Existence of a vector whose norm is 1 in a Banach space

Question

Given a finite dimensional Banach space V over reals, I have to show that there exists $v \in V$ such that $\|v\|=1$

At first, I thought that there's an identity element I in V.

And $\|I\|=1$.

But then I think it is not necesarily true all the time.

So now I'm thinking about constructing a Cauchy sequence converging to v while its norm converges to 1.

Or can I just say that we can arbitrarily choose such vector by manipulating its basis?

Thank you.

Answer 1

if $x \in V$ is non-zero then $||x|| \gt 0$, so set: $$ x' = \frac{x}{||x||} $$ it is now evident that $||x'|| = 1$