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furqannoor · Dec 8, 2009, 06:40 PM

Show that three of the vectors u=(1,1,-1), v=(-1,1,1), w=(1,3,-1), x=(1,1,0) form a basis for space and express the fourth vector as a linear combination of these basis vectors.

galactus · Dec 9, 2009, 07:10 AM

Show that three of the vectors u=(1,1,-1), v=(-1,1,1), w=(1,3,-1), x=(1,1,0) form a basis for space

In order to form a basis, they must span and be linearly independent.

The problem says show that three of the vectors form a basis. Which three? Here are the first three (u,v,w). Try the vector x in there somewhere and see if you can find a basis.

To show the set spans R^3, we have to show that an arbitrary vector $b=(b_{1},b_{2},b_{3})$ can be expressed as a linear combination

$b=c_{1}u+c_{2}v+c_{3}w$

of the vectors in what we can call S. Expressing this equation in terms of components gives us:

$(b_{1},b_{2},b_{3})=c_{1}(1,1,-1)+c_{2}(-1,1,1)+c_{3}(1,3,-1)$

$(b_{1},b_{2},b_{3})=(c_{1}-c_{2}+c_{3}, \;\ c_{1}+c_{2}+3c_{3}, \;\ -c_{1}+c_{2}-c_{3})$

Equate componets:

$c_{1}-c_{2}+c_{3}=b_{1}$

$c_{1}+c_{2}+3c_{3}=b_{2}$

$-c_{1}+c_{2}-c_{3}=b_{3}$

Thus, to show that S spans R^3, we have to show that the system above has a solution for all choices of $b=(b_{1},b_{2},b_{3})$ .

To prove S is linearly independent, we must show that the only solution of

$c_{1}u+c_{2}v+c_{3}w=0$

is $c_{1}=c_{2}=c_{3}=0$

So, we have

$c_{1}-c_{2}+c_{3}=0$

$c_{1}+c_{2}+3c_{3}=0$

$-c_{1}+c_{2}-c_{3}=0$

has only the trivial solution.

We can do this both by seeing if the determinant of the system is not equal to 0.

$Det\begin{vmatrix}1&-1&1\\1&1&3\\-1&1&-1\end{vmatrix}=0$

The determinant is 0, so it does not form a basis.

Try u,v, and x. I think they form a basis.