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Let $V = \mathbb{R}^4$. Consider the subspace

$$U = \{(a_1,a_2,a_3,a_4) \in \mathbb{R}^4 | a_1 +a_2 +a_3 = 0\} \;of\; V$$

Consider the elements $u_1 =(0,0,0,1)$ and $u_2 =(5,−2,−3,0)$ of $U$. Find another element $u_3 \in U$ such that $\{u1,u2,u3\}$ is a basis of $U$, and prove that it is indeed a basis.

I know the proof of a basis is that the elements must be linearly independent and spans the entire vector space, but how do you do this?

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  • $\begingroup$ Step one: Show that $U$ is three dimensional. Step two: find three vectors in $U$ such that they are linearly independent. Conclude that those three vectors form a basis for $U$. There are infinitely many correct answers here. Literally pick any other element of $U$ so that the three are linearly independent. $\endgroup$ – JMoravitz Apr 15 '16 at 17:38
  • $\begingroup$ Do I not need to do prove that the three elements make a span (as that is the main thing I am unsure of)? $\endgroup$ – HELP Apr 15 '16 at 17:44
  • $\begingroup$ Fact: If you have an $n$-dimensional space $V$, and you have a collection of $n$ linearly independent vectors in that space $v_1,v_2,\dots,v_n$, then $v_1,v_2,\dots,v_n$ not only span the space $V$, but act as a basis for it. $\endgroup$ – JMoravitz Apr 15 '16 at 17:46
  • $\begingroup$ interesting, thank you. $\endgroup$ – HELP Apr 15 '16 at 17:48
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$u_3 = (1,0,-1,0)$ would work. $u_3$ is in $U$, and it is independent from $u_1$ and $u_2$.

And to prove that the vectors span U in a hand-wavy way.
The dimension U is less than 4, since U is a subspace in $R^4$, and clearly there are vectors in $R^4$ that are not in U, and we have 3 independent vectors in U, the dimension of U must be 3 and these vectors span U.

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  • $\begingroup$ Why do we prove the span in a "hand-wavy way", is there no general proof to show something is a span? $\endgroup$ – HELP Apr 15 '16 at 17:51
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Finding $u_3$ shouldn't be a problem ( cf. JMoravitz's comment/ Doug M's example). Trying to dispel the "hand-wavy way" for proof of span:

Suppose $u_3 = (a_1, a_2, a_3, a_4)$. An arbitrary vector spanned by the above basis is a linear combination $\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3$ which results in coordinates: $(5\alpha_2 + \alpha_3 a_1, -2\alpha_2 + \alpha_3 a_2, -3\alpha_2 + \alpha_3 a_3, \alpha_1a_1 + \alpha_3a_4)$. Applying the defining condition: $5\alpha_2 + \alpha_3 a_1 -2\alpha_2 + \alpha_3 a_2 -3\alpha_2 + \alpha_3 a_3 = \alpha_3(a_1 + a_2 + a_3) = 0$. For $\alpha_3 \neq 0$ this implies: $a_1 + a_2 + a_3 = 0$. But that condition is already satisfied as $u_1 + u_2 + u_3 $ belongs to the span $\langle u_1, u_2, u_3\rangle$.

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