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I came across the matrix of the form $$ \begin{pmatrix}a_{1} & a_{2} & a_{3} & a_{4}\\ a_{2} & a_{5} & a_{6} & 0\\ a_{3} & a_{6} & 0 & 0\\ a_{4} & 0 & 0 & 0 \end{pmatrix}$$ and would like to know more about it's properties in general. Is there any general theory about symmetric matrices with zeros below the anti-diagonal, or anything related?

EDIT: I'm particularly interested for the implications for positive or negative definiteness.

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  • $\begingroup$ Note that this matrix is slightly better than symmetric, on each antidiagonal all of the entries are identical. Also, perhaps it would be helpful if you indicated the context where you found this matrix? $\endgroup$ – Travis Willse Apr 9 '15 at 12:12
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    $\begingroup$ Note that doing two line swaps (i.e. multiplication with a permutation matrix) this turns into a lower triangular matrix. $\endgroup$ – Beni Bogosel Apr 9 '15 at 12:16
  • $\begingroup$ the determinant is $d^4$, you can also get information about about eigenvalues of your matrix $\endgroup$ – Math137 Apr 9 '15 at 12:23
  • $\begingroup$ @Travis Thanks, my mistake, it should be no more than symmetric $\endgroup$ – Fedor Apr 9 '15 at 12:26
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Well, write J for the antidiagonal matrix, and T for the companion matrix of polynomial $x^n$. Then the matrix you wrote is just $J (d+cT+bT^2+aT^3)$. The full set of those matrices is $J \cdot k[T]$, where $k$ is your base field. It is isomorphic (as a right $k[T]$-module) to $k[T]/T^n$. What do you want more?

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  • $\begingroup$ I had to edit the question to make the matrix symmetric but not Hankel, making your answer obsolete. $\endgroup$ – Fedor Apr 9 '15 at 12:55
  • $\begingroup$ Also I edited the question to make a more specific inquiery $\endgroup$ – Fedor Apr 9 '15 at 12:56

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