My only comment is that I don't see why you are insisting on "orthonormal bases". The concept of "orthonormal" only applies to inner product spaces while the dimension theorem is true in any vector space. Just drop "orthonormal" and look at "bases".
I'm doing a question on proving dimension theorem. I've thought of case 1 and I'm not sure if it is the correct approach. Please help me check it and comment it.
I have to prove using the following facts.
Given facts :
Let U be an m-dimensional subspace of and let V be a k-dimensional subspace of U, where 0<k<m
a) Any orthonormal basis for V can be expanded to form an orthonormal basis for U
b) If , then U=V\bigoplusW
My proof:
Case 1 :
Let basis for U =
By Gram-Schmidt Process, we can find an orthonormal basis for U =
Let basis for U+V =
By Gram-Schmidt Process, we can find an orthonormal basis for U+V =
U is a subspace in U+V
By part (a) in the given fact, we know that can be expanded to
Let basis for V =
By Gram-Schmidt Process, we can find an orthonormal basis for V =
dim V=m-k
dim U=k
dim (U+V) = m
and here come the result
Actually I know that using basis is simplier. But my tutor said if we do this question using orthonormal basis approach, things will be easier in proving the linear independence of basis for U+V.
Now I try to use different ways to prove. This time I try to prove is subspace of U, V and U+V. Also, U and V are subspaces of U+V. So we can use part (a) to show we can use orthonormal basis for U and V to expand to form orthonormal basis for U+V.
i.e. Let orthonormal basis for =
Orthonormal basis for U =
Orthonormal basis for V
Orthonormal basis for U+V =
Then how can we prove dimension of basis for U+V is n+k+m? First we should prove there are n+k+m elements in the basis for U+V. But now we only know the dimension of orthonormal basis for U+V. What can we do?