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

  • 1
    $\begingroup$ 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

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.

| cite | improve this answer | |
  • 3
    $\begingroup$ 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

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 $$\lambda=2-2\cos{x}=4\sin^2{\dfrac{x}{2}}=4\sin^2{\dfrac{j\pi}{2(n+1)}}$$

| cite | improve this answer | |

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.