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Do we always have an inner product for a finite dimensional vector space $V$ over a field $k$ such that $V$ is a Hilbert space? Thank you very much.

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No. Part of the requirements of an inner product space is that it be over the real or complex numbers. Certainly any vector space over some other field would be a counterexample. –  kahen Apr 5 '13 at 13:42
    
For any basis $\{\vec{v}_1, \dots, \vec{v}_n\}$, one can define an inner product by $\langle \vec{v}_i, \vec{v}_j \rangle = \delta_{ij}$. Completeness is not an issue in finite dimensions. –  Sammy Black Apr 5 '13 at 13:42
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As @kahen says, you would need $k=\mathbb{R}$ or $k=\mathbb{C}$. –  Sammy Black Apr 5 '13 at 13:43
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A finite-dimensional real or complex vector space always has an inner product, but it has a lot of them - there is no "obvious" right choice for the inner product unless the vector space has some structure. –  Thomas Andrews Apr 5 '13 at 13:44
    
@kahen Maybe convert to an answer, including Sammy Black's comment? (or vice versa) –  Julian Kuelshammer Apr 5 '13 at 15:59

1 Answer 1

up vote 2 down vote accepted

Since it's part of the definition of an inner product space that it must be over the real or complex numbers, any vector space over some other field will not admit an inner product.

Barring that obvious obstruction, the situation where one's vector space $V$ is over $\mathbb K$ ($=\mathbb R$ or $\mathbb C$) is quite simple: For any basis $v_1,v_2,\dotsc,v_n$ one has an inner product given by

$\qquad \displaystyle\langle v_i,v_j\rangle = \delta_{ij} = \begin{cases} 1 & \text{if }i=j \cr 0 & \text{if }i\neq j\end{cases}\ $ extended using sesquilinearity to all of $V$.

Some related things you might find enlightening:

  • Norms Induced by Inner Products and the Parallelogram Law: If a norm satisfies the parallelogram law, does it come from an inner product? Yes, but it takes a surprising amount of work to actually show that this is the case.

  • Any two norms on a finite dimensional linear space (vector space over $\mathbb R$ or $\mathbb C$) are equivalent. If you search Math.SE you're likely to find several questions and answers related to this. It's very common to prove this in real analysis courses.

  • Suppose $V$ is a finite dimensional normed linear space and $K \subset V$ is compact , absolutely convex and spans $V$, then there is a norm on $V$ such that $K$ is the unit ball w.r.t. that norm (and per the above equivalent to the given norm on $V$). It's not too difficult to come up with an idea that could work—just draw a picture of the situation in $\mathbb R^2$.

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