# Finite-dimensional subspaces of normed vector spaces are direct summands

Here is a problem in functional analysis from Folland’s book:

If $\mathcal{M}$ is a finite-dimensional subspace of a normed vector space $\mathcal{X}$, then there is a closed subspace $\mathcal{N}$ such that $\mathcal{M}\cap \mathcal{N} = 0$ and $\mathcal{M}+\mathcal{N} = \mathcal{X}$.

I tried the following approach:
I am trying to define a projection map $\pi_{\mathcal{M}}$ from $\mathcal{X}$ to $\mathcal{M}$, which would be continuous and hence taking the inverse of any closed set would give a closed set in $\mathcal{X}$. I am confused about what the projection map would be. Please suggest some approach.

#### Solutions Collecting From Web of "Finite-dimensional subspaces of normed vector spaces are direct summands"

Let $\{e_1, …, e_n\}$ be a basis for $\mathcal M$. Every $x \in \mathcal M$ has then a unique representation $x = \alpha_1(x)e_1 +…+ \alpha_n(x)e_n$.
Each $\alpha_i$; is a continuous linear functional on $\mathcal M$ (a linear map from finite dimensional space is always continuous) which extends to a member of $\mathcal X^*$, by the Hahn-Banach theorem ($\mathcal X^*$ is the dual of $\mathcal X$). Let $\mathcal N$ be the intersection of the null spaces of these exten­ sions. Then $\mathcal X = \mathcal M\oplus \mathcal N$.