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Clément de Seguins Pazzis
Clément de Seguins Pazzis
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The result is true indeed. Here is a rather technical solution. I work by induction over $k$, thus denoting by $S_k$ the above $(2k)\times(2k)$ matrix. Let us transform the matrix through the following series of row and column operations: $C_{2k} \leftarrow C_{2k}+s_1 C_{2k-1}$ and then, for odd $j$ from $3$ to $2k-3$,
$L_k \leftarrow L_k+s_{k-j}L_j$ for $k$ from $j+1$ to $2k$.

After those operations, the matrix obtained from $S_k$ has the following form: $$S'_k=\begin{bmatrix} 0 & D \\ T & ? \end{bmatrix} $$ where $D=\begin{bmatrix} 1 & 0 \\ s_1 & 0 \end{bmatrix}$, the odd-labelled columns of $T$ are $\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$, ..., $\begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$, and $T$ is equivalent to $S_{k-1}$.

By the induction hypothesis $T$ has rank $k-1$ and hence its column space is spanned by the above vectors. To conclude that $S$$S_k$ has rank $k$ we then need to prove that the even-labelled entries in the last column of $S'_k$ are all zero.

Let $p \in \{2,...,k\}$. By careful examination, one finds that the $(2p,2k)$-entry of $S'_k$ equals $$s_{2p}+\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}$$ where $A_p$ is the set of all lists $(i_1,\dots,i_\ell)$ of positive integers whose sum is $2p$ and in which $i_1$ and $i_\ell$ are odd and the other entries are even. Now, we show by induction that this sum equals $s_{2p}$. First of all, $\theta : (i_1,\dots,i_\ell) \mapsto (i_\ell,\dots,i_1)$ is an involution on $A_p$, and as $\mathbb{F}$ has characteristic $2$ we deduce that $$\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}= \sum_{(i_1,\dots,i_\ell) \in A_p,(i_1,\dots,i_\ell)=(i_\ell,\dots,i_1)} s_{i_1} s_{i_2}\cdots s_{i_l}.$$

Next, if we have a symmetric list $(i_1,\dots,i_{2j})=(i_{2j},\dots,i_1)$ in $A_p$ with an even number of entries and $j \geq 2$, we use equality $s_{i_j}s_{i_{j+1}}=s_{i_j}^2=s_{2i_j}$ to see that it defines the same product has the symmetric list $(i_1,\dots,i_{j-1},2i_j,i_{j+2},\dots,i_{2j})$ with $2j-1$ entries. Pairing lists in this manner and - if $p$ is odd - noting that $s_p^2=s_{2p}$, we find that $$\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}= \begin{cases} s_{2p}+\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} & \text{if $p$ is odd} \\ \sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} & \text{if $p$ is even.} \end{cases} $$$$\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}= \begin{cases} s_{2p}+\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} & \text{if $p$ is odd} \\ \sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} & \text{if $p$ is even,} \end{cases} $$ where $B_p$ isdenotes the subset of $A_p$ consisting of the symmetric lists $(i_1,\dots,i_{2j+1})$ in which $i_{j+1}=2t$ for some odd integer $t$. For such a list, we see that the product of the $s_{i_t}$'s is the same as the one for the list $(t,i_1,\dots,i_{j-1},2i_1,i_{j+1},\dots,i_{2p+1},t)$, whence we obtain that $$\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l}= \sum_{(i_1,\dots,i_\ell) \in C_p} s_{i_1} s_{i_2}\cdots s_{i_l},$$ where $C_p$ denotes the set of all symmetric lists $(i_1,\dots,i_{2j+1})$ in which $i_{j+1}=2i_1$.

IfIf $p$ is odd then $C_p$$B_p$ is empty and we obtain the claimed resultare done. Assume

Assume now that $p=2q$ for some integer $q$. Then, $(i_1,...,i_{2j+1}) \mapsto (i_1,...,i_{j-1},i_j,i_1)$$(i_1,...,i_{2j+1}) \mapsto (i_1,...,i_{j-1},i_j,(i_{j+1})/2)$ maps $C_p$$B_p$ bijectively onto $A_q$, and and using $s_{2i_1}=s_{i_1}^2$$s_{i_{j+1}}=s_{(i_{j+1}/2)}^2$ we deduce that $$\sum_{(i_1,\dots,i_\ell) \in C_p} s_{i_1} s_{i_2}\cdots s_{i_l} =\Bigl(\sum_{(i_1,\dots,i_\ell) \in A_q} s_{i_1} s_{i_2}\cdots s_{i_l}\Bigr)^2.$$$$\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} =\Bigl(\sum_{(i_1,\dots,i_\ell) \in A_q} s_{i_1} s_{i_2}\cdots s_{i_l}\Bigr)^2.$$ By induction, we deduce that $$\sum_{(i_1,\dots,i_\ell) \in C_p} s_{i_1} s_{i_2}\cdots s_{i_l} =(s_{2q})^2=s_{2p},$$$$\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} =(s_{2q})^2=s_{2p},$$ QED.

The result is true indeed. Here is a rather technical solution. I work by induction over $k$, thus denoting by $S_k$ the above $(2k)\times(2k)$ matrix. Let us transform the matrix through the following series of row and column operations: $C_{2k} \leftarrow C_{2k}+s_1 C_{2k-1}$ and then, for odd $j$ from $3$ to $2k-3$,
$L_k \leftarrow L_k+s_{k-j}L_j$ for $k$ from $j+1$ to $2k$.

After those operations, the matrix obtained from $S_k$ has the following form: $$S'_k=\begin{bmatrix} 0 & D \\ T & ? \end{bmatrix} $$ where $D=\begin{bmatrix} 1 & 0 \\ s_1 & 0 \end{bmatrix}$, the odd-labelled columns of $T$ are $\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$, ..., $\begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$, and $T$ is equivalent to $S_{k-1}$.

By the induction hypothesis $T$ has rank $k-1$ and hence its column space is spanned by the above vectors. To conclude that $S$ has rank $k$ we then need to prove that the even-labelled entries in the last column of $S'_k$ are all zero.

Let $p \in \{2,...,k\}$. By careful examination, one finds that the $(2p,2k)$-entry of $S'_k$ equals $$s_{2p}+\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}$$ where $A_p$ is the set of all lists $(i_1,\dots,i_\ell)$ of positive integers whose sum is $2p$ and in which $i_1$ and $i_\ell$ are odd and the other entries are even. Now, we show by induction that this sum equals $s_{2p}$. First of all, $\theta : (i_1,\dots,i_\ell) \mapsto (i_\ell,\dots,i_1)$ is an involution on $A_p$, and as $\mathbb{F}$ has characteristic $2$ we deduce that $$\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}= \sum_{(i_1,\dots,i_\ell) \in A_p,(i_1,\dots,i_\ell)=(i_\ell,\dots,i_1)} s_{i_1} s_{i_2}\cdots s_{i_l}.$$

Next, if we have a symmetric list $(i_1,\dots,i_{2j})=(i_{2j},\dots,i_1)$ in $A_p$ with an even number of entries and $j \geq 2$, we use equality $s_{i_j}s_{i_{j+1}}=s_{i_j}^2=s_{2i_j}$ to see that it defines the same product has the symmetric list $(i_1,\dots,i_{j-1},2i_j,i_{j+2},\dots,i_{2j})$ with $2j-1$ entries. Pairing lists in this manner and - if $p$ is odd - noting that $s_p^2=s_{2p}$, we find that $$\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}= \begin{cases} s_{2p}+\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} & \text{if $p$ is odd} \\ \sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} & \text{if $p$ is even.} \end{cases} $$ where $B_p$ is the subset of $A_p$ consisting of the symmetric lists $(i_1,\dots,i_{2j+1})$ in which $i_{j+1}=2t$ for some odd integer $t$. For such a list, we see that the product of the $s_{i_t}$'s is the same as the one for the list $(t,i_1,\dots,i_{j-1},2i_1,i_{j+1},\dots,i_{2p+1},t)$, whence we obtain that $$\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l}= \sum_{(i_1,\dots,i_\ell) \in C_p} s_{i_1} s_{i_2}\cdots s_{i_l},$$ where $C_p$ denotes the set of all symmetric lists $(i_1,\dots,i_{2j+1})$ in which $i_{j+1}=2i_1$.

If $p$ is odd then $C_p$ is empty and we obtain the claimed result. Assume now that $p=2q$ for some integer $q$. Then, $(i_1,...,i_{2j+1}) \mapsto (i_1,...,i_{j-1},i_j,i_1)$ maps $C_p$ bijectively onto $A_q$, and using $s_{2i_1}=s_{i_1}^2$ we deduce that $$\sum_{(i_1,\dots,i_\ell) \in C_p} s_{i_1} s_{i_2}\cdots s_{i_l} =\Bigl(\sum_{(i_1,\dots,i_\ell) \in A_q} s_{i_1} s_{i_2}\cdots s_{i_l}\Bigr)^2.$$ By induction, we deduce that $$\sum_{(i_1,\dots,i_\ell) \in C_p} s_{i_1} s_{i_2}\cdots s_{i_l} =(s_{2q})^2=s_{2p},$$ QED.

The result is true indeed. Here is a rather technical solution. I work by induction over $k$, thus denoting by $S_k$ the above $(2k)\times(2k)$ matrix. Let us transform the matrix through the following series of row and column operations: $C_{2k} \leftarrow C_{2k}+s_1 C_{2k-1}$ and then, for odd $j$ from $3$ to $2k-3$,
$L_k \leftarrow L_k+s_{k-j}L_j$ for $k$ from $j+1$ to $2k$.

After those operations, the matrix obtained from $S_k$ has the following form: $$S'_k=\begin{bmatrix} 0 & D \\ T & ? \end{bmatrix} $$ where $D=\begin{bmatrix} 1 & 0 \\ s_1 & 0 \end{bmatrix}$, the odd-labelled columns of $T$ are $\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$, ..., $\begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$, and $T$ is equivalent to $S_{k-1}$.

By the induction hypothesis $T$ has rank $k-1$ and hence its column space is spanned by the above vectors. To conclude that $S_k$ has rank $k$ we then need to prove that the even-labelled entries in the last column of $S'_k$ are all zero.

Let $p \in \{2,...,k\}$. By careful examination, one finds that the $(2p,2k)$-entry of $S'_k$ equals $$s_{2p}+\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}$$ where $A_p$ is the set of all lists $(i_1,\dots,i_\ell)$ of positive integers whose sum is $2p$ and in which $i_1$ and $i_\ell$ are odd and the other entries are even. Now, we show by induction that this sum equals $s_{2p}$. First of all, $\theta : (i_1,\dots,i_\ell) \mapsto (i_\ell,\dots,i_1)$ is an involution on $A_p$, and as $\mathbb{F}$ has characteristic $2$ we deduce that $$\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}= \sum_{(i_1,\dots,i_\ell) \in A_p,(i_1,\dots,i_\ell)=(i_\ell,\dots,i_1)} s_{i_1} s_{i_2}\cdots s_{i_l}.$$

Next, if we have a symmetric list $(i_1,\dots,i_{2j})=(i_{2j},\dots,i_1)$ in $A_p$ with an even number of entries and $j \geq 2$, we use equality $s_{i_j}s_{i_{j+1}}=s_{i_j}^2=s_{2i_j}$ to see that it defines the same product has the symmetric list $(i_1,\dots,i_{j-1},2i_j,i_{j+2},\dots,i_{2j})$ with $2j-1$ entries. Pairing lists in this manner and - if $p$ is odd - noting that $s_p^2=s_{2p}$, we find that $$\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}= \begin{cases} s_{2p}+\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} & \text{if $p$ is odd} \\ \sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} & \text{if $p$ is even,} \end{cases} $$ where $B_p$ denotes the subset of $A_p$ consisting of the symmetric lists $(i_1,\dots,i_{2j+1})$ in which $i_{j+1}=2t$ for some odd integer $t$. If $p$ is odd then $B_p$ is empty and we are done.

Assume now that $p=2q$ for some integer $q$. Then, $(i_1,...,i_{2j+1}) \mapsto (i_1,...,i_{j-1},i_j,(i_{j+1})/2)$ maps $B_p$ bijectively onto $A_q$, and using $s_{i_{j+1}}=s_{(i_{j+1}/2)}^2$ we deduce that $$\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} =\Bigl(\sum_{(i_1,\dots,i_\ell) \in A_q} s_{i_1} s_{i_2}\cdots s_{i_l}\Bigr)^2.$$ By induction, we deduce that $$\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} =(s_{2q})^2=s_{2p},$$ QED.

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Clément de Seguins Pazzis
Clément de Seguins Pazzis
  • 1.5k
  • 12
  • 18

The result is true indeed. Here is a rather technical solution. I work by induction over $k$, thus denoting by $S_k$ the above $(2k)\times(2k)$ matrix. Let us transform the matrix through the following series of row and column operations: $C_{2k} \leftarrow C_{2k}+s_1 C_{2k-1}$ and then, for odd $j$ from $3$ to $2k-3$,
$L_k \leftarrow L_k+s_{k-j}L_j$ for $k$ from $j+1$ to $2k$.

After those operations, the matrix obtained from $S_k$ has the following form: $$S'_k=\begin{bmatrix} 0 & D \\ T & ? \end{bmatrix} $$ where $D=\begin{bmatrix} 1 & 0 \\ s_1 & 0 \end{bmatrix}$, the odd-labelled columns of $T$ are $\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$, ..., $\begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$, and $T$ is equivalent to $S_{k-1}$.

By the induction hypothesis $T$ has rank $k-1$ and hence its column space is spanned by the above vectors. To conclude that $S$ has rank $k$ we then need to prove that the even-labelled entries in the last column of $S'_k$ are all zero.

Let $p \in \{2,...,k\}$. By careful examination, one finds that the $(2p,2k)$-entry of $S'_k$ equals $$s_{2p}+\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}$$ where $A_p$ is the set of all lists $(i_1,\dots,i_\ell)$ of positive integers whose sum is $2p$ and in which $i_1$ and $i_\ell$ are odd and the other entries are even. Now, we show by induction that this sum equals $s_{2p}$. First of all, $\theta : (i_1,\dots,i_\ell) \mapsto (i_\ell,\dots,i_1)$ is an involution on $A_p$, and as $\mathbb{F}$ has characteristic $2$ we deduce that $$\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}= \sum_{(i_1,\dots,i_\ell) \in A_p,(i_1,\dots,i_\ell)=(i_\ell,\dots,i_1)} s_{i_1} s_{i_2}\cdots s_{i_l}.$$

Next, if we have a symmetric list $(i_1,\dots,i_{2j})=(i_{2j},\dots,i_1)$ in $A_p$ with an even number of entries and $j \geq 2$, we use equality $s_{i_j}s_{i_{j+1}}=s_{i_j}^2=s_{2i_j}$ to see that it defines the same product has the symmetric list $(i_1,\dots,i_{j-1},2i_j,i_{j+2},\dots,i_{2j})$ with $2j-1$ entries. Pairing lists in this manner and - if $p$ is odd - noting that $s_p^2=s_{2p}$, we find that $$\sum_{(i_1,\dots,i_\ell) \in A_p} s_{i_1} s_{i_2}\cdots s_{i_l}= \begin{cases} s_{2p}+\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} & \text{if $p$ is odd} \\ \sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l} & \text{if $p$ is even.} \end{cases} $$ where $B_p$ is the subset of $A_p$ consisting of the symmetric lists $(i_1,\dots,i_{2j+1})$ in which $i_{j+1}=2t$ for some odd integer $t$. For such a list, we see that the product of the $s_{i_t}$'s is the same as the one for the list $(t,i_1,\dots,i_{j-1},2i_1,i_{j+1},\dots,i_{2p+1},t)$, whence we obtain that $$\sum_{(i_1,\dots,i_\ell) \in B_p} s_{i_1} s_{i_2}\cdots s_{i_l}= \sum_{(i_1,\dots,i_\ell) \in C_p} s_{i_1} s_{i_2}\cdots s_{i_l},$$ where $C_p$ denotes the set of all symmetric lists $(i_1,\dots,i_{2j+1})$ in which $i_{j+1}=2i_1$.

If $p$ is odd then $C_p$ is empty and we obtain the claimed result. Assume now that $p=2q$ for some integer $q$. Then, $(i_1,...,i_{2j+1}) \mapsto (i_1,...,i_{j-1},i_j,i_1)$ maps $C_p$ bijectively onto $A_q$, and using $s_{2i_1}=s_{i_1}^2$ we deduce that $$\sum_{(i_1,\dots,i_\ell) \in C_p} s_{i_1} s_{i_2}\cdots s_{i_l} =\Bigl(\sum_{(i_1,\dots,i_\ell) \in A_q} s_{i_1} s_{i_2}\cdots s_{i_l}\Bigr)^2.$$ By induction, we deduce that $$\sum_{(i_1,\dots,i_\ell) \in C_p} s_{i_1} s_{i_2}\cdots s_{i_l} =(s_{2q})^2=s_{2p},$$ QED.