Is the norm of a basis element of $\mathbb{R}^n$ always equal to $1$?

Question

Is it possible to have a norm $\Vert \cdot \Vert$ such that $$\Vert e_k \Vert \neq 1$$

where $e_k$, $k = 1, \dots, n$ is one of the standard basis elements of $\mathbb{R}^n$?

Answer 1

Sure, define an inner product on $\mathbb{R}^n$ as follows:

$$\langle a_1e_1 + ... + a_ne_n , b_1e_1 + ... + b_ne_n \rangle = 2a_1b_1 + ... + 2a_nb_n $$

This is positive definite, linear in both parts, symmetric.

Then $||v|| = \sqrt{\langle v, v\rangle}$, and $||e_i|| = \sqrt{2}$

And as the comments implied, there is nothing particularly special about $\sqrt{2}$ in this case, this will work if you replace 2 by any positive real number in the definition of the inner product. Additionally, if instead of defining the inner product with respect to linear combinations of the standard basis, you defined it similarly but with respect to linear combinations of some other basis, you are likely to get that the norm of standard basis vectors are very different from $1$, and potentially not all equal.