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I do not know how to calculate the inverse of the matrix. \begin{vmatrix} 1 & 1 & 1 & \cdots & 1 &\\ \beta^{2m}_{1} & \beta^{2m-1}_{1} & \beta^{2m-2}_{1} & \beta^{m+1}_{1} &1\\ \beta^{2m}_{2} & \beta^{2m-1}_{2} & \beta^{2m-2}_{2} & \beta^{m+1}_{2} &1 \\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ \beta^{2m}_{m-1} & \beta^{2m-1}_{m-1} & \beta^{2m-2}_{m-1} & \beta^{m+1}_{m-1} &1 \\ \beta^{2m}_{m} & \beta^{2m-1}_{m} & \beta^{2m-2}_{m} & \beta^{m+1}_{m} &1 \end{vmatrix} where $m \geq 2$ and all $\beta_i>1$ are different.

amWhy
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Tao
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  • You have a Vandermonde matrix, google on this (i.e., Vandermonde matrix) and you will find a lot of information. – Chris Godsil Feb 28 '13 at 14:30
  • actually , it is not a Vandermonde matrix – Tao Feb 28 '13 at 14:34
  • Where did you find this matrix? There may not be a useful closed-form solution, in particular because when $m$ is a large number that matrix becomes fairly ill-conditioned to start with. I don't mean to imply that there is no closed form solution, but that formula is likely not going to be numerically stable. It may be better to try to invert that matrix numerically with preconditioned LU decomposition, or some other algorithm which can invert ill-conditioned matrices. – Eric Haengel Feb 28 '13 at 15:13
  • what is the exponent in each column, is it all integer values $1$ to $2m$? If so, it is a Vandermonde matrix. If not, can you specify the exponents? I am confused as you have the $\ldots$ (dots) in the first row but no other row... – adam W Feb 28 '13 at 15:22
  • m is an integer , I come across this matrix in dynamical system and fractal and if I calculate the inverse, then I will find a surprising result..... – Tao Feb 28 '13 at 15:33
  • in the first row there are all 1. – Tao Feb 28 '13 at 15:35
  • we should note that there is no $\beta^{m},,\beta^{m-1}\dots \beta$, so it is not a Vandermonde matrix. – Tao Feb 28 '13 at 15:38
  • I want to calculate the determinant firstly: \begin{vmatrix} 1 & 1 & \cdots & 1 &\ \beta^{2m-1}{1} & \beta^{2m-2}{1} & \beta^{m+1}{1} &1\ \beta^{2m-1}{2} & \beta^{2m-2}{2} & \beta^{m+1}{2} &1 \ \vdots & \vdots & \ddots & \vdots &\ \beta^{2m-1}{m-1} & \beta^{2m-2}{m-1} & \beta^{m+1}{m-1} &1 \ \beta^{2m-1}{m} & \beta^{2m-2}{m} & \beta^{m+1}{m} &1 \end{vmatrix} – Tao Feb 28 '13 at 15:43
  • For determinant, this might help. http://en.wikipedia.org/wiki/Schur_polynomial – polkjh Feb 28 '13 at 16:12
  • So if I understand, it is basically a sub-matrix of the Vandermonde? (also use the @ symbol followed by my username if you are responding to my question) – adam W Feb 28 '13 at 18:07

1 Answers1

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(This is not quite a solution, but it's getting close.)

We can write your matrix $A$ in partitioned form $$ A = \begin{pmatrix} a^T&1\\ B&a\end{pmatrix} $$ where $a$ is the vector with all entries equal to 1. There is a permutation matrix $P$ such that $$ AP = \begin{pmatrix} 1&a^T\\ a&B\end{pmatrix}. $$ Assume $B$ is invertible. $$ AP = \begin{pmatrix} 1&0\\ 0&B\end{pmatrix} \begin{pmatrix} 1&0\\ B^{-1}a&I\end{pmatrix} \begin{pmatrix} 1&a^T\\ 0&I-B^{-1}aa^T\end{pmatrix}. $$ Now $\det(I-MN)=\det(I-NM)$ for any matrices $M$ and $N$ and so if $M=B^{-1}a$ and $N=a^T$, then $$ \det(I-B^{-1}aa^T) = \det(1-a^TB^{-1}a) = 1 - a^TB^{-1}a. $$ Therefore $\det(AP) =\det(B) (1-a^TB^{-1}a)$. (Here I am effectively working out the Schur complement of $B$ in $AP$ explicitly.)

I think that in the question the matrix $B$ is a diagonal matrix times a Vandermonde matrix, and so we can write down its inverse using Lagrnage interpolating polynomials. I do not see any easy way to evaluate $a^TB^{-1}a$ though.

Note that we can compute the inverse of $AP$ using the factorization, provided we can invert $I-B^{-1}J$, where $J=aa^T$. For if $\beta:=a^TB^{-1}a$ and $\beta\ne1$, then it is easy to verify that $$ (I-B^{-1}J)(I+ \frac1{1-\beta} B^{-1}J) = I. $$

Chris Godsil
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