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Given $v1,v2,\ldots,vk,u,w$ in a vectorial space $V$, it is said that $x_1v_1+...+x_kv_k=u$ has a single solution and that $x_1v_1+...+x_kv_k=w$ has no solution.

How do you find the dimension of $\mathrm{Sp}(v_1,...,v_k,w)$?

From what I understand the latter says that $w$ is independent of $v_1,...,v_k$ so it can span, but I don't understand how to prove it and what is the connection of vector $u$.

Thank you very much for your help.

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  • $\begingroup$ I think you can first consider $W = Span(v_1,\dots,v_k)$ and say that $u \in W$. You can also say that ${v_1,\dots,v_k}$ form a basis of $W$ (that's because if they are a basis: $$\forall w\in W \exists! a_o,\dots,a_k |w=a_ov_1+\dots+a_kv_k.$$ Now you can say that $W$ is a $k$-dimensional space. Let's try to add $w$. It doesn't fit in $W$, so the dimension must grow up by one. $\endgroup$ – Alberto Andrenucci Jun 18 '17 at 13:09
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Saying that for some vector $u$, the linear system $x_1v_1+\dots x_k x_k=u$ has a unique solution means: 1) $u$ lies in the span $\langle v_1, \dots, v_k\rangle$ (existence of a solution) and 2) the vectors are linearly independent (uniqueness). Therefore they generate a subspace of dimension $k$.

On the other hand, saying that the linear system $x_1v_1+\dots x_k x_k=w$ has no solution means, as you noticed, $w$ is independent of $v_1,\dots, v_k$, so that the system of vectors $v_1,\dots, v_k,w$ is linearly independent, and thus generates a subspace of dimension $k+1$.

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  • $\begingroup$ thank you very much for your help $\endgroup$ – linearproblems Jun 18 '17 at 19:39

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