Do these vectors span $\mathbb{R}^{3}$? Does $\{(2,-1,4),(1,0,2),(3,1,5)\}$ span $\mathbb{R}^{3}$ (written exactly like this)?
I know the answer to the question but don't know how to set it up. Normally my teacher gives them in the form of vectors ($\vec v_1=\left[\begin{matrix}8\\2\\3\end{matrix}\right]$) but we have never been given this in the form of points.
Do you set the matrix up as:
$$A = \begin{pmatrix} 2 & -1 & 4 \\ 1 & 0 & 2 \\ 3 & 1 & 5\end{pmatrix}$$
or
$$A = \begin{pmatrix} 2 & 1 & 3 \\ -1 & 0 & 1 \\ 4 & 2 & 5\end{pmatrix}$$
I have tried both ways, and after RREF they both contain pivots in every row/column so I would answer the question with 'yes', I just would like to know the correct format of the matrix.
Thanks
 A: If the vectors span $\mathbb R^3$ when written as $\begin{bmatrix}2\\-1\\4\end{bmatrix}, \begin{bmatrix}1\\0\\2\end{bmatrix}, \begin{bmatrix} 3\\ 1\\ 5\end{bmatrix}$, then they also span $\mathbb R^3$ when written as $(2,-1,4),(1,0,2),(3,1,5)$, and vice versa. They are the same elements of $\mathbb R^3$ no matter whether they are written horizontally or vertically -- it's just a difference of typographical convenience, not of the underlying mathematics. So you're free to switch to the other notation if you find that helps your calculations.
There is a mathematical difference between 1×3 and 3×1 matrices -- though still not one that would make a difference as to whether your set is linearly independent -- but when the problem mentions $\mathbb R^3$ instead of $\mathbb R^{1\times 3}$ (or whatever your notation for a matrix space is), then it implies that what you have is meant as simply vectors and not single-row matrices.
A: Normally, this is solved like this, you take three vectors, and they do span $R^3$ if and only if they are linearly independent. So, you check linear independency. You solve equation $\alpha x + \beta y + \gamma z = 0$ and you will get three equations and three unknowns, if that has nontrivial solution, they do span $R^3$. You will get a system:
$2\alpha+ \beta+ 3\gamma = 0$\
$-\alpha+\gamma = 0$\
$4\alpha+ 2\beta+5\gamma=0$\
And if you get other solution for $\alpha, \beta, \gamma$ that three zeros then yes, otherwise no.
