# Vector space

1. Jan 29, 2009

### Dell

i am given 2 subspaces of R4
W=sp{(a-b,a+2b,a,b)|a,b$$\in$$R}
U=sp{(1,0,1,1)(-6,8,-3,-2)}
a homogenic system for W- system for a vector (x,y,z,t) belonging to W

i see the basis for W is : a(1,1,1,0)+b(-1,2,0,1),, i put these vectors into an extended matrix with (x y z t) on the other side, and after a series of elementary operations, i get
x+t-z=0
y-z-2t=0

next i am asked to find
a basis for W+U and W$$\cap$$U

forW$$\cap$$U
i find a homogenic system for U, and compre it with the system i found for W which comes to
x+t-z=0
y-z-2t=0
y+8z-8t=0
3t-4z+x=0
and i come to
t=1.5z
z=z
y=4z
x=-0.5z
so for W$$\cap$$U i get a basis (-0.5, 4, 1, 1.5)

for W+U i take the basis of each and check independace of all of them together, in which i get that all4 are independant, therefore the basis for W+U={(1,0,1,1)(-6,8,-3,-2)(1,1,1,0)(-1,2,0,1),}
if i perform elementary colum operations on them i can get to (1000)(0100)(0010)(0001), doesnt this mean that W+U is the whole vector space R4 ??

the final question is
find a vector other than the zero vector which is orthagonal to all the vectors in U+W
is this possible, since i found that W+U is the whole vector space R4 (supposing i was correct there)??

2. Jan 30, 2009

### Defennder

Yes it does. In fact a certain theorem tells you that if you can find a set of n linearly independent vectors spanning R^n then that set is a basis for R^n. EDIT: I just checked it and I realised that the vectors are in fact not linearly independent. You need to re-check that part.

EDIT: As above the vectors are not linearly independent.

Last edited: Jan 30, 2009
3. Jan 30, 2009

### kurnimaha

If W $$\cap$$ U $$\neq$$ {0} then the set {basis of U} $$\cup$$ {basis of W} must be linearly dependent.