# A question on inner products

• I
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
Homework Helper
Gold Member
2020 Award
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

PeroK
Homework Helper
Gold Member
2020 Award
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."

PeroK
Homework Helper
Gold Member
2020 Award
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

PeroK
Homework Helper
Gold Member
2020 Award
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 ?

PeroK
Homework Helper
Gold Member
2020 Award
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 ...

PeroK
Homework Helper
Gold Member
2020 Award
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

PeroK
Homework Helper
Gold Member
2020 Award
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

PeroK
Homework Helper
Gold Member
2020 Award
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.

PeroK
Homework Helper
Gold Member
2020 Award
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:
PeroK
Homework Helper
Gold Member
2020 Award
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 ?

PeroK