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I am given that $\mathbf{v}_1 = \begin{pmatrix}1\\ 3 \\ 1 \\ 1\\ \end{pmatrix}$ and $\mathbf{v}_2 = \begin{pmatrix}1\\ -1 \\ 6 \\ 2\\ \end{pmatrix}$ in $\mathbb{R^4}$. How would I describe $\mathrm{Span}\{\mathbf{v}_1,\mathbf{v}_2\}$? I know that $[\mathbf{v}_1\; \mathbf{v}_2]$ row reduces to $\begin{pmatrix}1 & 0\\ 0 & 1\\ 0 & 0\\ 0 & 0 \end{pmatrix}$, but is there a "nontrivial" way to actually describe $\mathrm{Span}\{\mathbf{v}_1,\mathbf{v}_2\}$ beyond saying that it is all of the linear combinations of $\mathbf{v}_1$ and $\mathbf{v}_2$? For example, is there a geometric description of $\mathrm{Span}\{\mathbf{v}_1,\mathbf{v}_2\}$? I know you can usually describe it as a plane in $\mathbb{R^3}$, but I am unsure of how to give a meaningful description in $\mathbb{R^4}$.

Edit: (is this right?) $\mathrm{Span}\{\mathbf{v}_1,\mathbf{v}_2\}$ is a plane in $\mathbb{R^4}$ through the origin, because neither vector in this problem is a multiple of the other.

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  • $\begingroup$ $\mathbb{R^2}$ is to $\mathbb{R^4}$ as $\mathbb{R^1}$ is to $\mathbb{R^3}$. $\endgroup$ – JonMark Perry Feb 5 '15 at 5:48
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It is still a plane. Two vectors can only form a plane (possibly degenerate).

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  • $\begingroup$ What do you mean by degenerate? $\endgroup$ – fancynancy Feb 5 '15 at 5:28
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    $\begingroup$ degenerate means that both vectors are proportional to each other, in which case the "plane" they span is only a line. Such line is called a degenerate plane. $\endgroup$ – Rogelio Molina Feb 5 '15 at 5:30
  • $\begingroup$ The vectors may be linearly dependent. In this situation, if there are merely two vectors, their span is the span of just one of them (the other belonging to that span) and defines a line, which is a degenerate plane. $\endgroup$ – user173897 Feb 5 '15 at 5:34
  • $\begingroup$ So $\mathrm{Span}\{\mathbf{v}_1,\mathbf{v}_2\}$ is just a plane in $\mathbb{R^4}$ through the origin, because neither vector in this problem is a multiple of the other. Is that right? $\endgroup$ – fancynancy Feb 5 '15 at 5:59

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