Proving (x,x) = 0 implies x = 0 in real vector space V

[steenis]

TL;DR Summary

A semidefinite inner product is also positive-definite

I have the followinq question:

Let $(,)$ be a real-valued inner product on a real vector space $V$. That is, $(,)$ is a symmetric bilinear map $(,):V \times V \rightarrow \mathbb{R}$ that is non-degenerate

Suppose, for all $v \in V$ we have $(v,v) \geq 0$

Now I want to prove that if $(x,x)=0$ then $x=0$ for $x \in V$

Can anybody help me ?

 

[PeroK]

Summary:: A semidefinite inner product is also positive-definite

I have the followinq question:

Let $(,)$ be a real-valued inner product on a real vector space $V$. That is, $(,)$ is a symmetric bilinear map $(,):V \times V \rightarrow \mathbb{R}$ that is non-degenerate

Suppose, for all $v \in V$ we have $(v,v) \geq 0$

Now I want to prove that if $(x,x)=0$ then $x=0$ for $x \in V$

Can anybody help me ?

Apart from $(v,v) \geq 0$, what other properties does $(,)$ have?

 

[steenis]

It is a symmetric bilinear map $(,):V \times V \rightarrow \mathbb{R}$ that is non-degenerate

 

[PeroK]

It is a symmetric bilinear map $(,):V \times V \rightarrow \mathbb{R}$ that is non-degenerate

Would it not be useful to translate those words into mathematical form?

By the way, the ethos on this site is to get you to do as much of the work yourself.

 

[steenis]

I think the words "symmetric", "bilinear", "map" , "$V \times V$", "$\mathbb{R}$", etc., are all mathematical terms, well known in linear algebra. So I do not know how to translate those mathematical words into "mathematical form."

 

[PeroK]

I think the words "symmetric", "bilinear", "map" , "$V \times V$", "$\mathbb{R}$", etc., are all mathematical terms, well known in linear algebra. So I do not know how to translate those mathematical words into "mathematical form."

A proof usually requires definitions to be translated into something that can be manipulated. For example: to prove that the square of every even number is an even number, you would translate "even number" into $n$, such that $n = 2k$ for some integer $k$.

That's what I mean by mathematical form.

 

[steenis]

Well, in this case it is given that "$(v,v) \geq 0$ for all $v \in V$". My question is, can anybpdy manipulate this statement into "if $(x,x)=0$ then $x=0$, for $x \in V$". In this context, $(,)$ is a symmetric bilinear map that is non-degenerate

 

[PeroK]

Well, in this case it is given that "$(v,v) \geq 0$ for all $v \in V$". My question is, can anybpdy manipulate this statement into "if $(x,x)=0$ then $x=0$, for $x \in V$". In this context, $(,)$ is a symmetric bilinear map that is non-degenerate

As I said, the ethos on this site is that you at least make some effort to produce a proof. The logic is that you should be able to make a start at least.

 

[steenis]

So try this. Let $x \in V$ such that $x \neq 0$ and $(x,x)=0$. The there must be a $w \in V$ with $(v,w) \neq 0$ otherwise $x=0$, because $(,)$ is non-degenerate. And here I am stuck Can anybody finish this ?

 

[PeroK]

So try this. Let $x \in V$ such that $x \neq 0$ and $(x,x)=0$. The there must be a $w \in V$ with $(v,w) \neq 0$ otherwise $x=0$, because $(,)$ is non-degenerate. And here I am stuck Can anybody finish this ?

I think the trick is to start looking at things like $(x + ay, x + ay)$, for any $a \in \mathbb R$ and $y \in V$.

Try to show that $(x, x) = 0$ implies $\forall y: \ (x, y) = 0$. Which is equivalent to denegeracy.

 

[steenis]

No, one step too far for me. I do not see it ...

 

[PeroK]

If you expand $(x + ay, x + ay)$, remembering that that is always $\ge 0$, then you get an inequality that holds for all real numbers $a$. The trick is to show that is impossible. Try that step.

 

[steenis]

Expanding $(x+ay,x+ay) \geq 0$ results in $a{^2} (y,y)+2a(x,y) \geq 0$, Substituting $a=(x,y) \ne 0$ gives
$(x,y)^2 (y,y)+2(x,y)^2 \geq 0$, so $(y,y) \geq -2$. here I am stuck again

 

[PeroK]

Expanding $(x+ay,x+ay) \geq 0$ results in $a{^2} (y,y)+2a(x,y) \geq 0$, Substituting $a=(x,y) \ne 0$ gives
$(x,y)^2 (y,y)+2(x,y)^2 \geq 0$, so $(y,y) \geq -2$. here I am stuck again

What about $a = -(x, y)$?

 

[steenis]

That results in $(y,y) \geq 2$, that doesnot make it more clear

 

[PeroK]

That results in $(y,y) \geq 2$, that doesnot make it more clear

That's essentially the contradiction you were looking for. $y$ was just some vector that had non-zero "inner product" with $x$. You can scale $y$ down arbitraily.

If you go through the proof, you might see where you can tweak it to make things clearer.

Also, I think you need to get the proof formalised with all the logic and assumptions straight.

Let me show you a neat trick was to use some elementary calculus:

You had $a{^2} (y,y)+2a(x,y) \geq 0$. Now if we define a function $f(a) = a{^2} (y,y)+2a(x,y)$, we can find its minimum and show that is less than $0$. This technique is useful and can be used to prove the Cauchy-Schwartz inequality. Note that for given $x, y$, $f(a)$ is a regular real-valued function.

And, of course, the minimum value of $a$ is quite informative.

 

[steenis]

not ready

 

[PeroK]

not ready

You must be able to find the minimum of a quadratic function!

 

[steenis]

There is an extremum at $a=\frac {-(x,y)} {(y,y)}$ which has to be a minimum. The value of $f$ at $a$ is $-(x,y)^2(y,y)$ which is always smaller than 0. You can give me the last step, please

 

[PeroK]

There is an extremum at $a=\frac {-(x,y)} {(y,y)}$ which has to be a minimum. The value of $f$ at $a$ is $-(x,y)^2(y,y)$ which is always smaller than 0. You can give me the last step, please

Unless $(x, y) = 0$, that contradicts that $(x + ay, x+ay) \ge 0$.

To summarise: what you've shown is that $(x, x) = 0$ implies $\forall y \in V: \ (x, y) = 0$, hence $x$ maps to the zero functional and the bilinear map is degenerate, unless $x = 0$. In other words, degeneracy is equivalent to having a non-zero $x$ with $(x, x) = 0$. Which is what you had to prove.

 

[steenis]

Thank you very much. Do you have a reference ?

 

[PeroK]

Thank you very much. Do you have a reference ?

I haven't seen this precise proof before. But, if you look for proofs of the Cauchy-Schwartz inequality you'll find a few techniques for these sorts of problems.