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Given a complex inner product on a finite-dimensional vector space, is there always a matrix $M$ such that $\langle x,y \rangle=y^*Mx$. What are the properties of such a matrix?

I saw on the wiki page that if the vector space is $\mathbb{C}^n$ then there is always such a matrix, which is Hermitian positive-definite. I was just wondering if the same could be said for other finite-dimensional spaces.

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  • $\begingroup$ A real symmetric matrix defines a (real, of course) inner product on a finite dimensional vector space iff it is definite positive, so yes: one can generalize in a way what you wrote above. $\endgroup$ – Timbuc Jan 12 '15 at 21:50
  • $\begingroup$ Well, every finite dimensional complex vector space is isomorphic to $\Bbb C^n$, so the wiki page says everything there is to be said, as far as I can tell. $\endgroup$ – Omnomnomnom Jan 12 '15 at 21:55
  • $\begingroup$ If $x$ belongs to an abstract inner product space (not $\mathbb C^n$) then what does it mean to multiply $x$ by $M$? $\endgroup$ – littleO Jan 12 '15 at 22:25
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Given an inner product space $V$ with basis $B = \{v_1,\dots,v_n\}$, we can write $$ \langle x,y \rangle = [y]_B^* [M]_B[x]_B $$ Here, $[x]_B$ is the coordinate vector of $x$ with respect to the basis $B$. The matrix $[M]_B$ can be defined by $$ [M]_B = \pmatrix{ \langle v_1,v_1 \rangle & \cdots & \langle v_1,v_n \rangle\\ \vdots & \ddots & \vdots\\ \langle v_n,v_1 \rangle & \cdots & \langle v_n,v_n \rangle } $$ The matrix $[M]_B$ must be positive definite, and every positive definite matrix defines an inner product in this fashion.

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