The strategy is to show that every norm is equivalent to the $\ell^1_n$ norm by a compactness argument on the unit sphere, and then use transitivity of equivalence. **Step 1: Fix a basis and a reference norm.** Let $\{e_1, \ldots, e_n\}$ be a basis for $V$ and define the $\ell^1_n$ norm by $\|v\|_{\ell^1_n} := \sum_{i=1}^n |v_i|$ for $v = \sum_{i=1}^n v_i e_i$. We show that any norm $\|\cdot\|$ on $V$ is equivalent to $\|\cdot\|_{\ell^1_n}$. **Step 2: Upper bound.** For any $v = \sum_{i=1}^n v_i e_i$, the triangle inequality and homogeneity give: \begin{align*} \|v\| = \left\|\sum_{i=1}^n v_i e_i\right\| \le \sum_{i=1}^n |v_i| \cdot \|e_i\| \le \left(\max_{1 \le i \le n} \|e_i\|\right) \sum_{i=1}^n |v_i| = C \|v\|_{\ell^1_n} \end{align*} where $C = \max_i \|e_i\| > 0$. **Step 3: Lower bound via compactness.** Consider the unit sphere $S_1 := \{v \in V : \|v\|_{\ell^1_n} = 1\}$ and the [function](/page/Function) $f: S_1 \to \mathbb{R}$ defined by $f(v) = \|v\|$. By Step 2, $|f(v) - f(w)| \le \|v - w\| \le C\|v - w\|_{\ell^1_n}$, so $f$ is Lipschitz [continuous](/page/Continuity) on $(S_1, \|\cdot\|_{\ell^1_n})$. [claim:Compactness of $S_1$] The unit sphere $S_1$ (and the closed unit ball $\overline{B}_1(0)$) in $\|\cdot\|_{\ell^1_n}$ is compact. [/claim] [proof] Let $(v^{(k)})_{k=1}^\infty \subset \overline{B}_1(0)$ be a [sequence](/page/Sequence). Write $v^{(k)} = \sum_{i=1}^n \lambda_i^{(k)} e_i$. Since $\sum_{i=1}^n |\lambda_i^{(k)}| \le 1$, each component sequence $(\lambda_i^{(k)})_k$ is bounded by $1$. By Bolzano-Weierstrass, extract a convergent subsequence for $\lambda_1$; from that subsequence extract a further convergent subsequence for $\lambda_2$; repeat $n$ times. The final subsequence converges in all components simultaneously, hence converges in $\|\cdot\|_{\ell^1_n}$. The [limit](/page/Limit) lies in $\overline{B}_1(0)$ since the ball is closed. Sequential compactness equals compactness in [metric spaces](/page/Metric%20Space). [/proof] Since $f$ is continuous on the compact [set](/page/Set) $S_1$, it attains its infimum: there exists $v^* \in S_1$ with $f(v^*) = \inf_{v \in S_1} \|v\|$. Since $v^* \ne 0$ (as $\|v^*\|_{\ell^1_n} = 1$), we have $A := f(v^*) = \|v^*\| > 0$. Therefore $\|v\| \ge A$ for all $v \in S_1$. For arbitrary $v \in V \setminus \{0\}$, the vector $v / \|v\|_{\ell^1_n}$ lies in $S_1$, giving $\|v\| / \|v\|_{\ell^1_n} \ge A$, i.e. $\|v\| \ge A \|v\|_{\ell^1_n}$. **Step 4: Combine.** We have $A\|v\|_{\ell^1_n} \le \|v\| \le C\|v\|_{\ell^1_n}$ for all $v \in V$. Applying this to any two norms $\|\cdot\|$ and $\|\cdot\|'$ (both equivalent to $\|\cdot\|_{\ell^1_n}$) and using transitivity of Lipschitz equivalence gives the result.