TITLE: Why is it true that every finite-dimensional inner product space is a Hilbert space?
QUESTION [10 upvotes]: A Hilbert space is, by definition, a complete inner product space. If $(V,|.|)$ is finite dimensional inner product space of dimension $n$ then it is (topologically) isomorphic to $\mathbb{R}^n$ which is of course complete. My instinct here is to say "and therefore, $V$ is also a Hilbert space". I'm not sure about this step however since completeness depends on the norm and the norm depends on the selected inner product (assuming the induced norm $||x|| = (x | x)^{1/2}$ is used). So, must we place some condition on the inner product or is the topological isomorphism between $\mathbb{R}^n$ enough to guarantee that V is Hilbert?

REPLY [4 votes]: As a matter of fact, the compactness of the unit sphere in a finite dimensional space can tell us that all norms on that space are equivalent. Therefore, in any norm (induced by inner product or otherwise), finite dimensional normed linear spaces are always complete.