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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 $x$. – Gerry Myerson Jun 28 '12 at 5:35
I really do no understand your hint, could you please elaborate a lilbit? – La Belle Noiseuse Jun 28 '12 at 5:51
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. – Gerry Myerson Jun 28 '12 at 6:13

Hint for 1: if $y^t x \ne 0$ you can take $B = u u^t$ where $u$ is a suitable multiple of $y$. If $y^t x = 0$ you can take $B = u u^t + \overline{x} \overline{x}^t$ where $u$ is a certain linear combination of $y$ and $\overline{x}$.

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dear sir, I really need more explanation, while doing by hand from ur hint, not able grasp or think what is going on, and how the matrix $B$ would be? – La Belle Noiseuse Jun 28 '12 at 6:31
Did you get the case $y^t x \ne 0$? For the other case, if you take $u = \alpha y + \beta \overline{x}$, what is $(u u^t + \overline{x} \overline{x}^t) x$? What should $\alpha$ and $\beta$ be to make that $y$? – Robert Israel Jun 28 '12 at 17:57

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