Let $V$ be a finite dimensional vector space with an inner product $\langle\;,\,\rangle$. I have to prove that any other inner product $\langle\;,\,\rangle'$ in V can be expressed as $\langle u,v\rangle'=\langle T(u),v\rangle$ where $T$ is a positve definite linear operator in $V$.

All I was able to prove is that $\langle u,v\rangle'=\langle T(u),v\rangle$ is in fact an inner product, but don't know where to start with the rest.

Thanks in advance.


Hint: let $\{e_i\}$ be the standard basis. What is $<Te_i,e_j>$ in terms of the entries of $T$?

  • $\begingroup$ $<T(e_i),e_j>=<a_1e_1+....+a_ne_n,e_j>=a_j$. But I don't know what to do next... $\endgroup$
    – Cath
    Oct 24 '13 at 21:13
  • $\begingroup$ Not quite - $Te_i = (t_{1i},t_{2i},\dots,t_{ni})^T$ i.e. the $i$th column of $T$. If you now take the inner product with $e_j$ you get the $t_{ji}$ entry of $T$. (sorry, I should have reversed the order of these to make it more natural). Anyway, now you can work backward. If there is such a matrix $T$, then you need $<Te_i,e_j> = <e_i,e_j>'$, so computing $<e_i,e_j>'$ tells you what the entries of $T$ should be. Now you just have to prove that $<Tu,v> = <u,v>'$ for arbitrary $u,v$ (expand them in a basis and use linearity), and show $T$ is positive definite. $\endgroup$
    – BaronVT
    Oct 24 '13 at 21:27

$V$ ($n$ dimensional) is Hilbert space because is isomorphic to $\Bbb R^n$ which is complete.

It's all about sesquilinear forms.By sesquilinear map we mean a map $φ$:$V*V->\Bbb C$ with properties: 1)The map is linear to the first variable,the map $x->φ(x,y):V->\Bbb C$ is linear. 2)The map is antilinear to the second variable,the map $x->\overline φ(x,y):V->\Bbb C$ is linear.

The inner product is a sesquilinear form from it's properties.

Because $V$ is finite every linear functional(especially every sesquilinear form) is bounded.

We now say that every sesquilinear form $φ$:$V*V->\Bbb C$ defines a unique bounded operator$T \in B(V)$ from the relation $φ(x,y)=<Tx,y>$ for every $x,y \in V$.

Proof: Let $x \in V$. Also let $f_x:V-> \Bbb C:y-> \overline φ(x,y)$.

The properties 1),2) of $φ$ show that $f_x$ is linear and bounded because $φ$ is bounded.

Because $V$ is Hilbert space,we can apply Riesz theorem that shows that exists a unique $z_x \in V$:$f_x(y)=<y,z_x>$ for every $y \in V$ and thus $<z_x,y>=\overline f_x(y)=φ(x,y)$ for every $y \in V$. The $z_x$ is unique because the relation stands for every $y \in V$.

Thus the map $T:x->z_x$ is well defined and we have that $<Tx,y>=φ(x,y)$ for every $x,y \in V$.

You can easily see that $T$ is linear and bounded and that $\lVert T\rVert=\Vert φ\rVert$.

Now if we consider $φ$ as the inner product we have that $T$ is positive because $<Tx,x>=<x,x>=\lVert x\rVert^2 >=0$

  • $\begingroup$ @Cath, check my proof please. thank you $\endgroup$
    – Haha
    Nov 24 '13 at 12:38

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