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Prove or disprove the following statement:

"If $(v_1,v_2,v_3,v_4)$ is a basis for the vector space ${\Bbb R}^4$ and $\textbf{W}$ is a subspace of ${\Bbb R}^4$ then some subset of $(v_1,v_2,v_3,v_4)$ forms a basis for $\textbf{W}$."

I have an intuition for this and I think it is false because it's saying that I can write $(v_1,v_2,v_3,v_4)$ as a linear combination of each other since the vectors form a basis. However, just because $\textbf{W}$ is in he subspace of ${\Bbb R}^4$ does not mean I can write is with linear combinations of $(v_1,v_2,v_3,v_4)$. There is no guarantee that I can completely span $\textbf{W}$ with $(v_1,v_2,v_3,v_4)$ right?

Is that even the right approach or is my thinking completely off? How would I actually do this mathematically?

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  • $\begingroup$ Your intution has not come out in the words you have written. Rewrite the second sentence starting with "However, just because...". $\endgroup$ – P Vanchinathan Feb 3 '18 at 7:40
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Your idea is right. A concrete counter example would be good.

Let $v_i = e_i$, the $i$-th standard unit vector.

Let $W = \operatorname{span}\{ v_1 + v_2\}$, then none of the $\{e_i\}$ is a basis for $W$.

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  • $\begingroup$ I'm just going to add step to be very concrete about this. So Essentially, you're saying that let $v_i=e_i= \begin{bmatrix} \vdots \\ \vdots \\ 0 \\ 1 \\ 0 \\ \vdots \\ \vdots \\ \end{bmatrix}$ and , $v_1=e_1= \begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ \vdots \\ \end{bmatrix}$ $v_2=e_2= \begin{bmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ \vdots \\ \end{bmatrix}$. However, if I add $e_1$ and $e_2$ it's impossible for me to get $e_i$ since I can't write $e_i$ as a linear combination of $e_1$ and $e_2$. Is this correct? $\endgroup$ – Future Math person Feb 3 '18 at 7:46
  • $\begingroup$ Clearly the dimension of $\operatorname{span} \{ v_1+v_2\}$ is one. hence if the subset of $\{ v_1, \ldots, v_4\}$ is a basis, it has to be of the form of $\{ e_i \}$, however, we can easily check that none of them can be a basis. since after all, none of them is inside $W$. $\endgroup$ – Siong Thye Goh Feb 3 '18 at 7:54
  • $\begingroup$ I understand this now. Thank you so much! $\endgroup$ – Future Math person Feb 3 '18 at 8:20
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Actually, since the set $\{v_1,v_2,v_3,v_4\}$ has $2^4=16$ subsets, there are only $16$ subspaces of $\mathbb{R}^4$ which have a basis which is a subset of $\{v_1,v_2,v_3,v_4\}$. Since $\mathbb{R}^4$ has infinitely many subspaces…

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