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I realize this is a novice question, but I'm stuck here. Suppose you have an ordered basis $B = (v_1, v_2, v_3, v_4)$ of $R^4$. My problem is how to determine a system of equations that defines, relatively to the basis $B$, the linear subspace of $R^4$ generated by the vectors $v_1-v_2+v_4$ and $-v_1+3v_3$.

Again, I realize this is a novice question, but I'd appreciate some insight into it... Thanks in advance.

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Although I already posted a complete answer, if this is a homework problem, please tag it with the [homework] tag; it helps to know when writing answers. – Arturo Magidin Mar 1 '11 at 6:03
@ArturoMagidin: this wasn't exactly homework, it was an exercise I came across in some extra reading, and got confused about. Although I suppose it does fit the "homework" category... Should add the tag nonetheless? Last but not least, thanks for the answer! – wmnorth Mar 2 '11 at 10:47
if it was not homework, but is an exercise you came across, you should mention that in the body; because it sure reads like homework (the tag is only for actual homework, so no, don't add it; but giving the context in the body would be a good thing). – Arturo Magidin Mar 2 '11 at 16:54

2 Answers 2

up vote 2 down vote accepted

Since you are working relative to the basis $B$, you may as well imagine that you are working over the standard basis and that the vectors you have are, as lhf points out, $(1,-1,0,1)$ and $(-1,0,3,0)$.

What is a system of equations that determines these vectors? To get a subspace, you'll want a homogeneous system of equations. And these two vectors should generate all solutions. So, first of all, your system should have four unknowns and two equations, since you want two parameters. The solutions will all look like $$\left(\begin{array}{c} x_1\\x_2\\x_3\\x_4\end{array}\right) = \left(\begin{array}{r} 1\\-1\\0\\1\end{array}\right)r + \left(\begin{array}{r}-1 \\0\\3\\0\end{array}\right)s$$ with $r$ and $s$ arbitrary.

What system is this? Well, \begin{align*} x_1 & = r - s\\ x_2 &= -r\\ x_3 &= 3s\\ x_4 &= r. \end{align*} Equivalently, plugging in $x_4$ for $r$ and $\frac{1}{3}x_3$ for $s$, we get \begin{align*} x_1 +\frac{1}{3}x_3 - x_4 &= 0\\ x_2 +x_4 &= 0. \end{align*} You'll note that the two vectors you have, $(1,-1,0,1)$ and $(-1,0,3,0)$, are solutions. And any solution is a linear combination of these two. So this gives the system you want in terms of coordinate vectors relative to $B$.

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Relative to $B$, the given vectors have coordinates $(1,-1,0,1)$ and $(-1,0,3,0)$.

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