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I am having trouble to understand why this given vector spans $ \mathbb R^3$.

$V=\left\{\begin{bmatrix}-4\\-6\\0\end{bmatrix},\begin{bmatrix}2\\-1\\-4\end{bmatrix}\right\}$

I understand that a non-zero determinant shows that the given set spans the space. However, this is not the square matrix. Therefore, I can not use determinant. I searched online and found this thread.

How to tell if a set of vectors spans a space?

According to this thread, we can use row reductions to calculate the rank of the given matrix. If it is equal to space, we can say that it spans this. I tried to row reduce this and end up with

$\begin{bmatrix}1\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix}$

This shows that the rank is $2$. Therefore, please explain how to arrive at the solution and why it spans $\mathbb R^3$. I know there are lots of questions about spanning sets. I am still kinda confused about calculating spans. I use the following procedure:

$1)$ if the matrix is square, non-zero determinant shows span. If not, rank tells span.

Please guide me is this correct way? or should I use any other procedure? Thanks. please explain in detail.

Community
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user298815
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    your set Vector doesn't span $\Bbb R^3$ – janmarqz Dec 15 '15 at 05:48
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    Following @janmarqz's comment, you need at least 3 vectors to span $\mathbb R^3$, but that condition is certainly not sufficient. – user217285 Dec 15 '15 at 05:49
  • A spanning set of a vector space should have cardinality at least the dimension of the vector space. Moreover, if the given set spans $\Bbb{R}^{3}$, then we can solve two equations with three unknowns. – Yes Dec 15 '15 at 05:50
  • It will, however, span a subspace of $\Bbb R^3$, namely the subspace $span\left{ \left[\begin{smallmatrix} -4\-6\0\end{smallmatrix} \right], \left[\begin{smallmatrix} 2\-1\-4 \end{smallmatrix}\right] \right}$ – JMoravitz Dec 15 '15 at 05:53
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    Why are people downvoting this post? OP has clearly shown his/her (extensive) research, even to the point where OP has actually finished the problem and shown that we do not have spanning. – user217285 Dec 15 '15 at 07:44

1 Answers1

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The set $\{(−4, −6, 0), (2, −1, −4)\}$ does not span $\mathbb R^3$. The set contains only two vectors, so its span is at most $2$-dimensional. In fact, since $(−4, −6, 0)$ and $(2, −1, −4)$ are linearly independent, the span is exactly $2$-dimensional.

Chris Culter
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