How to compute distance with respect to inner product? Suppose $A$ is the invertible matrix : 
$$A=
\begin{pmatrix}
   2 & 0 & 1\\
   0 & 1 & -1\\
   1 & 0 & 1
\end{pmatrix}$$
We know that the function given by $\langle u,v\rangle = (Au)\cdot (Av) $ is an inner product on $\mathbb{R}^3$. Compute the distance between $(1,0,0)$ and $(0,1,0)$ with respect to this inner product. 
How would you go about solving this?  
 A: Notice that $Ae_i$ is the $i$-th column of $A$ so
$$\|e_1-e_2\|^2 = \|A(e_1 - e_2)\|_2^2 = \|Ae_1 - Ae_2\|_2^2 = \left\|\begin{pmatrix} 2 \\ -1 \\ 1\end{pmatrix}\right\|^2 = 2^2 + (-1)^2 + 1^2 = 6$$
and therefore $\|e_1-e_2\| = \sqrt6$.
A: If $x=(1,0,0)$ and $y=(0,1,0)$, the distance between them is given by $\| x-y\|$. And recall that, the norm is defined by $\| v\|=\sqrt{\langle v,v\rangle}$. Thus
$$\| x-y\|=\| (1,-1,0)\|=\sqrt{\langle (1,-1,0),(1,-1,0)\rangle }=\cdots$$
A: An inner product $\langle\cdot\rangle$ defines a norm by $\|x\|^2= \langle x,x\rangle$, or equivalently, $\|x\|=\sqrt{\langle x,x\rangle}$. A norm defines a distance by $d(x,y)=\|x-y\|$.
So the distance is $$\|(1,0,0)-(0,1,0)\|= \|(1,-1,0)\|= \sqrt{\langle (1,-1,0),(1,-1,0)\rangle} = \sqrt{A(1,-1,0)\cdot A(1,-1,0)}$$
Now compute the matrix product etc.
A: Hint: How would you usually calculate the distance between the two points? Can you formulate that in terms of the standard inner product? Now do the calculation with this new inner product instead of the standard one.
