If I have vectors $\vec v_1=\left[\begin{matrix}0\\0\\-3\end{matrix}\right]$,$\vec v_2=\left[\begin{matrix}0\\-3\\9\end{matrix}\right]$, and $\vec v_3=\left[\begin{matrix}4\\-2\\-6\end{matrix}\right]$, do these vectors span $\mathbb R^3$? I'm pretty sure they don't because if we put them into matrix form, i.e: $\left[\begin{matrix}0&0&4\\0&-3&-2\\-3&9&-6\end{matrix}\right]$ then there's a pivot in the right most column meaning there is no solution for the system. However, there is a pivot in every column which would mean that $\vec v_1$,$\vec v_2$, and $\vec v_3$ are all in the span of $\mathbb R^3$. Which rule has "the upper hand" in determing the answer to the question?

  • $\begingroup$ Why don't you just compute the determinant of this matrix? If you have already seen the determinant, of course... $\endgroup$ – Julien Feb 10 '13 at 21:44
  • $\begingroup$ @julien I actually have not seen the determinant in this use, although now I am curious, what does the determinant tell us? $\endgroup$ – TheHopefulActuary Feb 10 '13 at 21:44
  • $\begingroup$ Ah! Then Gaussian elimination is the right way. $\endgroup$ – Julien Feb 10 '13 at 21:46
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    $\begingroup$ So do Gaussian elimination to obtain an upper-triangular matrix. Then look at the diagonal c0efficients. If at least one is $0$, the answer is no. If they're all nonzero, the answer is yes. Finally, note that the question of your title is not the first question of your post. This second one is the right question here. The answer to the title is trivially yes. $\endgroup$ – Julien Feb 10 '13 at 21:48
  • $\begingroup$ You have to do row swaps so the rightmost column is not the first pivot. Once you do the swaps then the first column (the leftmost one) is the first pivot. Actually, just one swap; the first and the third. $\endgroup$ – James S. Cook Feb 10 '13 at 21:50

They do, and you can see this by reducing this to row echelon form. But this is easy! If you switch the top and bottom rows, you get the matrix $$\begin{bmatrix} -3 & 9 & -6\\ 0 &-3 &-2\\ 0 &0 &4 \end{bmatrix}$$ which is already in row echelon form! Since none of the rows consist of all zeros, the rank of your matrix must be 3. I.e. the span of your three vectors has dimension 3. But any dimension 3 subspace of $\mathbb{R}^3$ must necessarily be all of $\mathbb{R}^3$---this is because $\mathbb{R}^3$ is itself a dimension 3 vector space. Thus, your three vectors do in fact span $\mathbb{R}^3$.

  • $\begingroup$ But I thought that if the right most column had a pivot then there is no solution and the system is inconsistent. Does what you said only hold because it's not augmented? $\endgroup$ – TheHopefulActuary Feb 11 '13 at 3:09
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    $\begingroup$ Yup. This guy can be viewed as the matrix part of a homogeneous system, and a homogeneous system always contains at least one solution---namely, the zero vector. $\endgroup$ – Avi Steiner Feb 11 '13 at 4:28

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