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.


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 $B$.

  • $\begingroup$ Will i always be able to get rid of those scalars($r$ and $s$ in your example) in order to get the final equations? $\endgroup$ – jscherman Sep 8 '17 at 21:20
  • 1
    $\begingroup$ @jscherman: yes, because they are only different names for the free variables. $\endgroup$ – Arturo Magidin Sep 9 '17 at 4:48

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


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