# direct sum of subspaces

• October 1st 2011, 03:16 AM
Juneu436
direct sum of subspaces

Let $V= \sum U_k$ where $V$ is a vector space and $U_1,...,U_j$ are the subspaces of $V$.
Use the fact that $U_1 \cap U_2 =0,..., (U_1+...+U_{j-1}) \cap U_j=0$ to prove that $V$ is the direct sum of the subspaces $U_1,...,U_j$.

• October 1st 2011, 04:59 AM
Juneu436
Re: direct sum of subspaces
EDIT: sorry made a mistake in my working.

Does anyone have any ideas on how to prove it?

Thanks
• October 1st 2011, 05:48 AM
Deveno
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)?
• October 1st 2011, 06:31 AM
Juneu436
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)?

$dim(U_1+ U_2)=dim(U_1)+dim(U_2)-dim(U_1 \cap U_2)$
So
$dim(U1+U2+....+Uj)= [\sum_{k=1}^j dim(U_k)] - dim((U_1+...+U_{j-1}) \cap U_j)$

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

Is this correct? Where to from here?

Thanks
• October 1st 2011, 06:38 AM
Deveno
Re: direct sum of subspaces
if dim(V) = n, what must any n-dimensional subspace be?
• October 1st 2011, 06:54 AM
Juneu436
Re: direct sum of subspaces
Quote:

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

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

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

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

Does this complete the proof?