Show that the set of all vectors in R^n that are orthogonal to both u & v is a subspace of R^n -linear algebra

Okay, go back to the definition of "subspace": a subset is a subspace if (i) 0 is in the subspace, and (ii) if whenever x and y are in the subset and a and b are scalars, then ax+by is in the subset.

(i) Clearly 0 is orthogonal to anything, so in particular it's orthogonal to u and v.

(ii) Let x and y each be in the set of vectors orthognal to u and v, and a and b be scalars. Then (denoting dot-product by *)

(ax+by) * u =

a (x*u) + b(y*u) =

a(0) + b(0) =

0.

Similarly, (ax+by)*v = 0, so ax+by is in the subset.

Therefore the subset of vectors orthogonal to u and v is a subspace.

the three regulations of subspaces are that they ought to incorporate the 0 vector, they're additive, and that they ought to have the capacity to be greater via a consistent. that's between the given regulations of subspaces. An occasion: study right here as vertical columns The set U = [x, x- y, y] is a subset of R3 u1 = [x1, x1-y1, y1] and u2 = [x2, x2-y2,y2] u1 + u2 = [x1+x2, (x1+x2)-(y1+y2), y1+y2] desire that facilitates, i'm a splash rusty on my linear algebra. this is been a mutually as.

Well, since u and v are linearly independent, they form a basis of a subspace S in R^n.

I think there is a theorem that says S perp (the othogonal complement of S) is also a subspace of R^n.