Prove that if w⃗ is orthogonal to each of the vectors in the set S = {⃗v1, ⃗v2, . . . , ⃗vm}, then w⃗ is also orthogonal to each of the vectors in the subspace W = SpanS of V .

Answer 1

This question isn't very clear, but making necessary reasonable assumptions, here we go.

Assume by way of a conditional proof that w is orthogonal to the vectors in S (which is to say the inner product with each of them is equal to zero by definition of orthogonal).

Let u be an arbitrary vector in Span(S).

Then u = a₁v₁ + ... + a_mv_m, where the a's are elements of the field the vector space V is over.

Now <w,u> = <w,a₁v₁ + ... + a_mv_m> = <w,a₁v₁> + ... + <w,a_mv_m> = a₁<w,v₁> + ... + a_m<w,v_m> = a₁*0 + ... + a_m*0 = 0.

The first equality comes from substitution.

The second and third are both properties of linearity, and inner products are linear.

the fourth equality comes from the assumption that w is orthogonal to the vectors in S.

Since u was an arbitrary vector in Span(S), the above is true for all elements of W = Span(S).

By condition proof, if w is orthogonal to the vectors in S, then it is orthogonal to all of the vectors in W.