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

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• steenis
In summary: 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...In summary, this conversation shows that the inner product ##(,)## is symmetric, bilinear, and non-degenerate, has the property that if ##(x,x) = 0## then ##x = 0## for ##x \in V##, and can be translated into mathematical form.
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 ?

steenis said:
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?

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

steenis said:
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.

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."

steenis said:
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.

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

steenis said:
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.

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 ?

steenis said:
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.

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

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.

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

steenis said:
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)##?

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

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

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

Last edited:
steenis said:
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.

Thank you very much. Do you have a reference ?

steenis said:
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.

## 1. How can I prove that (x,x) = 0 implies x = 0 in a real vector space V?

The proof for this statement involves using the properties of a real vector space, specifically the property of the zero vector. Let's assume that (x,x) = 0. This means that the inner product of x with itself is equal to 0. Since the inner product is a real number, it must be greater than or equal to 0. Therefore, the only way for the inner product to be equal to 0 is if x is the zero vector.

## 2. Is this statement true for all real vector spaces?

Yes, this statement is true for all real vector spaces. The proof is based on the properties of a real vector space, which are universal for all vector spaces. Therefore, this statement holds true for any real vector space V.

## 3. Can this statement be extended to complex vector spaces?

No, this statement only applies to real vector spaces. In complex vector spaces, the inner product can be equal to 0 even if the vector is not the zero vector. This is because the inner product in complex vector spaces involves complex conjugation, which can result in a non-zero inner product for non-zero vectors.

## 4. What is the significance of this statement in linear algebra?

This statement is significant because it shows the relationship between the inner product and the zero vector in a real vector space. It also highlights the importance of understanding the properties of a vector space in order to make conclusions about its vectors. This statement can also be used in various proofs and applications in linear algebra.

## 5. Can you provide an example to illustrate this statement?

Sure, let's consider the real vector space V = {(x,y) | x, y ∈ ℝ}. In this space, the inner product of two vectors (x,y) and (a,b) is defined as (x,y) · (a,b) = xa + yb. Now, let's take the vector (1,0). The inner product of (1,0) with itself is (1,0) · (1,0) = 1·1 + 0·0 = 1. Since the inner product is not equal to 0, we can conclude that (1,0) is not the zero vector. Therefore, this example shows that (x,x) = 0 does not imply x = 0 in this real vector space V.

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