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I have read that Hilbert spaces generalize Euclidean spaces to infinite dimensions.

This suggests to me that any finite dimensional Hilbert space is isomorphic to a space with the norm $||v||=\sqrt {v_1^2 +...+v_n^2}$. Is this true? If not, then it seems it is wrong to say that Hilbert spaces generalize Euclidean spaces to infinite dimensions (i.e. they would be more general than that).

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  • $\begingroup$ Are you familiar with Gram-Schmidt orthonormalization? Given an arbitrary basis of a finite dimensional inner product space, the procedure generates an orthonormal basis, i.e. one with respect to which the inner product is the usual Euclidean one (the dot product). Given such a basis, it should be clear how to get an isometry (the natural notion of isomorphism for inner product spaces) from any finite dimensional inner product space to $\mathbb{R}^n$ equipped with the dot product. $\endgroup$ – jawheele Apr 18 at 13:42
  • $\begingroup$ @jawheele, do you mean hilbert space rather than inner product space? I can't see how that would be an isomorphism from an arbitrary finite inner product space that is incomplete, to $R^n$ with the dot product. $\endgroup$ – user56834 Apr 22 at 8:18
  • $\begingroup$ The existence of the above isomorphism is actually a proof that every finite dimensional inner product space is complete, and it is true more generally that every finite dimensional normed vector space is complete. $\endgroup$ – jawheele Apr 22 at 14:05
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If we are given an inner product on a finite dimensional space, there will always be a change of basis that turns the given inner product into the Euclidean one.

By definition the product $(\cdot,\cdot)$ can be represented as a symmetric positive-definite matrix $A$ such that $(x,y) = x^T A y$. Now by the spectral theorem (which you probably learnt in Linear Algebra) the matrix $A$ has an orthogonal basis of eigenvectors. In other words there is an orthogonal matrix $M$ such that $J = M^TAM$ is diagonal. That means $(x,y) = x^T M^TJ My = (Mx)^TJ(My)$.

Exercise: Use positive-definiteness to prove $J$ has only positive diagonal entries $\lambda_1,\lambda_2,\ldots,\lambda_n$.

It follows there is a further diagonal matrix $L$ such that $J=LIL = L^TIL$. That gives $$(x,y) = (Mx)^TL^TIL(My)= (LMx)^T(LMy)$$ Put another way, for each $x,y$ we have

$$\big(\,(LM)^{-1}x,(LM)^{-1}y\, \big) = x^T y$$

That means the given product is the Euclidean product with respect to the basis with transition matrix $LM$.

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It is true in multiple ways. You have to distinguish between algebraic and tological isomorphy.

Let us consider vector spaces over $\mathbb{R}$, the case for $\mathbb{C}$ is analogous. First, realize, that any real vector space $V$ of dimension $n$ is (algebraically) isomorphic to $\mathbb{R}^n$. To see the isomorphism, one can choose an arbitrary base $$ b_1, ..., b_n \in V$$ of $V$ and then define the mapping $$ V \ni v \mapsto (\text{coordinates of v with respect to $b_1,...,b_n$}) \in \mathbb{R}^n \quad (\dagger).$$ It is not hard to show, that this is a bijection as well as a vector space homomorphism.

So every finite dimensional real vector space V is algebraically isomorphic to $\mathbb{R}^n$.

Morever, every Hilbert space of dimension $n$ is topologically isomorphic to $\mathbb{R}^n$ (where the topology on $H$ is induced by the scalar product of $H$). To see this, realize that $(\dagger)$ is a continuous bijection with continuous inverse and therefore a topological isomorphism.

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Any Euclidean space is Hilbert space, so Hilbert spaces generalize Euclidean. But Euclidean space isn't just a finite dimensional linear space, it's a finite dimensional linear space with additional structure - either inner product, or norm that satisfies parallelogram law.

Such structure can be introduced on any finite dimensional space, so any finite dimensional space is isomorphic to some Euclidean as linear space.

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  • $\begingroup$ I'm not sure what linear spaces have to do with this? Do you mean this: en.wikipedia.org/wiki/Linear_space_(geometry)? It seems like a very different formalization than norm/metric/inner product based formalizations. $\endgroup$ – user56834 Apr 18 at 14:59
  • $\begingroup$ I meant usual vector spaces (they are also called linear spaces sometime). Did you mean something different in question text? ("any finite dimensional linear spaces") $\endgroup$ – mihaild Apr 18 at 15:04
  • $\begingroup$ whoops that should have been "hilbert space". Typo, sorry! $\endgroup$ – user56834 Apr 18 at 16:40

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