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Let $V$ be a finite dimensional vector space over $\Bbb R$.

Let $A\subset V$ with the property that whenever $c_i$ is a sequence of scalars where $\sum_{i=1}^n c_i=1$ and $\{v_1,v_2,\dots ,v_n\}$ is a set of vectors in $A$ then $\sum _{i=1}^n c_iv_i\in A$.

Show that $A=x_0+W$ for some $x_0\in V$ and some subspace $W$ of $V$.

Since $\dim V<\infty$ so $A$ has a basis say $\{b_1,b_2,\dots ,b_n\}$ then any $a(\in A)=\sum c_ib_i\implies a-\sum c_ib_i=0.$

I could not proceed further.Getting no idea how to choose $x_0,V$ .Some hints needed.

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    Why downvote?Care to explain – 2017-01-19
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    I didn't downvote. Note that $A$ is not necessarily a linear subspace, so you can't start by taking a basis. – 2017-01-19

1 Answers 1

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Show that if $x_0 \in A$, then $S= A - \{x_0\}$ is a subspace.

Try before looking:

Suppose $v \in S$, then $v=a-x_0$ for some $a \in A$. Then $\lambda(a-x_0)+x_0 \in A$ (scalars sum to one) hence $\lambda(a-x_0) = \lambda v\in S$. Similarly, if $v_k = a_k -x_0 \in S$ for $k=1,2$, then $a_1-x_0+a_2-x_0 + x_0 \in A$, hence $a_1-x_0+a_2-x_0 = v_1+v_2 \in S$. Hence $S$ is a subspace and $A = S+ \{x_0\}$.

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    Nice answer!I have solved it without looking at the hidden region,thank you – 2017-01-19