# Proof of a theorem about finite dimensional normed linear spaces

Recently I was reading a proof of the following proposition,

Any two norms on a finite dimensional normed linear space are equivalent.

To prove this result the author used the following result without proof,

Result. Let $X$ be a finite dimensional normed linear space of dimension $n$. Suppose that $\mathcal{B}:=\{u^{(1)},\ldots,u^{(n)}\}$ be a basis of $X$ and $x\in X$. Furthermore, suppose that $x=\alpha_1u^{(1)}+\ldots+\alpha_nu^{(n)}$. If $\lVert\cdot\rVert$ be any norm on $X$ then prove that, $$\lVert x\rVert\ge d\left(\alpha_ju^{(j)},X_j\right)$$ for all $j\in \{1,\ldots,n\}$ where, $$X_j=\operatorname{span}\{u_i:i\ne j\}$$and $$d\left(\alpha_ju^{(j)},X_j\right)=\inf\{\lVert \alpha_ju^{(j)}-z\rVert:z\in X_j\}$$

I tried to prove this result but couldn't. Can anyone give me a proof of the above theorem?

Observe that clearly $z=-(x - \alpha_ju^{(j)})\in X_j$, and $$\|\alpha_ju^{(j)} - z\| = \| x\|$$ which by the definition of the infimum and the distance means that $\|x\|\geq d(\alpha_ju^{(j)},X_j)$ :)
• I don't understand why $z=-(x-\alpha_ju^{(j)})$. Can you elaborate it? – user 170039 Apr 14 '17 at 8:07
• This is just the definition of $z$; I'm picking this particular $z$ to show the inequality you want – amakelov Apr 14 '17 at 8:08
• I don't understand. Your equation implies that $z=-(a_1u^{(1)}+\ldots+a_{j-1}u^{(j-1)}+a_{j+1}u^{(j+1)}+\ldots+a_{n}u^{(n)})$. Why should $z$ be of this particular form? – user 170039 Apr 14 '17 at 8:15
• It is because I chose it this way. If that helps you can just forget about $z$ and think about the vector $x-\alpha_ju^{(j)}$. This vector is an element of $X_j$, which implies that the distance between this vector and $\alpha_ju^{(j)}$ is lower bounded by the distance between $\alpha_ju^{(j)}$ and the whole subspace $X_j$ – amakelov Apr 14 '17 at 8:17