrangle$. We have $ker p_{C}cap langle Wrangle=langle Bsetminus
Wranglecaplangle Wrangle=0$, because otherwise the set $B$ is linearly dependent.
Thus the restriction $p_{C}|langle Wrangle$ of the map $p_{C}$ on the set $langle Wrangle$
is injective.
Put $K'=(kK)^{k-1}$. Define a map $f$ from the subset $D^k$ of points of the set $[0, K']^k$ with all
integer coordinates to $langle Wranglecap mathbb Z^Nsubset mathbb R^N$ as follows.
Let $d=(d_1,dots,d_k)in D^k$. Put $f(d)=p_C(dv)$, where $dv=d_1v_1+dots d_kv_k$.
Since $d_iin [0, K']$ and $v_iin [0,K]^N$ for each $iin [k]$, each coordinate of a vector
$dv$ (and, hence, of the vector $f(d)=p_C(dv)$ too) is at most $kK'K$. Since
$$|C|=|B_0setminus (Bsetminus W)|=|B_0|-|Bsetminus W|=|B_0|-(|B|-|W|)=
N-(N-|W|)=|W|le k-1,$$
$|f(Q)|le (kK'K+1)^{k-1}$. We have $|D^k|>|f(Q)|$, because $(1+(kK)^{k-1})^{frac 1{k-1}}>(1+(kK)^k)^{frac 1{k}}$,
because when $a>1$ is a constant and $x>0$ a function $(1+a^x)^{frac 1x}$ decreases.
Therefore the function $f$ is not injective. So there exist distinct elements $d=(d_1,dots,d_k)$
and $d'=(d'_1,dots,d'_k)$ of $D^k$ such that $p_C(dv)=f(d)=f(d')=p_C(d'v)$. Since $dv$ and $dv'$
belong to $langle Wrangle$ and the restriction $p_{C}|langle Wrangle$ is injective,
$dv=d'v$. It remains to put $f_i=d_i-d'_i$ for each $iin [k]$.$square$
Remark that for each $B_n$, the sequence $(lambda_1,dots,lambda_{n+1})$ is determined up to a multiplication by $(-1)$. The lemma implies that $|lambda_i|le (n+1)^n$ for each $i$, so $P_nle (n+1)^{n(n+1)}$.
References
[L] Serge Lange, Algebra, Addison-Wesley, 1965 (Russian translation, Moskow, Mir, 1968).
answered Jan 24 at 3:59
Alex RavskyAlex Ravsky
41.1k32282
edited Jan 24 at 13:05
answered Jan 24 at 3:59
Alex RavskyAlex Ravsky
41.1k32282
answered Jan 24 at 3:59
Alex RavskyAlex Ravsky
41.1k32282
answered Jan 24 at 3:59
Alex RavskyAlex Ravsky
41.1k32282
41.1k32282
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