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Euclidean Inner Products

steenis

Well-known member
MHB Math Helper
Jul 30, 2016
255
I have the following (small) problem:
Let $ ( , ):V \times V \rightarrow \mathbb{R} $ be a real-valued non-degenerate inner product on the real vector space $V$.
Given, for all $v \in V$ we have $(v,v) \geq 0$
Now prove that if $(x,x)=0$ then $x=0$ for $x \in V$, that is, prove that the inner product is Euclidean.
I think it is easy, but I cannot find it. Thank you.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
An inner product requires positive definiteness.
That is, if $x\ne 0$ then $(x,x)>0$.
It follows from the linearity axiom that $(0,0)=(0\cdot x,0)=0\cdot(x,0)=0$, so that the desired property follows.

A pseudo-Euclidean space drops the requirement that the bilinear form is positive definite.
Such a bilinear form is not an inner product though.
Did you perhaps mean such a bilinear form instead of an inner product?
 
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steenis

Well-known member
MHB Math Helper
Jul 30, 2016
255
I am sorry, but I do not understand your answer, can you explain how the desired property follows, please? How do you use the hypothesis?

My definition comes from
Rosén - Geometric Multivector Analysis (2019), p.5, where an Euclidean inner product is an inner product with the additional property that $(v,v) \gt 0$ for all $0 \ne v \in V$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
Check out condition 3 in the definition of an inner product on wiki and the first Elemental Property that follows, which is what you want to prove.

It looks as if Rosén's definition of an inner product is different from the one on wiki.
If so, could you quote it please?
 
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steenis

Well-known member
MHB Math Helper
Jul 30, 2016
255
It would follow if "positive definiteness" is given, but "positive definiteness" is not given, look at the hypothesis carefully. In fact, "positive definiteness" is what we want to prove as you stated correctly in your former post.

Definition of Rosén
An inner product on a real vector space $V$ is a symmetric bilinear map $( , ):V \times V \rightarrow \mathbb{R}$ that is non-degenerate.
An Euclidean inner product is an inner product with the additional property that $(v,v) \gt 0$ for all $0 \ne v \in V$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
Perhaps you can also quote Rosén's definition of non-degenerate?

Suppose we use wiki's definition of non-degenerate as it applies to finite dimensional vector spaces.
Then we have: if $(x,y)=0$ for all $y\in V$, then $x=0$.

I thought about it a bit, and came up with the following proof inspired by a proof of the Cauchy-Schwarz inequality.
Perhaps there is a simpler proof, but I didn't find one yet.

Proof by contradiction

Suppose there is a $v\in V$ such that $v\ne 0$ and $(v,v)=0$.
Then if follows from non-degeneracy that there must be an $u\in V$ such that $(v,u)\ne 0$.

Then for any $\lambda\in \mathbb R$ we have $(v-\lambda u,v-\lambda u)=\lambda^2(u,u)-2\lambda(u,v)\ge 0$.
Let $\mu>0$.
Substitute $\lambda=\mu(u,v)$ to find $(\mu(u,v))^2(u,u)-2\mu(u,v)(u,v)=\mu^2(u,v)^2(u,u)-2\mu(u,v)^2\ge 0\implies \mu(u,u)-2 \ge 0 \implies (u,u)\ge \frac 2\mu$.
This implies that $(u,u)$ is greater than any real number, which is a contradiction. Qed.
 
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steenis

Well-known member
MHB Math Helper
Jul 30, 2016
255
If a bilinear map $(,):V \times V \rightarrow \mathbb{R}$ is non-degenerate if
for $x \in V$ we have $(v,x)=0$ for all $v \in V$ then we must have $x=0$
and if
for $y \in V$ we have $(y,v)=0$ for all $v \in V$ then we must have $y=0$

I think I understand your proof, thank you.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
I've still been thinking about your problem statement, which is interesting. Thank you for posting it.

I believe it boils down to a variant of the triangle inequality.
In an euclidean space we have that the length of a segment of a triangle must be between the sum and the difference of the other two segments.
However, in a pseudo-euclidean space, the triangle inequality (or the Cauchy-Schwarz inequality) does not necessarily hold.
Still, the same reasoning can be applied.
To prove the Cauchy-Schwarz inequality, we typically look at the projection of a vector $u$ onto another vector $v$ and see what happens.
In this case we can look at the projection of some vector $u$ that has a non-zero inner product with a vector $v$ that has "length" zero.
That is, if we have $(v,v)=0$ for some non-zero vector $v$, and if we have $(u,v)\ne 0$ for some vector $u$, which is required for non-degeneracy, then we can look at what happens.
The 'normal' projection of vector $u$ onto $v$ is given by $\pi_v(u)=v-\frac{(v,u)}{(u,u)}u$.
So we can take a look at $\pi_v(u)$ and see what happens.
Turns out that it leads to a contradiction with the requirement that $(\pi_v(u),\pi_v(u))\ge 0$.
Of course we could have that $(u,u)=0$, so that we couldn't apply this, but in that case, we can get a contradiction when we evaluate $(v-(v,u)u, v-(v,u)u)$ instead.
 
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steenis

Well-known member
MHB Math Helper
Jul 30, 2016
255
(1) I shouldnot have mentioned the term euclidean inner product, it leads to confusion, sorry about that.

(2) I can follow your reasoning. I think it is a more elegant proof than the one you gave in your former post.

(3) I supposed it was an small problem, but there is more to it than I thought.
 
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