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Suppose the finite dimensional (real or complex) vector space $V$ has two inner products $\langle ,\rangle_1$ and $\langle,\rangle_2$. If we choose a basis, $\{v_1,\ldots,v_n\}$, each of these inner products is represented by a matrix $A_1$ and $A_2$, respectively. The usual proof that there is a linear isomorphism $T:V\longrightarrow V$ such that $$ \langle v,w\rangle_1=\langle Tv,Tw\rangle_2 $$ resorts to finding an orthonormal basis for each inner product.

I was trying to prove this without resorting to that. My naive first attempt was to see what equations result if we let $T(v_i)=\sum c_jv_j$ and see if it was obvious the system of equations could be solved. This was messy.

Upon investigation online I read that not all infinite dimensional vector spaces can be given an orthogonal basis. Here I mean an orthogonal subset such that each vector is a finite linear combination. This made me think that finding an orthogonal basis might be necessary for finding an isomorphism that respects the inner products.

I have three questions. First, is there a nice proof that such a $T$ exists in the finite dimensional case that doesn't first find orthogonal bases? Second, is there an infinite dimensional vector space with two inner products that cannot be related by such a $T$? Does such an example have to be complete?

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  • $\begingroup$ There is no unique choice for $T$. You would expect that an arbitrary choice (e.g. choice of basis) has to be made at some point. $\endgroup$
    – wj32
    Commented Oct 2, 2013 at 3:58

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For the finite dimensional case. For $k\in \{1\dots n\}$ we know that $\langle v_k,.\rangle_1$ is a linear functional. We know that there exists a unique vector $w_k$ s.t. $\langle v_k,.\rangle_1=\langle w_k,.\rangle_2$. (You know how to find those vectors with the Gram matrix? That alone is rather nasty.)

Expanding all $w_k$ in the given basis of the $v_i$ we can compute all $\langle v_i,v_j\rangle_1$ in terms of $\langle v_i,v_j\rangle_2$, where $G_k^{-1}$, the inverse of the Gram matrix of $(\langle v_i,v_j\rangle_k)$, $k\in\{1,2\}$ are involved. That's even nastier than the above calculation, but we're done.

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  • $\begingroup$ The Gram matrix was what I referred to as "messy" in the post. $\endgroup$
    – J126
    Commented Oct 2, 2013 at 11:43

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