Finding Basis for Subspaces of R4

  • Thread starter Thread starter Dell
  • Start date Start date
  • Tags Tags
    Basis Subspaces
Dell
Messages
555
Reaction score
0
i am given 2 subspaces of R4
W=sp{(a-b,a+2b,a,b)|a,b\inR}
U=sp{(1,0,1,1)(-6,8,-3,-2)}
and am asked to find:
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\capU

forW\capU
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\capU 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), doesn't 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)??
 
Physics news on Phys.org
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), doesn't this mean that W+U is the whole vector space R4 ??
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 realized that the vectors are in fact not linearly independent. You need to re-check that part.

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)??
EDIT: As above the vectors are not linearly independent.
 
Last edited:
If W \cap U \neq {0} then the set {basis of U} \cup {basis of W} must be linearly dependent.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top