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In a textbook is the question:

Find a basis for the subspace $$V=\{(x_1,x_2,x_3,x_4):x_1+2x_2+x_3+x_4=0, 3x_1+6x_2+4x_3+x_4=0\}.$$

They say that $V$ is a subspace of $\mathbb{R^4}$ and are able to find a spanning set by solving the system of homogeneous equations getting: $$\{\begin{bmatrix}-2 \\ 1 \\ 0 \\ 0 \end{bmatrix},\begin{bmatrix}-3 \\ 0 \\ 2 \\ 1 \end{bmatrix}\}.$$ They then show that these vectors are linearly independent and so it forms a basis for $V$.

I thought though that if you're in $\mathbb{R^4}$ then you need at least 4 vectors to span the space, and 4 to form a basis. Here though we only have two. Could someone please tell me where my logic went wrong and why these indeed span the subspace and form a basis?

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  • $\begingroup$ They ask you to find a basis for a subspace, not for the entire ${\bf R}^4$. $\endgroup$ – avs Aug 11 '16 at 23:18
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You do indeed need at least 4 vectors to span $\mathbb{R}^4$. However, here we're not trying to span $\mathbb{R}^4$, we're trying to span $V$.

For a more concrete example: think about the subspace $W=\{(a, b): a=0\}$ of $\mathbb{R}^2$ (basically, $W$ is the $y$-axis). Clearly $W$ is spanned by a one-element set - e.g., $\{(0, 1)\}$ - even though it takes two vectors to span $\mathbb{R}^2$. Does this help?

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  • $\begingroup$ I think I understand now, thanks :) $\endgroup$ – user342661 Aug 12 '16 at 3:00
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Noah's answer is definitely sufficient, but I'm going to try to add a bit of algebraic insight for you. Write $v=(1,2,1,1)$ and $w=(3,6,4,1)$. Let $f,g: \mathbb{R}^4 \rightarrow \mathbb{R}$, $f:x \mapsto v \cdot x$ and $g: x \mapsto w \cdot x$.

$f,g$ are linear maps and $\mathbb{R}$ is a field, so they are either surjective or trivial. Note that we have $V= \ker(f) \cap \ker(g)$. If $\dim(V)=4$, then $\mathbb{R}^4/V$ is trivial and both $f,g$ must be the zero map. Hopefully you can see this is clearly not the case.

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