Question

Is there a difference between orthogonal basis and orthogonal complement of a subspace

Answer 1

I need help finding the orthogonal subspace W of R5 spanned by vectors, V1,V2,V3,V4,V5

Answer 2

I know how to find the orthogonal basis using the gram-schmidt process

Answer 3

the orthogonal basis of a plane are 2 independant vectors in the plane whose dot product is 0; the orthogonal compliment is the setup such that every vector in the plane is orthogonal to a vector ... usually called the normal vector ... such that

a(x-xo)+b(y-yo)+c(z-zo) = 0

Answer 4

so one in r5 would be a(t-t0)+b(u-u0)+c(v-v0)+d(w-w0)+c(z-z0)=0

Answer 5

that would be a suitable dot product yes :)

but it might be simpler to say for higher dimensions, the we need to find the basis of the nullspace of A

Answer 6

so AX=0

rref (A) * (x1,x2,x3,x4,x5)=0

Answer 7

\[<-2,1,0,-1,-2>,<1,3,1,-2,-4>,<3,2,1,-1,-2>,<-1,4,1,-3,-6>,<-4,2,0,-2,-4>\]

Answer 8

so after i find the null space of A would tht be my orthogonal component to the subspace W of r5

Answer 9

the basis of the nullspace is the orthogonal compliment yes

Answer 10

dunno if we would have to tweak the null basis tho to be orthogonal itself ...

http://ltcconline.net/greenl/courses/203/Vectors/orthogonalComplements.htm

Answer 11

i get (1,0,-1,1,2) (0,1,1,1,0)

Answer 12

rref{[-2,1,3,-1,-4},{1,3,2,4,2},{0,1,1,1,0},{-1,-2,-1,-3,-2},{-2,-4,-2,-6,-4}}
\[x_1=1x_3-1x_4-2x_5\\
x_2=-1x_3-1x_4+0x_5\\
x_3=1x_3+0x_4+0x_5\\
x_4=0x_3+1x_4+0x_5\\
x_5=0x_3+0x_4+1x_5\\\]
\[\begin{pmatrix}
1&-1&-2\\
-1&-1&0\\
1&0&0\\
0&1&0\\
0&0&1\\
\end{pmatrix}\]

Answer 13

http://www.wolframalpha.com/input/?i=rref%7B%7B-2%2C1%2C3%2C-1%2C-4%7D%2C%7B1%2C3%2C2%2C4%2C2%7D%2C%7B0%2C1%2C1%2C1%2C0%7D%2C%7B-1%2C-2%2C-1%2C-3%2C-2%7D%2C%7B-2%2C-4%2C-2%2C-6%2C-4%7D%7D

Answer 14

might have to dbl chk to see if i remembered that correctly

Answer 15

well i got the same rref as the one you did, but did you then turn that into a system of equations and let x3,x4,x5 be variables and wrote everyting in terms of x1 and x2 and do parametric stuff

Answer 16

yes

i took and defined all the components of X in terms of the free variables

Answer 17

there example at the bottom of the link used the vectors as row vectors

Answer 18

nullspace of A^T rrefs to orthro comp of column vectors

Answer 19

so maybe it should have gone this route

rref{{-2,1,0,-1,-2},{1,3,1,-2,-4},{3,2,1,-1,-2},{-1,4,1,-3,-6},{-4,2,0,-2,-4}}

http://www.wolframalpha.com/input/?i=rref%7B%7B-2%2C1%2C0%2C-1%2C-2%7D%2C%7B1%2C3%2C1%2C-2%2C-4%7D%2C%7B3%2C2%2C1%2C-1%2C-2%7D%2C%7B-1%2C4%2C1%2C-3%2C-6%7D%2C%7B-4%2C2%2C0%2C-2%2C-4%7D%7D

<1,2,-7,0,0>, <1,-5,0,-7,0>,<2,-10,0,0,-7>

Answer 20

oh

Answer 21

how do u verify

Answer 22

lol, thats a good question :)

if we go back to the basics, how would you find the solution to Ax = 0?

Answer 23

spose W = (a1,a2,a3,...,an); then W^T is such that a1 * x = 0, a2*x=0, a3*x=0, ..., an*x=0

Answer 24

so it looks to me like if we take our vectors and represent them as the rows of the matrix A, then Ax = 0 represents the solution

Answer 25

http://www.wolframalpha.com/input/?i=kernel%7B%7B-2%2C1%2C0%2C-1%2C-2%7D%2C%7B1%2C3%2C1%2C-2%2C-4%7D%2C%7B3%2C2%2C1%2C-1%2C-2%7D%2C%7B-1%2C4%2C1%2C-3%2C-6%7D%2C%7B-4%2C2%2C0%2C-2%2C-4%7D%7D

seems to verify it on my end