Can two rows of a symmetric Pascal matrix (in general) be ever linear dependent?
Is the rank of a symmetric Pascal matrix always equal to the number of rows (m)?
I have a symmetric Pascal matrix where m=4 and n=4 and should count its rank.
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2A $4\times 4$ Symmetric Pascal Matrix has rank $4$. So, its rows are never linearly independent. – Debashish Jun 19 '14 at 09:53
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Thanks for an answer... is this generally valid for a symmetric Pascal matrix of any size? – Kozuch Jun 19 '14 at 09:56
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1Thats what I am thinking now ... Nice question. – Debashish Jun 19 '14 at 09:57
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The symmetric pascal matrix $S$ has $LU$ factorization where $L$ is the lower triangular pascal matrix, and $U$ is the upper triangular matrix. Both $L$ and $U$ have all ones on the diagonal. In particular, the determinants of these are each 1. Thus, using $\det(L)\det(U)=\det(LU)=\det(S)=1$ we can conclude that $S$ is invertible and must have rank=#-columns=#-rows.
In response to your question about dependence of rows: It is equivalent to consider the columns (since it is symmetric). By the above reasoning, no two rows/columns of $S$ can be linear combinations of each other. In fact no $m$ rows or columns (where $1\leq m \leq n$=number of rows = number of columns) can be linearly dependent.
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