proof that dim V + dim V perp = n
I am trying to prove that the dimension of a subspace V of R^n + the dimension of the orthogonal complement of that subspace V is equal to n for any subspace V in R^n.
So i assumed the basis for the subspace V consisted of vectors v1, v2,..., vk. now i extend the basis so that the new basis is a basis for R^n. This new basis is v1, v2, ..., vk, w1, w2, ..., w_m. where m + k = n. So in order to prove this theorem i have to show that w1, w2, ..., w_m forms a basis for the orthogonal complement of V. What i have trouble showing is that these vectors span the orthogonal complement of V. i have that any vector in R^n let's say u can be expressed as a linear combination of v1, v2, ..., vk, w1, w2, ..., w_m since that is a basis for R^n. so u = c1v1 + ... +ckvk + d1w1 + ... + d_m(w_m). i tried taking the dot product on both sides with an arbitrary vector w in the orthogonal complement of V and i got u * w = w * (d1w1 + ...+ d_m(w_m)) since w is orthogonal to every vector in V so w * v_i = 0. now i'm stuck and i'm not sure how to continue on.
I can show that w1, ... w_m are linearly independent since v1, v2, ..., vk, w1, w2, ..., w_m form a basis which means that the only linear combination of them that forms the 0 vector is for all constants to be 0. Then since i know that v1, ... vk is also a basis, the only linear combination of those vectors that equal 0 is for all the constants to equal 0 as well. and therefore the only linear combination of w1, w2, ..., w_m that equal 0 is all the constants being 0 so those vectors are independent.
now all i need to show is that those vectors span the orthogonal complement of V and the proof will be done. is my approach to showing that the vectors span correct? if anyone could give a hint or 2 to help me move on in this problem i'd greatly appreciate it.