Let $\mathbf{x} = (x_1, \dots, x_n)$ and $\mathbf{y} = (y_1, \dots, y_n)$ be two vectors in $\mathbb{R}^n$. Show that $$\langle \mathbf{x}, \mathbf{y} \rangle = 2\left(\sum_{i=1}^nx_iy_i \right) - \sum_{i=1}^{n-1}(x_iy_{i+1}+x_{i+1}y_i)$$ defines an inner product.

The linearity and conjugate conditions are elementary, but how would I show positive-definiteness? That is, showing that: $$\langle \mathbf{x},\mathbf{x} \rangle = 2 \left(\sum_{i=1}^nx_i^2 \right) - \sum_{i=1}^{n-1}(2x_ix_{i+1}) \ge 0,$$ and $$\langle \mathbf{x},\mathbf{x} \rangle = 0 \iff \mathbf{x}=0. $$ I'm stuck on how to manipulate the expressions.

Please help!

  • $\begingroup$ Cauchy-Schwarz is your friend $\endgroup$ – eepperly16 Nov 19 '17 at 18:24
  • 2
    $\begingroup$ For the first part, you could use that $0 \leq (x-y)^2=x^2+y^2-2xy$. Try to rewrite the definition of the IP. $\endgroup$ – User123456789 Nov 19 '17 at 18:29
  • $\begingroup$ @eepperly16 Ordinarily that would be ideal, but without assuming Cauchy-Schwarz and working just from first principles is what's desired. $\endgroup$ – Roz Nov 19 '17 at 18:35

$2 \left(\sum_{i=1}^nx_i^2 \right) - \sum_{i=1}^{n-1}(2x_ix_{i+1})=x_1^2+x_n^2+\sum_{i=1}^{n-1}(x_i-x_{i+1})^2\geq0$


I find it easier and neat the following approach:

Let $ A \in \mathbb{M}_n(\mathbb{R})$ defined as $$ A(m,n) \doteq \begin{cases} 1 &,& n = m+1 \\ 0 \end{cases} $$

So for $ x = (x_1, ..., x_n)$ we have $Ax = (x_2, ..., x_n, 0)$.

Then we can rewrite the inner product $ \langle \cdot, \cdot \rangle $ in terms of the standard inner product $ \langle \cdot, \cdot \rangle_{\mathbb{R}^n} $ as

$$\begin{align} \langle x, y \rangle &= 2 \langle x, y \rangle_{\mathbb{R}^n} - \langle x, Ay \rangle_{\mathbb{R}^n} - \langle Ax, y \rangle_{\mathbb{R}^n} \\ &= \langle x - Ax, y - Ay \rangle_{\mathbb{R}^n} \end{align}$$

and take advantage of their already established properties.

Thus the linearity, the symmetry and the positive-definiteness follows immediately.

I'll leave you to check that $\langle x, x \rangle = 0 \iff x=0$.


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