1. ## [SOLVED] Problem about an inner product

Let $\displaystyle \alpha _1=(1,2)$ and $\displaystyle \alpha _2 =(3,4)$. Let $\displaystyle \text{( , )}$ be an inner product in $\displaystyle \mathbb{C}^2$ such that :
$\displaystyle (\alpha _1, \alpha _1)=1$
$\displaystyle (\alpha _1, \alpha _2)=1+2i$
$\displaystyle (\alpha _2, \alpha _2)=\frac{1}{2}$
Calculate $\displaystyle (\alpha , \beta) \forall \alpha$ and $\displaystyle \beta \in \mathbb{C}^2$.
My attempt : I tried to figure out how is definied the inner product but I didn't find. After all I'm not even sure I should start the problem this way. If I should, then it's not an obvious thing to do... I need help on this.

2. Originally Posted by arbolis
Let $\displaystyle \alpha _1=(1,2)$ and $\displaystyle \alpha _2 =(3,4)$. Let $\displaystyle \text{( , )}$ be an inner product in $\displaystyle \mathbb{C}^2$ such that :
$\displaystyle (\alpha _1, \alpha _1)=1$
$\displaystyle (\alpha _1, \alpha _2)=1+2i$
$\displaystyle (\alpha _2, \alpha _2)=\frac{1}{2}$
Calculate $\displaystyle (\alpha , \beta) \forall \alpha$ and $\displaystyle \beta \in \mathbb{C}^2$.
My attempt : I tried to figure out how is definied the inner product but I didn't find. After all I'm not even sure I should start the problem this way. If I should, then it's not an obvious thing to do... I need help on this.
let's show the elements of $\displaystyle \mathbb{C}^2$ by $\displaystyle (x,y), \ x,y \in \mathbb{C}.$ and the inner product by $\displaystyle <\alpha \mid \beta>, \ \alpha, \beta \in \mathbb{C}^2.$ for any

$\displaystyle a, b \in \mathbb{C}$ let $\displaystyle f(x,y)=-2x+\frac{3y}{2}, \ \ g(x,y)=x-\frac{y}{2}.$ see that $\displaystyle (x,y)=f(x,y)\alpha_1 + g(x,y)\alpha_2.$ thus for any

$\displaystyle x_1,y_1,x_2,y_2 \in \mathbb{C}$ we have:

$\displaystyle <(x_1,y_1) \mid (x_2,y_2)>=<f(x_1,y_1)\alpha_1+g(x_1,y_1)\alpha_2 \mid f(x_2,y_2)\alpha_1+g(x_2,y_2)\alpha_2>$

$\displaystyle =f(x_1,y_1)\overline{f(x_2,y_2)}<\alpha_1 \mid \alpha_1>+f(x_1,y_1)\overline{g(x_2,y_2)}<\alpha_1 \mid \alpha_2> +$

$\displaystyle g(x_1,y_1)\overline{f(x_2,y_2)} \ \overline{<\alpha_1 \mid \alpha_2>}+g(x_1,y_1)\overline{g(x_2,y_2)}<\alpha_ 2 \mid \alpha_2>,$

where "overline" means the complex conjugate. you should be able to finish the solution very easily now.

I must say it's hard for me to understand what you've done in some points.
You said
for any $\displaystyle a,b \in \mathbb{C}$
but since then you don't use $\displaystyle a$ and $\displaystyle b$. (but maybe I'll have to use them to finish the problem, so it's not a big deal).

But I have no idea of
see that $\displaystyle (x,y)=f(x,y)\alpha_1 + g(x,y)\alpha_2$
. I've tried to see it writing it in my sheet but I didn't success to understand.
We know that $\displaystyle f(x,y)=-2x+\frac{3y}{2}$ and that $\displaystyle \alpha _1=(1,2)$, so how do you multiply them? I then thought I'd have to substitute $\displaystyle x$ for the first coordonate of $\displaystyle \alpha_1$ and $\displaystyle y$ for the second one... but this makes no sense.
So when I'll understand that part I'll try to go further and finish it alone.
By the way I didn't see Calculus III (multivariable calculus) but I guess it doesn't matter.

4. Today I could ask a helper at university... he told me a hint : $\displaystyle \alpha _1$ and $\displaystyle \alpha _2$ is a basis of $\displaystyle \mathbb{C}^2$.
So I could solve the problem. (Well I think I did it correctly!)
I wrote that any $\displaystyle \alpha$ can be writen in the form $\displaystyle x\alpha _1 +y \alpha _2$ and wrote something similar for $\displaystyle \beta$. I ended with a not so beautiful expression (I had to deal with properties of inner product) as answer... but they don't ask it to be nice.