Why is it easy to calculate $\operatorname{rank}(A)=n$? when I read a paper with matrices methods, and found a difficult problem.

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

The author say it is easy to calculate that $\operatorname{rank}(A)=n$. I have found that for $n\le 5$ it is true, but for general $n$, I can't prove it.
$$A=P+iQ$$
$$P=\begin{bmatrix}
2&1&1&\cdots&1\\
1&4&2&\cdots& 2\\
1&2&6&\cdots& 3\\
\cdots&\cdots&\cdots&\cdots&\cdots\\
1&2&3&\cdots& 2n
\end{bmatrix},Q=\begin{bmatrix} 
2&2&3&\cdots& n\\
2&4&3&\cdots &n\\
3&3&6&\cdots& n\\
\cdots&\cdots&\cdots&\cdots&\cdots\\
n&n&n&\cdots& 2n\end{bmatrix}$$
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 $$\begin{align*}A^HA&=(P-iQ)(P+iQ)=P^2+Q^2+i(PQ-QP)=\\&=\binom{p}{Q}^T\cdot\begin{bmatrix}
I& iI\\
-iI & I
 \end{bmatrix} \binom{P}{Q}\end{align*}$$
 A: Notice that $A=S^1+iS^2$ and $S^1,S^2$ are real symmetric matrices or order $n$. Now define $u_{i}^t=(0\ldots,0,1,\ldots,1)$, where the first $i-1$ entries are $0$ and the last $n-i+1$ entries are $1$. Notice that $u_i$ is a column vector in $\mathbb{C}^n$ and $u_i^t$ is a row vector. Thus, $u_iu_i^t$ is a matrix of order $n$.
Let $\{e_1,\ldots,e_n\}$ be the canonical basis of $\mathbb{C}^n$. Notice that $e_i$ is a column vector in $\mathbb{C}^n$.
Let $B=\sum_{i=1}^nu_iu_i^t$. Notice that $B$ is a positive semidefinite real symmetric matrix of order $n$. 
Let us prove that the entries of $B$ are $B_{jk}=\min\{j,k\}$. 
Next, $B_{jk}=e_j^tBe_k=\sum_{i=1}^ne_j^tu_iu_i^te_k$.
Now, $e_j^tu_iu_i^te_k=1$, if $i\leq j$ and $i\leq k$ and $0$ otherwise. Thus, $e_j^tu_iu_i^te_k=1$ if $i\leq\min\{j,k\}$ and $0$ otherwise.
Then  $B_{jk}=\sum_{i=1}^ne_j^tu_iu_i^te_k=\sum_{i=1}^{\min\{j,k\}}1=\min\{j,k\}$.
Notice that $S^1_{ij}=\min\{j,k\}$ if $i\neq j$ and $S^1_{jj}=2j$.
Thus, $S^1-B$ is a diagonal matrix with diagonal $(1,2,\ldots,n)$. 
Since $S^1-B$ is a positive definite real symmetric matrix and $B$ is positive semidefinite then $S^1=(S^1-B)+B$ is positive definite. 
If the column vector $v\in\ker(A)$ then $Av=0$ then $Av\overline{v}^t=0_{n\times n}$ and  $tr(Av\overline{v}^t)=tr(S^1v\overline{v}^t+iS^2v\overline{v}^t)=0$. 
Since $S^1$ and $S^2$ are real symmetric matrices then $tr(S^iv\overline{v}^t)\in\mathbb{R}$, $i=1,2$. Thus, $tr(S^iv\overline{v}^t)=0$, $i=1,2$. But $S^1$ is positive definite. Thus, $v=0$ and $A$ has null kernel. Thus $A$ has rank $n$.
A: I am not adding anything new here. Just merging your work and Daniel's answer into a somewhat more concise form.
Write $A=P+iQ$, where $P,Q$ are real symmetric matrices. Now consider the matrix $J$ in your question. Actually you have calculated $JPJ^T$ incorrectly. The matrix product should be a real symmetric tridiagonal matrix whose main diagonal is $(2,4,6,\ldots,2n)$ and whose super- and sub- diagonals are $(-1,-2,\ldots,-(n-1))$. So, by Gershgorin disc theorem and the real symmetry of $JPJ^T$, this matrix product (and hence $P$, by matrix congruence) is positive definite.
Now comes Daniel's argument. Since $P,Q$ are real symmetric, both $x^\ast Px$ and $x^\ast Qx$ are real numbers for any complex vector $x$. It follows that if $Ax=0$, we must have $x^\ast Px=x^\ast Qx=0$. Yet, as $P$ is positive definite, $x$ must be zero. Hence $A$ is nonsingular.
