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 $\displaystyle dim(U+V)=dimU+dimV-dim(U\bigcap{V})$ using the following facts.

Given facts :

Let U be an m-dimensional subspace of $\displaystyle \Re^n$ and let V be a k-dimensional subspace of U, where 0<k<m

a) Any orthonormal basis $\displaystyle {v_{1},v_{2},...,v_{k}}$ for V can be expanded to form an orthonormal basis $\displaystyle {v_{1},v_{2},...,v_{k},v_{k+1},...,v_{m}}$for U

b) If $\displaystyle W=Span(v_{k+1},v_{k+2}...,v_{m})$, then U=V\bigoplusW

My proof:

Case 1 : $\displaystyle U\bigcap{V}={0}$

Let basis for U = $\displaystyle {x_{1},...,x_{k}}$

By Gram-Schmidt Process, we can find an orthonormal basis for U =$\displaystyle {u_{1},...,u_{k}}$

Let basis for U+V = $\displaystyle {x_{1},...,x_{k},x_{k+1},...,x_{m}}$

By Gram-Schmidt Process, we can find an orthonormal basis for U+V =$\displaystyle {u_{1},...,u_{k},u_{k+1},...,u_{m}}$

U is a subspace in U+V

By part (a) in the given fact, we know that $\displaystyle {u_{1},...,u_{k}}$can be expanded to $\displaystyle {u_{1},...,u_{k},u_{k+1},...,u_{m}}$

Let basis for V = $\displaystyle {x_{k+1},...,x_{m}}$

By Gram-Schmidt Process, we can find an orthonormal basis for V =$\displaystyle {u_{k+1},...,u_{m}}$

dim V=m-k

dim U=k

$\displaystyle dim(U\bigcap{V})=0$

dim (U+V) = m

and here come the result