Inner products on $V$ for which $(e_1, e_1) = 1$ and $(e_2, e_2) = 1$. I've got the following question:
Let $V$ be a $2-$dimensional real vector space with basis $\{e_1, e_2\}$. Describe all the inner
products $(−, −)$ on $V$ for which $(e_1, e_1) = 1$ and $(e_2, e_2) = 1$.
I don't really know where to go with this.
 A: An inner product is positive-definite. So $(x,x)>0$ for all nonzero $x$. Writing $\mu=(e_1,e_2)$, a place to start is to figure out what values of $\mu$ are consistent with positive-definiteness. We could write out $x=ae_1+be_2$ and compute $(x,x)$, however we know that $(\lambda x,\lambda x)=\lambda^2(x,x)$ so we can normalize $x$ however we see fit. Moreover, as we already know $(e_1,e_1)>0$ and similarly for any multiple of $e_1$, we may assume the $e_2$ component is nonzero, i.e. $b\ne 0$, so we may in fact normalize to ensure $b=1$, i.e. write $x=ae_1+e_2$. Write out the condition $(x,x)>0$ in this case, and by completing the square figure out what $\mu$ must be.
When you arrive at a range of $\mu$ values, see that they all induce inner products. (Hint:  if $A$ is any invertible map and $\langle -,-\rangle$ is an inner product, then so is $\langle A-,A-\rangle$. Show that your $(-,-)$ is in fact $\langle A-,A-\rangle$ where $A$ is a map that fixes $e_1$ but sends $e_2$ to $x$ and $\langle-,-\rangle$ is standard.)
A: To describe an inner product, you have to define $(u,v)$ for any $u,v \in V$. Since $V$ has the bases $e_1, \, e_2$, this means you must be able to say what 
$$
(\alpha e_1 + \beta e_2, \gamma e_1 + \delta e_2)
$$
is for any $\alpha, \beta, \gamma, \delta$, in terms of these coefficients. By inner product properties, this becomes
$$
\alpha \gamma (e_1,e_1) + (\alpha \delta + \beta \gamma) (e_1,e_2) + \beta \delta (e_2,e_2)
$$
and you already know what the first and the last term are. This means that you must find all possible values of $(e_1,e_2)$ that such an inner product could have. 
The Cauchy-Schwarz-Bunyakovsky inequality tells you what these possible values are.
