Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $V=\mathbb{F}$ is an inner product space over itself, is it true that $\langle1,1\rangle=1$ ?

If its true then I believe this follows from linearity, however, I was unable to use the linearity of the inner product to prove this.

share|improve this question
3  
Vector spaces don't have units... –  Grumpy Parsnip Nov 8 '12 at 14:17
    
Is $V$ one dimensional? If not how is the element $1 \in V$ defined? –  P.. Nov 8 '12 at 14:17
    
@JimConant - I will update in a second –  Belgi Nov 8 '12 at 14:17
    
@Pambos - yes, I updated the question, In my setting I view the field as a vector space over itself –  Belgi Nov 8 '12 at 14:18
1  
@coffeemath - I don't agree, by definition $\langle v,v\rangle=\overline{\langle v,v\rangle}\implies\langle v,v\rangle\in\mathbb{R}$ where $<$ is defined –  Belgi Nov 8 '12 at 14:53
show 3 more comments

1 Answer

up vote 7 down vote accepted

Not in general. The definition of the inner product is $\langle\cdot,\cdot \rangle:V\times V\rightarrow \mathbb{F}$ such that:

  1. $\langle x,y \rangle=\langle y,x \rangle^*$,

  2. $\langle a x,y \rangle=a\langle x,y \rangle$,

  3. $\langle x+y,z \rangle=\langle x,z \rangle+\langle y,z \rangle$,

  4. and $\langle x,x \rangle\ge 0$.

From these you cannot deduce $\langle 1,1 \rangle=1$, because you can define $\langle x,y \rangle=a x y^*$ with $a>0$ and this is an inner product. Actually in finite dimensional vector spaces inner products are commonly defined by a positive definite matrix which doesn't have to be the identity.

EDIT: Example $V=\mathbb{R}^n$. Let $A$ be a positive definite matrix then we may define $$\langle \cdot,\cdot \rangle :V\times V\rightarrow \mathbb{R}, (x,y) \mapsto y^TAx$$

and because $A$ is positive definite we have $x^TAx>0$, $\forall x\in V\backslash\{0\}$, so $\langle \cdot,\cdot \rangle$ is an inner product in $\mathbb{R}^n$.

EDIT2: Above I implicitly assumed that $A$ is symmetric, if this is not the case we may define $\langle x,y\rangle=y^TA_sx$ with $A_s=\frac{1}{2}\left(A+A^T\right)$.

share|improve this answer
4  
< and > are symbols for the relation "greater" and "smaller" you should use \langle and \rangle in $\LaTeX$. –  Asaf Karagila Nov 8 '12 at 14:33
    
thanks for the tip –  tst Nov 8 '12 at 14:39
    
tst: I guess a positive definite 1X1 matrix in this case is a matrix $[a]$ where $a>0$. If so, does the definition via positive matrices turn out to be your example $[x,y]=axy$? [not doubting your example, just not knowing the matrix inner product definition.] –  coffeemath Nov 8 '12 at 14:47
    
coffemath: Yes, it is exactly as you wrote it. I edited my answer to make it more clear. I hope this helps. –  tst Nov 8 '12 at 14:56
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.