Question

Let S be subspace of R^3 spanned by x=(1,-1,1)^T

a) Find a basic for S^ʇ

Answer 1

you want to find the null space for the 1 x 3 matrix [ 1 -1 1].

Answer 2

or rather, you want to find a basis for the null space.

Answer 3

so we find the null space then find the basic of that null space

Answer 4

yeah. Its a little odd since the matrix is 1 x 3, but it will work out the same.

Answer 5

Think of it this way, we know two vectors are perpendicular when their dot product is zero. So the vector (x,y,z) is perpendicular to (1,-1,1) when x-y+z=0. But this is the same as solving the matrix equation:\[A\vec{x}=0\]when A is the matrix [ 1 -1 1]

Answer 6

so x be any vector that makes right side equal to 0

Answer 7

like [0 0 0] ?

Answer 8

while that is true, you are looking for non-zero vectors. The zero vector will always be a solution to Ax = 0, thats called the trivial solution.

Answer 9

since we have 1 variable but 3 equation; wouldn't it likely be inconsistent?

Answer 10

Since there is only one equation, but three variables, we can make two of the variables free, and see what the last one has to be.

Answer 11

1 x1 - x2 +x3=0

x1= a

x2=b

a-b+x3=0
-a+b=x3

Answer 12

thats the right idea. Just make them numbers now so we can grab a basis. Like x1=1 and x2=0, or something like that.

Answer 13

There is a more straightforward way of doing this, which i'll post in a bit, but this isnt bad either.

Answer 14

x1=1
x2==0
-a+b=x3
-1+0=x3
x3=-1

Answer 15

So the vector (1, 0, -1) is one of your basis vectors. Now to get another that is independent of this one, make x1 = 0, and x2 = 1. See what x3 is, and that will be the second.

Answer 16

x1=0
x2=1
1=x3

0 1 1

Answer 17

that works. As you can see, the vector (1,-1,1) is perpendicular to both of the vectors you obtained, so it will be perpendicular to any linear combination of those vectors, ie the span of them. That subspace will be perpendicular to the span of (1,-1,1).

Answer 18

Here is another method. Even though the numerical value of the vectors are different, they do span the same space. So no worries.

Answer 19

quick question what does that upside down T next to S mean?

Answer 20

O.o i was assuming it meant perpendicular <,<

Answer 21

or orthogonal, like this:
http://en.wikipedia.org/wiki/Orthogonal_complement

Answer 22

yes , orthogonal complement

Answer 23

oh ok, we are still good then lol.

Answer 24

thank you so much, joe; helpful like always

Answer 25

Linear Algebra rules >.> lolol. yw!