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Let $X$ be a set, $V$ a vector space over a field $K$ and $F(X,V)$ the $K$-Vector space of all linear transformations from $X$ to $V$.

$F_{y,0}=\{f \in F(X, V ) \mid f (y) = 0\}$; $0F$ is the zero element.

Let $W$ be a linear subspace of $F(X,V)$.

1.$F(X, V ) = F_{y,0} + W,$

2.$W \cap F_{y,0} = 0F $

My approach was to say that W would be the set $W = F(X, V )\ F_{y,0}$.

However I think my approach is way to easy and to be honest i wouldn't know how to start to prove my answer, can anybody help?

Sigma Algebra
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Esko Mühlemann
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  • Surely $0_F\in F_{y,0}$ for every $y$, so I don't see how $F(X,V)\setminus {F_{y,0}}$ can be a linear subspace (it doesn't contain zero). – MPW Oct 17 '24 at 19:26
  • In your description $W$ is a direct addendum of the vector subspace $F_{y,0}$. If you assume $X$ to be a finite set and $V$ to be finite dimensional then $F(X,V)$ is a finite dimensional vector space and $W$ can be constructed by taking a basis of $F_{y,0}$ and by completing it to a basis of $F(X,V)$.\ In the infinite-dimensional case the same holds but you need the axiom of choice, see for example https://math.stackexchange.com/q/86762/1405738. – AleK3 Oct 17 '24 at 19:36
  • @AlessandroFrassineti thanks for the Link! So if I understand correctly I would have to expand the basis of Fy,0 to the basis of F(X,V)? Thus meaining that Fy,0 ∈ W. – Esko Mühlemann Oct 18 '24 at 05:28
  • No, then you take as a basis of $W$ only the vectors you added. In this way you get $W\cap F_{y,0} = \left\lbrace 0 \right\rbrace$ and $F(X,V) = W \oplus F_{y,0}$. – AleK3 Oct 18 '24 at 07:35

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