Show that any two norms on a finite dimensional vector space $V$ over the set of real numbers are equivalent. I know that the question has already an answer. But, I am trying to do it in a different way:-
I am using the fact that any two norms on $\mathbb{R}^n$ are equivalent.
Let us assume that the $dim V=n$. Let $\mathbb B$ be that basis of $V$ containing $n$ independent vectors which span $V$. Let $\mathbb B=\{\alpha_1,\alpha_2,.....,\alpha_n\}$.
Now, any vector $v$ can be written as $c_1\alpha_1+c_2\alpha_2+.....c_n\alpha_n.$
Then map $v$ to $(c_1,c_2,.....,c_n)$. Do it for all vectors $v\in V$. We will get that $\mathbb R^n\cong V$. Since any two norms on $\mathbb R^n$ are equivalent we get that any two norms on $V$ are equivalent. Is the argument correct?
Thanks!
 A: Not quite correct. For you must show that a given norm on $V$, when transported to $R^n$ by your mapping, becomes a norm on $R^n$, and vice versa. 
To show that not everything transports nicely, consider taking a orthonormal basis of $V$ and transforming it to $R^n$ by a linear map $L$. Will $L(b_1), \ldots, L(b_n)$ still be orthonormal? No. Not in general. (Example: under the map 
$(x, y) \mapsto (2x, y)$, the standard basis for 2-space, which is orthonormal, maps to $(2,0), (0, 1)$, which is not. [Using the standard inner product in both cases.]
Now as it happens, if $U: V \to \mathbb R$ is a norm and $L: W \to V$ is an isomorphism, than $U \circ L$ is a norm on $W$, and this is the missing lemma that you need to complete your argument. Can you see how to prove it? 
A: I will give you an alternative proof - the answer is already accepted, but I'm posting this to help people in the future. This proof is not hard to follow if you have a fairly simple knowledge of linear algebra and metric spaces (+ equivalence relations from Algebra). 
THEOREM: Any two norms in a normed vector space of finite dimension are equivalent.
Proof: First, let's remember that two norms (let's say, $||.||_1$ and $||.||_2$) are equivalent if there exists $c_1>0$ and $c_2>0$ such that $$c_1||u||_2 \le ||u||_1 \le c_2||u||_2.$$ Now, let $V$ be a vector space of finite dimension and ${e_1, e_2,...}$ a basis for $V$. Given $v \in V$, $v = \sum_{i=1}^{n}{x_i e_i}, \space x_i \in \mathbb{R}$, in a unique manner. We define $||v||_* := \sqrt{\sum_{i=1}^{n}{x_i^2}} $ and putting the notation of $V_* = (V, ||.||_*)$ (normed vector space, therefore metric space with that norm as metric), then we define $$\phi: \mathbb{R^n} \rightarrow V_*$$ $$(x_1, x_2, ..., x_n) \mapsto \sum_{i=1}^{n}{x_i e_i} = v. $$
Let $S^{n-1}:= {x \in \mathbb{R^n}; |x| = 1}$ the sphere of radius $1$ in $\mathbb{R^n}$, which is closed and bounded, therefore compact (Heine-Borel - https://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem).
Let's then define $S = \phi(S^{n-1})$, then with $v \in S$, we have that $$\sum_{i=1}^{n}{x_i^2} = 1 \Rightarrow ||v||_* = 1.$$
Since $\phi$ is an isometry (check this!), then $\phi$ is continuous (because is Lipschitz, particularly a weak contraction, therefore continuous) and therefore $\phi(S^{n-1}) = S$ is compact in $V_*$ (continuous function maps compact set to compact set). Let $||.||$ be any norm in $V$. Then it suffices to show that $||.||$ and $||.||_*$ are equivalent (and use transitivity from the relation of equivalence).
From here on, we must prove that $||.||$ and $||.||_*$ are equivalent!
Well, first we have that $$||v|| = ||\sum_{i=1}^{n}{x_i e_i}|| = \sum_{i=1}^{n} |x_i|||e_i|| \le \left( \sum_{i=1}^{n} |x_i|^2 \right )^{1/2}\left( \sum_{i=1}^{n} ||e_i||^2 \right )^{1/2}, $$ with the last passage coming from Hölder's inequality with $p=q=2$. Remembering that we had the basis ${e_i}$ fixed, we put $\left( \sum_{i=1}^{n} ||e_i||^2 \right )^{1/2} = c_1$ and, also, $\left( \sum_{i=1}^{n} |x_i|^2 \right )^{1/2} = ||v||_*$. Then we have the first inequality, $$||v|| \le c_1||v||_*, \space \forall v \in V.$$
We now construct the function $$||.||: V_* \rightarrow \mathbb{R}$$ $$v \mapsto ||v||$$ 
and $||v|| \leq ||v||_*$ (the inequality we just found), so $||.||$ is Lipschitz.
Also, $S \subset V_*$ is compact, then with $||.||:S \rightarrow \mathbb{R}$ restricted to the compact S, the norm given by $||.||$ attains a maximum $v_0 \in S$ (Extreme Value Theorem, https://en.wikipedia.org/wiki/Extreme_value_theorem) such that $$||v_0|| \le ||v|| \space \forall v \in S $$. Well, let's then put $v_0 = c_2$, and since $v \ne 0 \in V_* \Rightarrow \frac{v}{||v||_*}$, we have the following: $$||v_0|| \le  || \frac{v}{||v||_*} || \space \forall v \ne 0, \space v \in V_*, $$ that is, $||v_0|| \le \frac{1}{||v||_*}||v|| \space \forall v\ne 0, \space v \in V_*$ or, even better, $$c_2 ||v||_* \le ||v|| \space \forall v \ne 0, v \in V_*.$$ 
What about when $v = 0$? (try checking this too, the answer is below)

When $v = 0$, the null vector, from the definition of the norm, $||v||_* = 0 = ||v||$,  
 

and then we have the second inequality. That completes the proof.
