Thread: Orthogonal sets

Is it possible to find a orthogonal set that is not linearly independent?

Originally Posted by icanc

Is it possible to find a orthogonal set that is not linearly independent?

An orthogonal subset of an euclidean space is not linearly independent if and only if .

More generally, if are orthogonal vectors in an inner product space (we need the inner product to be able to define "orthogonal"), let be a linear combination of the vectors that is equal to the 0 vector. Take the inner product of both sides of that with each vector in turn. For example, the inner product of each side with [itex]v_1[/tex] would be . The right side is obviously 0 while on the left side, every product except [itex]v_1\cdot\v_1[/itex] is 0. You did not say "orthonormal" so we cannot assume but it is some non-zero number (as FernandoRevill says, assuming the set does NOT contain the 0 vector), p, so we have and must conclude that .

I am still a little confused. So, let's say that [3, 1, 1], [-1, 2, 1], and = [-1/2, -2, 7/2]. They are all orthogonal, I checked. Could you give me a similar example that would make the set not linearly independent?

sure, just add the 0-vector [0,0,0].

the point being, if we have an orthogonal set in R^3 that is all non-zero vectors, say v1,v2,v3. then consider any linear combination:

u = av1 + bv2 + cv3. if this is 0, then <v1,u> = 0, since <v1,0> = 0 (here <x,y> is just the dot product in R^3).

but <v1,u> = <v1,av1> + <v1,bv2> + <v1,cv3> = a<v1,v1> + b<v1,v2> + c<v1,v3> = a<v1,v1> (by orthogonality).

but since <v1,v1> is not 0 (because v1 is not 0), we must have a = 0.

now show that <v2,u> = 0 implies b = 0, and that <v3,u> = 0 implies c = 0, showing {v1,v2,v3} is linearly independent.

the only assumption we make is that none of the v's are 0-vectors, so any orthogonal set of non-zero vectors in R^3 is linearly independent.

Originally Posted by Deveno

sure, just add the 0-vector [0,0,0].

the point being, if we have an orthogonal set in R^3 that is all non-zero vectors, say v1,v2,v3. then consider any linear combination:

u = av1 + bv2 + cv3. if this is 0, then <v1,u> = 0, since <v1,0> = 0 (here <x,y> is just the dot product in R^3).

but <v1,u> = <v1,av1> + <v1,bv2> + <v1,cv3> = a<v1,v1> + b<v1,v2> + c<v1,v3> = a<v1,v1> (by orthogonality).

but since <v1,v1> is not 0 (because v1 is not 0), we must have a = 0.

now show that <v2,u> = 0 implies b = 0, and that <v3,u> = 0 implies c = 0, showing {v1,v2,v3} is linearly independent.

the only assumption we make is that none of the v's are 0-vectors, so any orthogonal set of non-zero vectors in R^3 is linearly independent.

Thanks a lot, this helps a ton. One more question. Does adding the zero vector still make the set orthogonal?

yes, because the 0-vector is orthogonal to every other vector.

Originally Posted by Deveno

yes, because the 0-vector is orthogonal to every other vector.

Got it, thanks again.