How do we show this matrix has full rank? 
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 $$A^HA=(P-iQ)(P+iQ)=P^2+Q^2+i(PQ-QP)=\binom{P}{Q}^T\cdot\begin{bmatrix}
I& iI\\
-iI & I
 \end{bmatrix} \binom{P}{Q}$$

OK, let me try again, maybe I'll get it right this time. I'll show that $P$ is positive definite. This will imply the claim because if $(P+iQ)(x+iy)=0$ with $x,y\in\mathbb R^n$, then $Px=Qy$, $Py=-Qx$, and by taking scalar products with $x$ and $y$, respectively, we see that $\langle x, Px \rangle = -\langle y, Py\rangle$, which implies that $x=y=0$. Here I use that $Q$ is symmetric.
Let me now show that $P>0$. Following math110's suggestion, we can simplify my original calculation as follows: Let
$
B=B_n = P -\textrm{diag}(1,2,\ldots , n)$. For example, for $n=5$, this is the matrix
$$ B_ 5= \begin{pmatrix} 1 & 1 & 1  & 1 & 1\\
1 & 2 & 2  & 2 & 2\\
1 & 2 & 3 & 3 & 3\\
1 & 2 & 3 & 4 & 4\\
1 & 2 & 3 & 4 & 5
\end{pmatrix} .
$$
I can now (in general) subtract the $(n-1)$st row from the last row, then the $(n-2)$nd row from the $(n-1)$st row etc. This confirms that $\det B_n=1$. Moreover, the upper left $k\times k$ submatrices of $B_n$ are of the same type; they equal $B_k$.
This shows that $B>0$, by Sylvester's criterion, and thus $P>0$ as well.
