when I see some maxtrix paper,and Met with difficulties problem to me.

define matriex $A=(a_{jk})_{n\times n}$,where $$a_{jk}=\begin{cases} j+k\cdot i&j<k\\ k+j\cdot i&j>k\\ 2(j+k\cdot i)& j=k \end{cases}$$ where $i^2=-1$.

the author say it is easy calculation we have $Rank(A)=n$.But I calculation sometimes,and can't prove it.

I met with the following difficulty reading the paper Li, Rong Xiu "The properties of a matrix order column" (1988):

Define the matrix $A=(a_{jk})_{n\times n}$, where $$a_{jk}=\begin{cases} j+k\cdot i&j<k\\ k+j\cdot i&j>k\\ 2(j+k\cdot i)& j=k \end{cases}$$ and $i^2=-1$.

The author says it is easy to show that $rank(A)=n$. I have proved for $n\le 5$, but I couldn't prove for general $n$.

Following is an attempt to solve this problem: let $$A=P+iQ$$ where $$P=\begin{bmatrix} 2&1&1&\cdots&1\\ 1&4&2&\cdots& 2\\ 1&2&6&\cdots& 3\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ 1&2&3&\cdots& 2n \end{bmatrix},Q=\begin{bmatrix} 2&2&3&\cdots& n\\ 2&4&3&\cdots &n\\ 3&3&6&\cdots& n\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ n&n&n&\cdots& 2n\end{bmatrix}$$

and define $$J=\begin{bmatrix} 1&0&\cdots &0\\ -1&1&\cdots& 0\\ \cdots&\cdots&\cdots&\cdots\\ 0&\cdots&-1&1 \end{bmatrix}$$ then we have $$JPJ^T=J^TQJ=\begin{bmatrix} 2&-2&0&0&\cdots&0\\ -2&4&-3&\ddots&0&0\\ 0&-3&6&-4\ddots&0\\ \cdots&\ddots&\ddots&\ddots&\ddots&\cdots\\ 0&0&\cdots&-(n-2)&2(n-1)&-(n-1)\\ 0&0&0&\cdots&-(n-1)&2n \end{bmatrix}$$ and $$A^HA=(P-iQ)(P+iQ)=P^2+Q^2+i(PQ-QP)=\binom{P}{Q}^T\cdot\begin{bmatrix} I& iI\\ -iI & I \end{bmatrix} \binom{P}{Q}$$

I met with the following difficulty reading the paper Li, Rong Xiu "The properties of a matrix order column" (1988):

Define the matrix $A=(a_{jk})_{n\times n}$, where $$a_{jk}=\begin{cases} j+k\cdot i&j<k\\ k+j\cdot i&j>k\\ 2(j+k\cdot i)& j=k \end{cases}$$ and $i^2=-1$.

The author says it is easy to show that $rank(A)=n$. I have proved for $n\le 5$, but I couldn't prove for general $n$.

Following is an attempt to solve this problem: let $$A=P+iQ$$ where $$P=\begin{bmatrix} 2&1&1&\cdots&1\\ 1&4&2&\cdots& 2\\ 1&2&6&\cdots& 3\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ 1&2&3&\cdots& 2n \end{bmatrix},Q=\begin{bmatrix} 2&2&3&\cdots& n\\ 2&4&3&\cdots &n\\ 3&3&6&\cdots& n\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ n&n&n&\cdots& 2n\end{bmatrix}$$

and define $$J=\begin{bmatrix} 1&0&\cdots &0\\ -1&1&\cdots& 0\\ \cdots&\cdots&\cdots&\cdots\\ 0&\cdots&-1&1 \end{bmatrix}$$ then we have $$JPJ^T=J^TQJ=\begin{bmatrix} 2&-2&0&0&\cdots&0\\ -2&4&-3&\ddots&0&0\\ 0&-3&6&-4\ddots&0\\ \cdots&\ddots&\ddots&\ddots&\ddots&\cdots\\ 0&0&\cdots&-(n-2)&2(n-1)&-(n-1)\\ 0&0&0&\cdots&-(n-1)&2n \end{bmatrix}$$ and $$A^HA=(P-iQ)(P+iQ)=P^2+Q^2+i(PQ-QP)=\binom{P}{Q}^T\cdot\begin{bmatrix} I& iI\\ -iI & I \end{bmatrix} \binom{P}{Q}$$

when I see some maxtrix paper,and Met with difficulties problem to me.

define matriex $A=(a_{jk})_{n\times n}$,where $$a_{jk}=\begin{cases} j+k\cdot i&j<k\\ k+j\cdot i&j>k\\ 2(j+k\cdot i)& j=k \end{cases}$$ where $i^2=-1$.

the author say it is easy calculation we have $Rank(A)=n$.But I calculation sometimes,and can't prove it.

I met with the following difficulty reading the paper Li, Rong Xiu "The properties of a matrix order column" (1988):

Define the matrix $A=(a_{jk})_{n\times n}$, where $$a_{jk}=\begin{cases} j+k\cdot i&j<k\\ k+j\cdot i&j>k\\ 2(j+k\cdot i)& j=k \end{cases}$$ and $i^2=-1$.

The author says it is easy to show that $rank(A)=n$. I have proved for $n\le 5$, but I couldn't prove for general $n$.

Following is an attempt to solve this problem: let $$A=P+iQ$$ where $$P=\begin{bmatrix} 2&1&1&\cdots&1\\ 1&4&2&\cdots& 2\\ 1&2&6&\cdots& 3\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ 1&2&3&\cdots& 2n \end{bmatrix},Q=\begin{bmatrix} 2&2&3&\cdots& n\\ 2&4&3&\cdots &n\\ 3&3&6&\cdots& n\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ n&n&n&\cdots& 2n\end{bmatrix}$$

and define $$J=\begin{bmatrix} 1&0&\cdots &0\\ -1&1&\cdots& 0\\ \cdots&\cdots&\cdots&\cdots\\ 0&\cdots&-1&1 \end{bmatrix}$$ then we have $$JPJ^T=J^TQJ=\begin{bmatrix} 2&-2&\cdots&0\\ -2&4&-3&\cdots\\ \cdots&\cdots&\cdots&\cdots\\ 0&\cdots&-(n-1)&2n \end{bmatrix}$$ and $$A^HA=(P-iQ)(P+iQ)=P^2+Q^2+i(PQ-QP)=\binom{P}{Q}^T\cdot\begin{bmatrix} I& iI\\ -iI & I \end{bmatrix} \binom{P}{Q}$$

I met with the following difficulty reading the paper Li, Rong Xiu "The properties of a matrix order column" (1988):

Define the matrix $A=(a_{jk})_{n\times n}$, where $$a_{jk}=\begin{cases} j+k\cdot i&j<k\\ k+j\cdot i&j>k\\ 2(j+k\cdot i)& j=k \end{cases}$$ and $i^2=-1$.

The author says it is easy to show that $rank(A)=n$. I have proved for $n\le 5$, but I couldn't prove for general $n$.

Following is an attempt to solve this problem: let $$A=P+iQ$$ where $$P=\begin{bmatrix} 2&1&1&\cdots&1\\ 1&4&2&\cdots& 2\\ 1&2&6&\cdots& 3\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ 1&2&3&\cdots& 2n \end{bmatrix},Q=\begin{bmatrix} 2&2&3&\cdots& n\\ 2&4&3&\cdots &n\\ 3&3&6&\cdots& n\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ n&n&n&\cdots& 2n\end{bmatrix}$$

and define $$J=\begin{bmatrix} 1&0&\cdots &0\\ -1&1&\cdots& 0\\ \cdots&\cdots&\cdots&\cdots\\ 0&\cdots&-1&1 \end{bmatrix}$$ then we have $$JPJ^T=J^TQJ=\begin{bmatrix} 2&-2&0&0&\cdots&0\\ -2&4&-3&\ddots&0&0\\ 0&-3&6&-4\ddots&0\\ \cdots&\ddots&\ddots&\ddots&\ddots&\cdots\\ 0&0&\cdots&-(n-2)&2(n-1)&-(n-1)\\ 0&0&0&\cdots&-(n-1)&2n \end{bmatrix}$$ and $$A^HA=(P-iQ)(P+iQ)=P^2+Q^2+i(PQ-QP)=\binom{P}{Q}^T\cdot\begin{bmatrix} I& iI\\ -iI & I \end{bmatrix} \binom{P}{Q}$$