# Is vector w in the subspace U which is in the span(S) containing orthogonal basis e_1,e_2,e_3

I am having a problem finding out how to check if a vector lie in a span or not. For the vectors in S, I have calculated through Gram-Schmidt the ortogonal basis, i.e.,$e_1=(1,1,0,0), e_2=(1/2,1/2,1,0), e_3=(-2/3,2/3,2/3,1)$ for

$S=(v_1=[1,1,0,0]^T,v_2[1,0,1,0]^T,v_3=[0,1,1,1]^T)$

and they are the subspace U=span(S).

Does the vector $w=[1,1,1,1]^T$ lie in $span(S)$?

To see if a vector lies in the span of a set ($S$ in this case), you would determine if it is a linear combination of the spanning vectors. To do this, one effective way is to put all of the vectors in a matrix equation, and row reduce. So it would look something like this: $$\begin{bmatrix}v_1&v_2&v_3&|&w\end{bmatrix}=\begin{bmatrix}1&1&0&|&1\\1&0&1&|&1\\0&1&1&|&1\\0&0&1&|&1\end{bmatrix}$$ This matrix consists of the vectors which span $U$ as the columns, and the $w$ vector is on the far right. This equation is the augmented matrix of the system $[v_1\,v_2\,v_3]x=w$. So if this system has a consistent solution, then $w$ is in the $span(S)$, since it can be written as a combination of the spanning vectors.

• When I tried that and got [\begin{bmatrix} 1&0&0&|&0\\ 0&1&0&|&0\\ 0&0&1&|&0\\ 0&0&0&|&1 \end{bmatrix} ] and that can't be a consistent solution as the last row ends up with 0=1
– S.n
Commented Dec 7, 2016 at 17:47
• @S.n I got something a bit different, however there is still an inconsistency in my solution as well (so the end entries in the far right column don't really matter, as long as there is a nonzero entry in the last column). So really the conclusion is what?
– Dave
Commented Dec 7, 2016 at 18:01
• Just to clarify, is it correct that this vector w does not lie in the span(S)?
– S.n
Commented Dec 7, 2016 at 18:02
• @S.n Yes, because there is an inconsistent solution in that system, so one cannot make $w$ out of a linear combination of $v_1,v_2,v_3$.
– Dave
Commented Dec 7, 2016 at 18:04