# span of s looks like geometrically?

• Apr 6th 2013, 08:56 PM
yangx
span of s looks like geometrically?
$\mathbf{S} = \begin{bmatrix}1 \\2 \\-1 \end{bmatrix}\begin{bmatrix}1 \\3 \\1 \end{bmatrix}\begin{bmatrix}2 \\7 \\4 \end{bmatrix}$

after RREF
$\mathbf{S} = \left(\begin{array}{ccc}1& 0& -1\\0& 1& 3\\0& 0& 0\\\end{array} \right)$

I know S is linearly dependent because the third vector is a combination of the first two.

Describe in words what the span S looks like geometrically?

I think it is...
The span S is a plane that contain the vectors $\begin{bmatrix}1 \\2 \\-1 \end{bmatrix}\begin{bmatrix}1 \\3 \\1 \end{bmatrix}$ and the zero vector.

Is this correct?

Find an implicit description for the span of S?
would this just be
$\begin{bmatrix}x \\y \\z \end{bmatrix} = t\begin{bmatrix}1 \\-3 \\1 \end{bmatrix}$

after a bit of reading, above would the the explicit description of span s. The implicit description would be equation of the plane.
• Apr 6th 2013, 10:19 PM
chiro
Re: span of s looks like geometrically?
Hey yangx.

Given a span of two vectors, the span will be the plane with a normal of the cross product of the two vectors.