TITLE: Are all finite dimensional hilbert spaces isomorphic to spaces with Euclidean norms?
QUESTION [6 upvotes]: 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).

REPLY [11 votes]: 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 $LM$.