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 $(n\times n)$


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

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  • $\begingroup$ 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

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 $X_{ij}=v_iv_j$.


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