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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 $$A^HA=(P-iQ)(P+iQ)=P^2+Q^2+i(PQ-QP)=\binom{P}{Q}^T\cdot\begin{bmatrix} I& iI\\ -iI & I \end{bmatrix} \binom{P}{Q}$$

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    $\begingroup$ You might want to write the tridiagonal matrix more suggestively using \ddots: $$JPJ^T=J^TQJ=\begin{bmatrix} 2&-2&0&\cdots&0\\ -2&4&-3&\ddots\qquad&0\\ 0&-3 &6&\ddots\qquad&0\\ \vdots&&\ddots\qquad&\ddots\ \qquad&\vdots\\ 0&\cdots&-(n-2)&2(n-1)&-(n-1)\\ 0&\cdots&0&-(n-1)&2n \end{bmatrix}$$ $\endgroup$ – Wolfgang Jan 1 '15 at 21:20
  • $\begingroup$ I believe the calculation for $JPJ^t$ (and thus the whole identity) is not correct. $\endgroup$ – Christian Remling Jan 3 '15 at 5:22
  • $\begingroup$ Sorry for the meaningless edits; I misclicked. $\endgroup$ – Christian Remling Jan 3 '15 at 22:33
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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.

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    $\begingroup$ Oh,It's Nice! Thank you,+1,Use your idea,I found $P=B+\text{diag}{(1,2,3,\cdots,n)}$,where $B=(b_{ij}),b_{i,j}=\min{(i,j)}$,if we prove $B$ is postive definite,also solve this problem too $\endgroup$ – math110 Jan 3 '15 at 6:33
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I use Christian Remling idea,In fact,I can find the matrix $$B_{ij}=\min{\{i,j\}}$$eigenvalue is $$\dfrac{1}{4\sin^2{\dfrac{j\pi}{2(n+1)}}},j=1,2,\cdots,n$$ proof: then we have $$B=\begin{bmatrix} 1&1&1&\ddots&1&1\\ 1&2&2&\ddots&\ddots&2\\ 1&2&3&3&\ddots&3\\ \vdots&\ddots&\ddots&\ddots&\ddots&\cdots\\ 1&\vdots&\ddots&\ddots&n-1&n-1\\ 1&2&\cdots&\cdots&n-1&n \end{bmatrix} $$ It is easy have $$C=B^{-1}=\begin{bmatrix} 2&-1\\ -1&2&-1\\ 0&\ddots&\ddots&\ddots\\ \vdots&\cdots&-1&2&-1\\ 0&\cdots&\cdots&-1&1 \end{bmatrix}$$ and consider $$b_{n}=|\lambda C-I|=\begin{vmatrix} \lambda-2&1&\cdots&\cdots&0\\ 1&\lambda-2&1&\cdots&0\\ \vdots&\ddots&\ddots&\ddots&\vdots\\ \cdots&\cdots&1&\lambda-2&1\\ 0&\cdots&\cdots&1&\lambda-2 \end{vmatrix} $$ so $$b_{n+1}=(\lambda-2)b_{n}-b_{n-1},b_{1}=\lambda-2,b_{2}=(\lambda-2)^2-1$$ let $\lambda-2=-2\cos{x}$, then $$b_{n+1}=-2\cos{x}\cdot b_{n}-b_{n-1},b_{1}=-2\cos{x},b_{2}=4\cos^2{x}-1$$ and induction have $$b_{n}=(-1)^n\cdot\dfrac{\sin{(n+1)x}}{\sin{x}}=0\Longrightarrow x=\dfrac{j\pi}{n+1},j=1,2,\cdots,n$$ so we $B^{-1}$ with eigenvalue is $$\lambda=2-2\cos{x}=4\sin^2{\dfrac{x}{2}}=4\sin^2{\dfrac{j\pi}{2(n+1)}}$$

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A modest introductory step only. The following partial algebraization might be useful: the present matrix is given by:

  • $\quad a_{kk}\ :=\ 2\cdot(k\ +\ i\cdot k)$
  • $\quad a_{km}\ :=\ \min(k\ m)\ +\ \imath\cdot\max(k\ m)$

for $\,\ k\,\ m=1\ldots n\,\ $ and $\,\ k\ne m.\ $ However, we may equivalently consider a matrix obtained from the given one by multiplying all entries by $\ 1-i.\ $ We obtain a matrix $\ (b_{mk})\ $ as follows:

  • $\quad b_{kk}\,\ :=\,\ 4\cdot k$
  • $\quad b_{km}\,\ :=\,\ (k+m)\ +\ \imath\cdot|k-m|$

for $\,\ k\,\ m=1\ldots n\,\ $ and $\,\ k\ne m$.

Good luck, and I will try to continue too.

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