# Bases and inner products

I am not quite sure what this question is asking for:

Given $f(\vec{x})=x^2+xy+y^2+yz+z^2+xz$, find a basis for the corresponding inner product on $\mathbb R^3$.

(I was told that there is an obvious orthogonal basis -- obtainable by inspection)

I don't even know what "corresponding inner product" means...

Thanks.

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A positive-definite quadratic form is essentially an inner product. To get an explicit expression for the symmetric bilinear form, look up the polarisation formula. – Zhen Lin Dec 2 '11 at 0:19
Thanks, @ZhenLin, but I still don't quite understand this, would you mind elaborating? – Owen H. Dec 2 '11 at 0:31
@ZhenLin: What does it mean to find a basis for an inner product? – Owen H. Dec 2 '11 at 0:39
The problem is a bit vague, but note $f(\vec{x}) = \vec{x}^T A \vec{x}$ where $A = \left[\begin{matrix} 1 & 1/2 & 1/2\\1/2 & 1 & 1/2\\1/2 & 1/2 & 1\end{matrix}\right]$. Given positive semidefinite matrix $B$, one can define an inner product: $\langle \vec{x}, \vec{y} \rangle_B = \vec{x}^T B \vec{y}$. See the connection? – p.s. Dec 2 '11 at 0:51
I think "find a basis for an inner product" means "find a basis B for the vector space such that B is orthonormal with respect to the inner product." – Gerry Myerson Dec 2 '11 at 3:26

I'm not sure how much to give away. The matrix being introduced is $$M \; = \; \left( \begin{array}{rrr} 1 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 1 \end{array} \right) ,$$
which is symmetric, and an orthogonal basis is indeed pretty easy to find. If required, one may then normalize to get an orthonormal basis. If we take a column vector $x$ and its transpose $x^T,$ then the value of the quadratic form at the point $x$ is just $$f(x) = x^T \; M \; x.$$ The inner product of two vectors $x,y$ is given by $$\langle x,y \rangle = \frac{1}{2} \left(f(x+y) - f(x) - f(y)\right) = x^T M y = y^T M x$$ which is one way to write polarization.
Here is something not mentioned. If you take all three entries in $\vec{x}$ to be integers, the result $f(\vec{x})$ is a nonnegative integer. Not all positive integers show up, though. A number $n \geq 0$ has a "representation" $n=f(x)$ if and only if $n$ is not of the form $4^k (16 m + 14).$ Go figure.
Thanks, Will. Is the "basis" simply the columns of $M$? I don't know what the definition of a basis of an inner product is... Also, very interesting last point! – Owen H. Dec 2 '11 at 0:45
Though the $x,y,z$ may or may not be real. – Owen H. Dec 2 '11 at 1:16