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Question: Let $n$ be a positive integer and $ H_n=\{A=(a_{ij})_{n×n}\in M_n(K) : a_{ij}=a_{rs} \text{ whenever } i +j=r+s\}$. Then what is $\dim H_n$?


Proof:

For $n=2$

$H_2=\begin{pmatrix} a_{11}& a_{12}\\a_{12}&a_{22}\end{pmatrix}$

Matrix of sum of indices $J_n$:

$J_2=\begin{pmatrix}\color{red}{2} & \color{red}{3} \\{3}&\color{red}{4}\end{pmatrix}$

$\dim(H_2) =3=2\cdot 2-1$


For $n=3$

$H_3=\left(\begin{array}{cc|c} a_{11}& a_{12}&a_{13}\\ \hline a_{12}&a_{13}&a_{23}\\a_{13}&a_{23}&a_{33}\end{array}\right)$

$J_3=\begin{pmatrix}\color{red}{2} & \color{red}{3} &\color{red}{4}\\ 3 &4&\color{red}{5}\\4&{5}&\color{red}{6}\end{pmatrix}$

$\dim(H_3) =5=2\cdot 3-1$


$J_n=\left(\begin{array}{ccc|c} \color{red}{2}& \color{red}{3}&\color{red}{\ldots}&\color{red}{n+1}\\ \hline 3&4&\ldots&\color{red}{n+2}\\4&5&\ldots&\color{red}{n+3}\\\vdots&\vdots&\ldots&\color{red}{\vdots}\\n+1&n+2&\ldots&\color{red}{n+n}\end{array}\right)$

$\dim(H_n) =2n-1$


$\dim(H_n) = |\{2,3,\ldots,2n\}|=2n-1$


How to make this proof rigorous by introducing partition of numbers?

From the work I have done so far, it is clear that I can find the basis and dimension. But I think the matrix of indices can simplify the problem easily. I need some nice tricks involving elementary number theory specially integer partitions on the matrix of indices.

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  • $\begingroup$ Taking $n=3$ as an example, I would try to prove that $$\begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1\end{pmatrix}$$ forms a basis for $H_3$. $\endgroup$
    – Feng
    Jul 20, 2022 at 12:10
  • $\begingroup$ Well. This is clear from $J_3$ .No new information. $\endgroup$ Jul 20, 2022 at 12:13
  • $\begingroup$ For general $n$, you can also use the same idea to find a basis for $H_n$, so you can prove rigorously $\mathrm{dim} H_n=2n-1$. Although I don't know if this idea fits your purpose: "introduccing a partition of numbers". Just a thought. $\endgroup$
    – Feng
    Jul 20, 2022 at 12:19
  • $\begingroup$ The only integer partitions involved here have two summands, and it's more properly 2 part compositions since, e.g., $a_{12}$ and $a_{21}$ are distinct. It seems unnecessary to invoke the theory of partitions. $\endgroup$ Aug 2, 2022 at 17:14

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Sums of indices can go from $2$ to $2n$. Define, for $2\le k\le n$ the matrix $M_k$ that has $1$ where $i+j=k$ and $0$ elsewhere.

It should be easy to prove that $\{M_k:2\le k\le 2n\}$ is a basis of your subspace. That's all, you don't need to count the number of nonzero entries in $M_k$.

Let's look at a typical matrix in the set when $n=3$. The possible sums are $2,3,4,5,6$ and we get $$ \begin{bmatrix} a & b & c \\ b & c & d \\ c & d & e \end{bmatrix}= a\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}+ b\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}+ c\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}+ d\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}+ e\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ This should give the idea of how to find the basis in the general case.

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    $\begingroup$ +1. I'm thinking the same way as yours. Maybe I'm not good at expressing my mind. $\endgroup$
    – Feng
    Jul 20, 2022 at 12:30
  • $\begingroup$ @LostinSpace I'd not say it's difficult in this case. First look at what happens for $3\times3$ matrices and make a conjecture. $\endgroup$
    – egreg
    Jul 28, 2022 at 9:15

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