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I think I came up with a solution, but I wonder if there's a "better" way to do it that my prof may be looking for.

Construct a basis for $\mathbb{R}^4$ that consists of the vectors $\left[\begin{array}{rrrr}1\\2\\3\\4\end{array}\right]$,$\left[\begin{array}{rrrr}1\\4\\6\\8\end{array}\right]$, and some of the standard basis vectors $\vec{e}_1$, $\vec{e}_2$, $\vec{e}_3$, $\vec{e}_4$ in $\mathbb{R}^4$

So what I did was constructed a $6\times 4$ matrix using the two given vectors and the four basis vectors as rows, then did gaussian elimination to find out that $\vec{e}_1$ is a linear combination of the two given...

$$2\left[\begin{array}{rrrr}1\\2\\3\\4\end{array}\right]-\left[\begin{array}{rrrr}1\\4\\6\\8\end{array}\right]=\left[\begin{array}{rrrr}1\\0\\0\\0\end{array}\right]$$

From there I could tell that if I picked any two of the remaining three basis vectors, I would end up with a linearly independent spanning set for $\mathbb{R}^4$, which is what I want... but I can't help but feel that my method is too hacky. Is there a more elegant way to do this?

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    $\begingroup$ I'd say that's precisely the way to go. Perhaps someone else will come along with a better suggestion, though. $\endgroup$ Sep 29, 2013 at 15:15
  • $\begingroup$ Correction: I did not read carefully enough. I thought you were describing the approach that Vadim took. $\endgroup$ Sep 29, 2013 at 15:25

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Instead of a $6\times 4$ matrix, make a $4\times 6$ matrix. Put this into row canonical form, and the columns with the pivots tell you which of the original vectors are in your basis. Be sure to put your two chosen vectors as the first two columns, so they will be included.

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