Re: direct sum of subspaces

EDIT: sorry made a mistake in my working.

Does anyone have any ideas on how to prove it?

Thanks

Re: direct sum of subspaces

you are given a clue as to where to start (a fact to use). since i am not an unkind person i wll give you another: what is dim(U1 + U2)?

what can you say therefore about dim((U1+U2)+U3),....,dim(U1+U2+....+Uj)?

Re: direct sum of subspaces

Quote:

Originally Posted by

**Deveno** you are given a clue as to where to start (a fact to use). since i am not an unkind person i wll give you another: what is dim(U1 + U2)?

what can you say therefore about dim((U1+U2)+U3),....,dim(U1+U2+....+Uj)?

$\displaystyle dim(U_1+ U_2)=dim(U_1)+dim(U_2)-dim(U_1 \cap U_2)$

So

$\displaystyle dim(U1+U2+....+Uj)= [\sum_{k=1}^j dim(U_k)] - dim((U_1+...+U_{j-1}) \cap U_j)$

Thus $\displaystyle dim(U_1+ U_2)=dim(U_1)+dim(U_2)$ and $\displaystyle dim(U1+U2+....+Uj)= \sum_{k=1}^j dim(U_k)$.

Is this correct? Where to from here?

Thanks

Re: direct sum of subspaces

if dim(V) = n, what must any n-dimensional subspace be?

Re: direct sum of subspaces

Quote:

Originally Posted by

**Deveno** if dim(V) = n, what must any n-dimensional subspace be?

If we let $\displaystyle W$ be a subspace of $\displaystyle V$ then $\displaystyle dim(W) \le n$ and if $\displaystyle dim(W)=n$ the $\displaystyle W=V$.

So any n-dimensional subspace must be $\displaystyle V$.

So in relation to my question, therefore $\displaystyle V$ is the direct sum of the subspace $\displaystyle U_1,...,U_j$.

Does this complete the proof?

Thanks for your help Deveno!