Let W be a subspace of a finite dimensional inner product sp

Let W be a subspace of a finite dimensional inner product space V and let {v_1, v_2, ...., v_k} be a basis for W with {u_1, u_2, ..., u_j} as basis for the orthogonal complement of W. Show that {v_1, v_2, ..., v_k, u_1, u_2, ..., u_j} is a basis for V. Suppose that W is a subspace of the Euclidean inner product space R^3 and let {(2, 3, 1)} be a basis for the orthogonal complement of W Assuming that (-3, 1, 3) W, find a basis for W.

Solution

We know that dim W +dim W^T =dimV

k+j= dim V

That is it is enough to show that the set

{ v1,v2,....vk,u1,u2,....uj} is linearly independent since it contains exactly dim V elements And without loss if generlity assune that {v1,v2,..vk} is orthonormal basis

For linearly independent ,let a1,a2,....ak,ak+1,....aj be scalars

Sych that a1v1+a2v2+......akvk+ak+1u1+.......aluj=0

Take product with v1 of both sides

a1=0 ,as v1.v1=1 and v1.v2=0 since we take orthonormal basis

And v1.ui =9 for all i = 1 to j ,since ui \'s are basis of orthogonal compliment if W

Similarly we can see that all scalars are zero

That means that set is linieay independent .

Hence proved,