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.
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.