![]() Is a projection of $E$ onto the span of the $y_k$, and $I-T$ will be a projection onto $F$ (unless I've messed something up, which is possible). If L is the lattice of subspaces of an f-dimensional vector space over lFq and. $$T:E\rightarrow E x\mapsto \sum_k \mu_k(x) y_k$$ Baclawski 2 has defined a (co)homology theory for a geometric lattice L. ![]() To show that the span is a subspace, we therefore only have to show that it is closed under addition and scalar multiplication. It is clear that the span of any set contains 0 (note our convention on empty sets). Finally, pick $y_k\in E$ with $y_k F=x_k$. Since by definition rad(V ) is orthogonal to all of V, any complementary subspace W to rad(V ) in the sense of usual linear algebra will give rise to an. For any subset S V, the set span(S) is a subspace of V. Then let $\mu_k$ be the composition of $E \rightarrow E/F$ with $x_k^*$. If F is of finite codimension in E, then by definition we can find a basis $\$. But then it's a perfectly reasonable question, and the answer is "yes". (As I was typing this, rpotrie got the same answer.)Īs for complementation: well, this only makes sense for closed finite codimension subspaces. ![]() So, use the axiom of choice to find a discontinuous linear functional, and you have found a codimension one subspace which isn't closed. ![]() Notice that its kernel is of codimension one. It's a standard result that a linear functional from a Banach space to the underlying field (real or complex numbers) is continuous if and only if the its kernel is closed. ![]()
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