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what is the pseudo-inverse of this type of matrix $$ A=\left[\begin{array} {C}a_0&0&b_1&b_2&\ldots&b_n \\ a_1&b_n&0&b_1&\ldots&b_{n-1} \\ a_2 &b_{n-1}&b_n&0&\ldots &b_{n-2} \\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots \\a_m&b_{n-m+1} & b_{n-m+2}&b_{n-m+3} &\ldots &0 \end{array}\right] $$

and are there any interesting properties of $A$?

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    $\begingroup$ Your matrix has $m+1$ rows and $n+2$ columns; are you assuming $m=n+1$? If not, it isn't a square matrix, and doesn't have an inverse. $\endgroup$ – Gerry Myerson Oct 6 '12 at 10:12
  • $\begingroup$ sorry I meant the pseudo-inverse $\endgroup$ – G.T. Oct 6 '12 at 10:16
  • $\begingroup$ If you meant the pseudo-inverse, please edit the body of the question so it says what you mean. Also, there's something funny going on in the bottom row, where the 2nd and 4th entries are equal. $\endgroup$ – Gerry Myerson Oct 6 '12 at 10:20
  • $\begingroup$ corrected as suggested, thanks! $\endgroup$ – G.T. Oct 6 '12 at 10:22

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