Isomorphisms of inner-product spaces

Solution 1:

1. If two vector spaces $V$ and $W$ are isomorphic, there are as many isomorphisms $V\to W$ as automorphisms of $W$. Indeed, is $Iso(V,W)$ is the set of isomorphisms $V\to W$, $Aut(W)$ is the set of all automorphisms of $W$ and $f_0\in Iso(V,W)$ is any isomorphism, the function $$a\in Aut(W)\mapsto a\circ f_0\in Iso(V,W)$$ is a bijection. It follows there are as many isomorphisms from $V$ to $W$ as there are automorphisms of $W$. A similar argument shows that there are as many such isomorphisms as there are automorphisms of $V$, too.

2. Let $\langle\mathord-,\mathord-\rangle_0$ be an inner product on a vector space $V$., and let as before $Aut(V)$ be the set of all automorphisms of $V$ and let $Inn(V)$ be the set of all inner products on $V$. Then for every $f\in Aut(V)$ there is an inner product $\langle\mathord-,\mathord-\rangle_f$ on $V$ such that for all $v,w\in V$ we have $$\langle v,w\rangle_f=\langle f(u),f(w)\rangle_0,$$ and the function $$f\in Aut(V)\mapsto \langle\mathord-,\mathord-\rangle_f\in Inn(V)$$ is surjective; in this way we obtain a description of all inner products on $V$. It is not injective, though.

   Indeed, two automorphisms $f$, $g\in Aut(V)$ have the same image, so that $\langle f(v),f(w)\rangle_0=\langle g(v),f(w)\rangle_0$ for all $v$, $w\in V$ if and only if the composition $f\circ g^{-1}$ preserves the original inner product $\langle\mathord-,\mathord-\rangle_0$, in the sense that $$\langle (f\circ g^{-1})(v),(f\circ g^{-1})(w)\rangle_0=\langle v,w\rangle_0$$ for all $v$, $w\in V$.

3. Suppose $V$ is a vector space with an inner product $\langle\mathord-,\mathord-\rangle$ and that $W$ is a vector space. Suppose, moreover, that $f:W\to V$ is an isomorphism of vector spaces. Then we can define a new inner product $\langle\mathord-,\mathord-\rangle'$, now on $W$, so that for all $v$, $w\in W$ we have $$\langle v,w\rangle'=\langle f(v),f(w)\rangle.$$ With respect to the inner products on $V$ and on $W$ that we now have, the map $f$ is an isomorphism of inner product spaces.

   This answers your 4th question.

4. I don't understand your 3rd question :-)

Solution 2:

Just a few fairly simple remarks should settle most questions.

• A finite dimensional inner product space always has at least one orthonormal basis.
• A linear map $V\to W$ between two finite dimensional inner product spaces is an isomorphism of inner product spaces if and only if the image of a (some fixed) orthonomal basis of $V$ is an orthonormal basis of $W$.
• If $V$ is a real vector space of dimension $n$, and $\def\B{\mathcal B}\B$ a basis of $V$, then there is a unique inner product on $V$ for which this basis is orthonormal. (This answers your last question) With $f_\B:V\to \Bbb R^n$ the map sending a vector $v$ to its coordinates in $\mathcal B$, this inner product is obtained from the by transport of structure via $f_\B$, in other words $\langle v,w\rangle_V=\langle f_\B(v),f_\B(w)\rangle_{\Bbb R^n}$ by definition. (By the previous point applied to $f_\B^{-1}$ the basis $\mathcal B$ is orthonormal if and only if $f_\B$ is an isomorhism of inner product spaces, and the equation used as definition states just that.)

So up to isomorphism $\Bbb R^n$ with the standard inner product is the unique $n$-dimensional inner product space. For two inner product spaces $V,W$ of dimension $n$, there are as many isomorphisms of inner product spaces as there are orthonomal bases in $W$ (or in $V$); the set is also in bijection with the set of automorphsims of the standard inner product space $\Bbb R^n$, which is the orthogonal group $O_n(\Bbb R)$. On a given space $V$ there are as many different (though isomorphic) inner products as there are cosets in $O_n(\Bbb R)\backslash GL_n(\Bbb R)$. (Both the sets of isomorphisms and of different inner products are infinite in general, so "as many" should be interpreted as "naturally in bijection with".) Explanation for the latter correspondence: fixing $\B$, the cooordinate map $V\to \Bbb R^n$ corresponding to any basis of $V$ is of the form $g\circ f_\B$ for some $g\in GL_n(\Bbb R)$, which defines an inner product on $V$, and $g_1,g_2\in GL_n(\Bbb R)$ define the same inner product iff the standard basis of $\Bbb R^n$ transported to $V$ by $(g_1\circ f_\B)^{-1}$ and then back to $\Bbb R^n$ by $g_1\circ f_\B$ give an orthonormal basis, which means $g_2\circ g_1^{-1}\in O_n(\Bbb R)$ of $g_2\in O_n(\Bbb R)g_1$. The set of inner products on $\Bbb R^n$ is also in bijection with the set of positive definite symmetric $n\times n$ matrices.