Let $A$ be a $n \times n$ matrix whose coeffiecients can be arranged in the following way $$ \begin{pmatrix} d_1 & d_1 & d_1 & d_1 & \cdots & d_1 \\ d_1 & d_2 & d_2 & d_2 &\cdots & d_2 \\ d_1 & d_2 & d_3 & d_3 & \cdots & d_3 \\ d_1 & d_2 & d_3 & d_4 & \cdots & d_4 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ d_1 & d_2 & d_3 & d_4 & \cdots & d_n \\ \end{pmatrix} $$ Does this class of matrices have a name? What is the spectrum of this matrix? The spectral radius in the case $0<d_1<d_2<d_3<\dots<d_n$ is for my purposes, but the whole spectrum in the general case would be interesting too.

  • $\begingroup$ Doing Gaussian elimination to get an upper triangular matrix. I.e., an $LU$ decomposition, you'll get diagonal entries of $U$ as $d_1, d_2-d_1, d_3-d_2-d_1, \dots, d_n-d_{n-1}-\cdots-d_1$. Maybe there is a hint here of some diagonalization process? $\endgroup$ – abnry Mar 21 '16 at 17:47
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    $\begingroup$ This is a semiseparable matrix. One thing that might help you in determining the spectrum is that the inverse of this matrix is tridiagonal. $\endgroup$ – J. M. isn't a mathematician Nov 14 '17 at 11:40

Some thoughts.

Your matrix is symmetric, thus has real eigenvalues.


$$H=\left(\begin{matrix} 1 & 0 & 0& \dots & 0 \\ -1 & 1 & 0 & \dots& 0 \\ \dots &\dots &\dots &\dots &\dots\\ 0 &\dots &0&-1 &1 \end{matrix}\right)$$

That is, $H$ is bidiagonal with $1$ on the diagonal and $-1$ on the subdiagonal.

You have $H A H^T = D$, with diagonal $D$. Also, the diagonal is made of $d_1, d_2-d_1, \dots, d_n-d_{n-1}$.

Thus, $A$ and $D$ are congruent. Then Sylvester's law of inertia tells you that there are as many positive (resp. negative, zero) eigenvalues as there are positive (resp. negative, zero) values in the diagonal entries of $D$. When (and only when) your inequalities hold, that means all eigenvalues are actually positive.

To show the above identity, subtract the row $n-1$ from row $n$, then row $n-2$ from row $n-1$, etc. Then do the same with columns.

This identity also shows that the determinant of $A$ is $d_1(d_2-d_1)\dots(d_n-d_{n-1})$.

Also, the trace of $A$ is $d_1+d_2+\dots+d_n$.

This is also the sum of eigenvalues of $A$, but when $d_1 \leq d_2 \leq \dots \leq d_n$, all eigenvalues are nonnegative, and their sum is greater than (or equal to) their maximum.

Therefore, in that case, the spectral radius of $A$ is not greater than $d_1+d_2+\dots+d_n$.

This value is only reached when there is only one nonzero eigenvalue, that is when $A$ (and therefore $D$) has rank one. This can happen only when there is an index $i<n$ and a positive real $t$ such that $d_1=d_2=\dots=d_{i}=0$ and $d_{i+1}=d_{i+2}=\dots=d_n=t$. Then the spectral radius, and only nonzero eigenvalue, is $(n-i)t$. Of course, this does not happen when your inequalities are strict (then all eigenvalues are nonzero).

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