0
$\begingroup$

I have a positive semi definite symmetric matrix $X$, $(n\times n)$.

let $X=vv^T$ s.t $\|v\|=1$.

I came to a point where I am stuck to show which is:

$v^TYv=\langle X,Y\rangle$ (How to show this equality?)- inner product is of symmetric matrices

and $Y$ is a symmetric matrix $(n\times n)$

$\endgroup$

closed as unclear what you're asking by Rory Daulton, user91500, Davide Giraudo, darij grinberg, Ken Jul 31 '15 at 13:39

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What does $< X, Y >$ mean? $\endgroup$ – chhro Jul 31 '15 at 11:16
  • $\begingroup$ Is it $\langle X,Y \rangle =tr(XY)$ ? $\endgroup$ – Svetoslav Jul 31 '15 at 11:18
2
$\begingroup$

If the inner product is $\langle X,Y\rangle = \operatorname{tr} (X^TY)= \operatorname{tr}(XY)$ ($X,Y$ are symmetric), then $v^TYv=\sum\limits_{i,j=1}^n Y_{ij}v_iv_j =\sum\limits_{i,j} Y_{ij}X_{ij} = \operatorname{tr}(XY)=\langle X,Y\rangle$, because you have that $X_{ij}=v_iv_j$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.