Are all finite dimensional hilbert spaces isomorphic to spaces with Euclidean norms?

Question

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).

Answer 1

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.

Answer 2

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.

Answer 3

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 $LM$.