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Could anyone give me a hint for these two

$1$ Let $x$ be a non-zero vector in $\mathbb{C}^n$ and $y$ be any vector in $\mathbb{C}^n$ then show that there exist a symmetric matrix $B$ such that $Bx=y$

$2$ Every symmetric non-singular matrix over $\mathbb{C}$ can be written as $P^tP$

thank you.

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    In the case where all entries of $x$ are nonzero, you can construct $B$ one row and column at a time. Then you have to think about how to handle zero entries in $x$.2012-06-28
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    I really do no understand your hint, could you please elaborate a lilbit?2012-06-28
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    Put an arbitrary symmetric $(n-1)\times(n-1)$ matrix in the upper left corner. For each of the first $n-1$ entries in the last column, there will be a unique choice consistent with $Bx=y$; use that choice, and also put it into the last row to keep symmetry. Then there's a unique choice for the last remaining entry of $B$. But Robert Israel's answer is better.2012-06-28

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