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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.

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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
@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
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 $A_s=\frac{1}{2}\left(A+A^T\right)$.

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< 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

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