Your very first step is incorrect. is NOT equal to .
Show that any map which preserves the inner product of , i.e., for which for all , is linear:
for all , and
Hint: Show that the vector has vanishing (squared) length
So first of course I used the hint, and the length squared is:
(because )
Now I want that to be zero, but I can't see how...
Hint, compare:
$\langle f(\alpha\mathbf{x} + \beta\mathbf{y}) - \alpha f(\mathbf{x}) - \beta f(\mathbf{y}), f(\alpha\mathbf{x} + \beta\mathbf{y}) - \alpha f(\mathbf{x}) - \beta f(\mathbf{y})\rangle$
with:
$\langle \alpha\mathbf{x} + \beta\mathbf{y} - \alpha\mathbf{x} - \beta\mathbf{y}, \alpha\mathbf{x} + \beta\mathbf{y} - \alpha\mathbf{x} - \beta\mathbf{y}\rangle = \langle \mathbf{0}, \mathbf{0}\rangle = 0$
using the fact that $\langle f(\mathbf{u}),f(\mathbf{v})\rangle = \langle \mathbf{u},\mathbf{v}\rangle$ for ANY two vectors $\mathbf{u},\mathbf{v}$.
You'll need to expand the first inner product using bilinearity, you should wind up with SIX terms.
No, it doesn't "all cancel out".
Continuing from what you have, we have:
$\langle f(\alpha\mathbf{x}+\beta\mathbf{y}),f(\alpha \mathbf{x}+\beta\mathbf{y})\rangle - \alpha\langle f(\alpha\mathbf{x}+\beta\mathbf{y}),f(\mathbf{x}) \rangle - \beta\langle f(\alpha\mathbf{x}+\beta\mathbf{y}),f(\mathbf{y}) \rangle
-\alpha\langle f(\mathbf{x}),f(\alpha\mathbf{x}+\beta\mathbf{y}) \rangle + \alpha^2\langle f(\mathbf{x}),f(\mathbf{x}) \rangle $
$+ \alpha\beta \langle f(\mathbf{x}),f(\mathbf{y}) \rangle - \beta \langle f(\mathbf{y}), f(\alpha\mathbf{x}+\beta\mathbf{y}) \rangle + \alpha\beta\langle f(\mathbf{y}),f(\mathbf{x}) \rangle + \beta^2\langle f(\mathbf{y}),f(\mathbf{y}) \rangle$
Which upon combining terms, and using the symmetry of the inner product becomes:
$ = \langle f(\alpha\mathbf{x}+\beta\mathbf{y}),f(\alpha \mathbf{x}+\beta\mathbf{y})\rangle + \alpha^2\langle f(\mathbf{x}),f(\mathbf{x}) \rangle + \beta^2\langle f(\mathbf{y}),f(\mathbf{y}) \rangle$
$- 2\alpha\langle f(\alpha\mathbf{x}+\beta\mathbf{y}),f(\mathbf{x}) \rangle - 2\beta\langle f(\alpha\mathbf{x}+\beta\mathbf{y}),f(\mathbf{y}) \rangle + 2\alpha\beta \langle f(\mathbf{x}),f(\mathbf{y}) \rangle$
These are the six terms I was referring to. Note how I have all the "f's" inside the inner product brackets. By the inner-product preserving property, we can now get rid of these, so this is equal to:
$ = \langle \alpha\mathbf{x}+\beta\mathbf{y},\alpha \mathbf{x}+\beta\mathbf{y}\rangle + \alpha^2\langle \mathbf{x},\mathbf{x} \rangle + \beta^2\langle \mathbf{y},\mathbf{y} \rangle$
$- 2\alpha\langle \alpha\mathbf{x}+\beta\mathbf{y},\mathbf{x} \rangle - 2\beta\langle \alpha\mathbf{x}+\beta\mathbf{y},\mathbf{y} \rangle + 2\alpha\beta \langle \mathbf{x},\mathbf{y} \rangle$
$= \langle \alpha\mathbf{x} + \beta\mathbf{y} - \alpha\mathbf{x} - \beta\mathbf{y}, \alpha\mathbf{x} + \beta\mathbf{y} - \alpha\mathbf{x} - \beta\mathbf{y} \rangle$
(here I just did the reverse of the expansion we did first)
$=\langle \mathbf{0},\mathbf{0} \rangle = 0$.
Since the square of the length of $f(\alpha\mathbf{x}+\beta\mathbf{y}) - \alpha f(\mathbf{x}) - \beta f(\mathbf{y})$ is zero, we conclude from the positive-definiteness of the inner product, that:
$f(\alpha\mathbf{x}+\beta\mathbf{y}) - \alpha f(\mathbf{x}) - \beta f(\mathbf{y}) = \mathbf{0}$, that is:
$f(\alpha\mathbf{x}+\beta\mathbf{y}) = \alpha f(\mathbf{x}) + \beta f(\mathbf{y})$, in other words: $f$ is linear.
Yea that's what I meant when I said they cancel out.
I'm used to using the 'dot product' instead of the inner product (but I guess they're the same thing). They never taught me the inner product (at university), but they expect me to just know it, or they just said it's like the dot product.
I expanded everything which gave me a long list of terms which just cancelled out,
ie. for the first term:
and so on with all the terms.
I think your way is better though, I should stop using this dot product since the question uses inner product. Thanks for the great explanation!
The "dot product" and an "inner product" are not quite the same thing. "Inner product" is the more general concept, and is characterized by these properties:
Given a vector space $V$ over a field $F$, where $F$ is a quadratically closed subfield of the complex numbers (typically either $\Bbb R$ or $\Bbb C$) we say a function:
$B:V \times V \to F$ is an inner product if:
1. $B(x,y) = \overline{B(y,x)}$ This is called conjugate-symmetry, and if $F \subseteq \Bbb R$, just symmetry, as then $\overline{z} = z$.
2. $B(-,y): V \to F$ for every $y \in V$ is linear (some presentations change this to linearity in the second argument).
Such a function is said to be sesquilinear, which in the real case ($F \subseteq \Bbb R$), means "bilinear".
3. $B(x,x) > 0$ and $B(x,x) = 0 \iff x = 0$ (this is called "positive-definiteness", and requires that $B(x,x)$ take its values in an ordered subfield of $F$).
The "dot-product" or "Euclidean inner product" is a particular KIND of inner product (one of many possible) on $\Bbb R^n$, given by:
$\displaystyle (x_1,x_2,\dots,x_n)\cdot (y_1,y_2,\dots,y_n) = \sum_{i = 1}^n x_iy_i$
It is relatively easy to check that the dot product satisfies conditions 1-3.
In the real case, it is common to associate a bilinear form on $\Bbb R^n$ with an $n \times n$ matrix $A$, in which case the associated bilinear function is:
$B(x,y) = x^TAy$
We say the bilinear form $B$ is symmetric if the matrix $A$ is, and likewise with positive-definite. The "usual dot product" is thus the bilinear form associated with the identity matrix.
Similar considerations hold in the complex case, but we talk of a sesquilinear form, and use the complex-conjugate transpose (or hermetian) instead of the transpose.