Can a parallelogram in $\mathbb R^2$ be a square in $(\mathbb R^2, (·, ·)_a)$?

Question

If we draw a parallelogram in $\mathbb R^2$, is it possible to define an alternative inner product $(·, ·)_a$ on $\mathbb R^2$ so that the parallelogram you drew is a square in $(\mathbb R^2,(·, ·)_a)$?

I wouldn't know where to begin with this. I'm assuming it has something to do with

$cos(\theta)$=$\frac{(x,y)}{\|x\|·\|y\|}$

but for the most part, I'm clueless.

All help is appreciated.

Answer 1

Just consider that you can find a linear transformation A from one V-space into another so that the parallelogram gets mapped to a square.

Now note that: $(x,y)\mapsto (Ax, Ay)_a$ with the normal inner product defines a new inner product of our original space.

edit:typo