Complements of a subspace

Question

Show that a non-trivial subspace $U$ of $V$ has two virtually disjoint complements iff $dim(U)\geq \frac{dim(V)}{2}$.

Definition 1:$S$ and $T$ are said to be virtually disjoint if $S\cap T=\{0\}$.

Definition 2: A subspace $T$ of $V$ is said to be complement of $S$ (a subspace of $V$) if $S\oplus T=V$ (i.e. $S+T$ is direct sum).

The if part is done, I am stuck with the only if part of the question. If possible try to find a suitable answer using Modular Law.

Any help will be appreciated.

Then I must not know what you mean by a "complement." Would you mind adding a definition of "complement" to the question? Thanks. — Robert Shore, Sep 10 '19 at 20:59

Exodd · Accepted Answer · 2019-09-10T21:14:46.740

1

for the only if, use the formula for the sum of subspaces, reported for example in here.

In fact, if $S,T$ are the disjoint complements, then $$ dim(S) + dim(T) = dim(S + T) \le dim(V) $$ but $dim(S) = dim(T) = dim(V) - dim(U)$ so $$ dim(U) = dim(V) - dim(S) \ge dim(V) - \frac{\dim(V)}2= \frac{\dim(V)}2. $$

edited Sep 10 '19 at 21:14

answered Sep 10 '19 at 21:04

Exodd

10,844

I was using this formula for the only if part but I was stuck. Kindly post the solution or provide some hints. – Chandramauli Chakraborty Sep 10 '19 at 21:08
@ChandramauliChakraborty edited – Exodd Sep 10 '19 at 21:14

Complements of a subspace

1 Answers1