# Span

• Feb 14th 2009, 02:10 PM
krepka
Span
Let W be the set of all (X1, X2, X3, X4, X5) in R5 which satisfy
2X1-X2+4/3X3-X4=0
X1+2/3X3-X5=0
9X1-3X2+6X3-3X4-3X5=0
Find a finite set of vectors which spans W

Does this mean I have to find the vectors such that
c1V1+c2V2+c3V3+c4V4+c5V5=(X1, X2, X3, X4, X5) for some uknown V? And if so, what's the best way to go about it, by row-reducing the matrix above?

Thanks.
• Feb 14th 2009, 03:17 PM
HallsofIvy
Quote:

Originally Posted by krepka
Let W be the set of all (X1, X2, X3, X4, X5) in R5 which satisfy
2X1-X2+4/3X3-X4=0
X1+2/3X3-X5=0
9X1-3X2+6X3-3X4-3X5=0
Find a finite set of vectors which spans W

Does this mean I have to find the vectors such that
c1V1+c2V2+c3V3+c4V4+c5V5=(X1, X2, X3, X4, X5) for some uknown V? And if so, what's the best way to go about it, by row-reducing the matrix above?

Thanks.

Since there is no "V" in what you give, I have to answer no! But if your question is "Do I have to find vectors (not necessarily 5) so that every (X1, X2, X3, X4, X5) can be written as a linear combination of them (c1V1+ c2V2+ ...)" then the answer is yes.

In fact, because you have 5 variables with 3 equations you will need at least 5- 3= 2 such vectors and the equations are dependent, up to 4. And I see now that math2009 has shown that they are NOT independent. 3 vectors are required

yes, you could row-reduce a matrix but I would recommend this: use the three equations to solve for three of the "X"s in terms of the other two, say solve for X3, X4, and X5 in terms of X1 and X2. Then setting X1= 1, X2= 0 will give you one vector in that set and setting X1= 0, X2= 1 will give you another. The choices of "1, 0" then "0, 1" ensure those vectors are independent so they will span the set (in fact they form a basis for the subspace).
• Feb 14th 2009, 04:32 PM
math2009
$\vec{x} \in W\ ,W\ is\ subspace\ of\ R^5\ ,A=\begin{bmatrix} 2&-1&\frac{4}{3}&-1&0 \\1&0&\frac{2}{3}&0&-1 \\9&-3&6&-3&-3 \end{bmatrix}\ , A\vec{x}=\vec{0},\ W=ker(A)$

We convert the problem "find W" to "find kernel of matrix A".

$rref(A)=\begin{bmatrix} 1&0&\frac{2}{3}&0&-1 \\0&1&0&1&-2 \\ 0&0&0&0&0 \end{bmatrix}\ , rank(A)=2,dim(ker(A))=5-rank(A)=3$, we need at least 3 linear independent vectors to span $W$

$W=\begin{bmatrix} -\frac{2}{3}m+t\\ -n+2t \\m \\n \\t \end{bmatrix}=m\begin{bmatrix} -\frac{2}{3}\\ 0 \\1 \\0 \\0 \end{bmatrix}+n\begin{bmatrix} 0\\ -1 \\0 \\1 \\0 \end{bmatrix}+t\begin{bmatrix} 1\\ 2 \\0 \\0 \\1 \end{bmatrix}=span(\begin{bmatrix} -\frac{2}{3}\\ 0 \\1 \\0 \\0 \end{bmatrix},\begin{bmatrix} 0\\ -1 \\0 \\1 \\0 \end{bmatrix},\begin{bmatrix} 1\\ 2 \\0 \\0 \\1 \end{bmatrix})$
• Feb 14th 2009, 06:18 PM
krepka
Thanks for the replies. Following your method, HallsofIvy, I got
(1,0,0,2,1);(0,1,0,-1,0); and (0,0,1,4/3,2/3). These are different from math2009's answers. Given that I don't yet know what "the kernel of A" is, I'm going to trust that the vectors I found form the basis for the subspace.

Thanks again.
• Feb 14th 2009, 08:47 PM
math2009
$\begin{bmatrix} 1\\ 0 \\0 \\2 \\1 \end{bmatrix}=2\begin{bmatrix} 0\\ -1 \\0 \\1 \\0 \end{bmatrix}+\begin{bmatrix} 1\\ 2 \\0 \\0 \\1 \end{bmatrix} \ , \
\begin{bmatrix} 0\\ 1 \\0 \\-1 \\0 \end{bmatrix}=-\begin{bmatrix} 0\\ -1 \\0 \\1 \\0 \end{bmatrix}$
, $\begin{bmatrix} 0\\ 0 \\1 \\ \frac{4}{3} \\ \frac{2}{3} \end{bmatrix}=\begin{bmatrix} -\frac{2}{3}\\ 0 \\1 \\0 \\0 \end{bmatrix}+\frac{4}{3} \begin{bmatrix} 0\\ -1 \\0 \\1 \\0 \end{bmatrix}+\frac{2}{3} \begin{bmatrix} 1\\ 2 \\0 \\0 \\1 \end{bmatrix}
$

It means we have the same solution.