# Can we speak of angles when our inner product isn't standard?

I'm wondering if we can speak of angles, specifically right angles when we live in an inner product space, but our product $\langle u, v\rangle$ is not the standard dot product that we all know and love, but rather some other obscure unknown product.

Specifically I'm looking at the first proof for Cauchy-Schwarz inequality proof found here https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality

Basically we have 3 vectors $u,v, z \in \mathbb F^n$ such that $u = z + v$ and $\langle z, v \rangle = 0$

The author then goes on to say that this means $u, v, z$ form a right triangle and so we can use the pythagorean theorem.

It's not perfectly clear to me why this is true. Our inner product is obscure. It is not necessarily true that if $\langle z, v \rangle = 0$ then $u \cdot v = 0$ right? And even if it did, in our world, the standard dot product "does not exist", so can we still use it to infer angles between vectors?

TLDR

$u = z + v$, $\langle z, v \rangle = 0$. The standard dot product does not exist. Can we use the pythagorean theorem? why? What does right angle even mean in our sense?

• Actually to be rigorous, when you say two vectors are orthogonal then you should always precise with respect to which inner product. – Surb Feb 8 '18 at 17:53
• Also note that by the cauchy schwartz inequality, every inner product allows you to define a “meaningful” notion of angle between two vecors. This definition may differs for different inner products on the same space. – Surb Feb 8 '18 at 17:55
• Finally if a triangle is the convex hull of three points and two of its edges are orthogonal wrt to a given inner product, the pythagor thm holds wrt to the norm induced by this innerproduct. – Surb Feb 8 '18 at 18:00
• I need to start thinking of pythagoras with respect to inner products rather than 90 degree angles. I'll get there. – Oria Gruber Feb 8 '18 at 18:04
• Well the key is to think about angles in terms of inner product – Surb Feb 12 '18 at 21:16

The answer lies in the relation between the Pythagorean theorem and the usual inner product.

$$\left<u+v, u+v\right>=\left<u, u\right>+2\left<u, v\right>+\left<v, v\right>$$

holds as soon as $\left<\cdot ,\cdot\right>$ is bilinear and symmetric. Therefore, when $\left<u, v\right>=0$, you have your Pythagorean theorem: $$\| u+v\|^2=\| u\|^2+\| v\|^2$$

• Ah so the Pythagorean theorem isn't as simple as "in a right triangle...". It's actually follows directly from a property of all inner products, regardless of angles, triangles, or the like. – Oria Gruber Feb 8 '18 at 17:53
• Exactly. This is part of the reason that makes inner products so powerful - they encode two fundamental geometric properties at once: distances and angles. – Arnaud Mortier Feb 8 '18 at 17:55
• Kinda makes me look at the Pythagorean theorem differently now. I used to hold it in high regard, but now it's as basic as "if $a = b + c$ and $c = 0$ then $a = b$ ", that hardly qualifies as a theorem. – Oria Gruber Feb 8 '18 at 17:57
• Hahaha don't go say that to Parseval. en.wikipedia.org/wiki/Parseval%27s_identity – Arnaud Mortier Feb 8 '18 at 17:59
• This is the Venn Diagram of theorems. – Oria Gruber Feb 8 '18 at 18:00

On real inner product vector spaces, you can define the angle $\alpha$ between two (nonzero) vectors $x$ and $y$ by $$\cos\alpha=\frac{\langle x,y\rangle}{\sqrt{\langle x,x\rangle\langle y,y\rangle}}$$ because the Cauchy-Schwarz inequality tells you precisely that the number in the right hand side belongs to $[-1,1]$. With this definition, we have $\alpha=\pi/2$ if and only if $\langle x,y\rangle=0$.

Right angles according to the “obscure” inner product need not be right angles with respect to the standard inner product, but it's of no consequence as long as we stick to the definition of angle with respect to the given inner product.

On the other hand, any “obscure” inner product on $\mathbb{R}^n$ can be written as $$\langle x,y\rangle_{?}=x^TB^TBy \tag{*}$$ for an invertible $n\times n$ matrix $B$. The mapping $f\colon x\mapsto Bx$ will then define an isomorphism of $\mathbb{R}^n$ with the standard inner product with the same space with the “obscure” one. Indeed, if $\langle x,y\rangle$ denotes the standard inner product, we have $$\langle x,y\rangle_?=\langle f(x),f(y)\rangle$$

Why is (*) possible? Consider the standard basis of $\mathbb{R}^n$, $\{e_1,\dots,e_n\}$ and form the matrix $$A=[\langle e_i,e_j\rangle_?]$$ (the Gram matrix of the “obscure” inner product). Such a matrix is symmetric (and positive definite), so it can be diagonalized with an orthogonal matrix: $A=U^TDU$ and the eigenvalues (the diagonal entries on the diagonal matrix $D$) are positive. If $\sqrt{D}$ is the diagonal matrix with positive entries on the diagonal such that $D=(\sqrt{D})^2$, then setting $B=\sqrt{D}U$ gives the required form.