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.

  • $\begingroup$ 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, 2015 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. $\endgroup$ Jan 12, 2015 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, 2015 at 22:25

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .