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Let $V$ be an $n$-dimensional $\mathbb{K}$-vector space with basis $\{e_i\}_{i=1}^n$, and consider the map $\phi:V\to \mathbb{K}$ given by $e_i\mapsto 1$. Let $K=\ker\phi$ and set $N$ to be the subspace of vectors whose coefficients are all equal. I'd like to show that $V=K\oplus N$ (provided $char(\mathbb{K})\nmid n$).

So far I've shown that $N\cap K=0$, but how can I write a general $v\in V$ as a sum of elements in $K$ and $N$?

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Is $\phi$ also assumed to be linear? – GEdgar Sep 19 '12 at 20:21
@GEdger: Yes! I forgot to include that – Bey Sep 20 '12 at 3:14
up vote 2 down vote accepted

Let $v=a_1e_1+\cdots+a_ne_n$, so $\phi(v)=a_1+\cdots+a_n$. Let $x=\frac{a_1+\cdots+a_n}{n}$, $u=(a_1-x)e_1+\cdots+(a_n-x)e_n$ and $w=xe_1+\cdots+xe_n$. Then $v=u+w$ and $u\in K$ while $w\in N$.

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Thanks! There's a small typo in your $w$; the first term should be $xe_1$, not $x_e1$ :) – Bey Sep 19 '12 at 19:57
@Bey Thanks for pointing that out. – Alex Becker Sep 19 '12 at 20:15
Alternatively, one may show $N\cap K = \{0\}$ (NOT $=\emptyset$ as is written in the original question) and determine the dimensions of $K$ and $N$. – Oliver Braun Sep 19 '12 at 20:30
Oh whoops. Thanks for that – Bey Sep 20 '12 at 3:10

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