0
$\begingroup$

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.

  • 0
    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

1 Answers 1

1

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 $\overline{x}$.

  • 0
    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 $y$?2012-06-28