linear algebra - Proof that $W^ {\perp\perp}=W$ (in a finite . . . You'll need to complete a few actions and gain 15 reputation points before being able to upvote Upvoting indicates when questions and answers are useful What's reputation and how do I get it? Instead, you can save this post to reference later
A question related to $S^ {\perp}$ and closure of span of $S$ You'll need to complete a few actions and gain 15 reputation points before being able to upvote Upvoting indicates when questions and answers are useful What's reputation and how do I get it? Instead, you can save this post to reference later
image of adjoint equals orthogonal complement of kernel As a side remark: If you know exact sequences and that taking duals is exact (actually left-exactness suffices, which is a general property of the $\hom$-functor), then you can show $\im (T^*)= (\ker T)^\perp$ as follows: The exact sequence $$ 0\longrightarrow \ker T\longrightarrow V\xrightarrow T\im T\longrightarrow 0 $$ gives an exact
Double orthogonal complement is equal to topological closure It remains to prove that $ (A^\perp)^\perp $ is the smallest one Suppose $ F $ is another closed subspace containing $ A $, then $ A^\perp\supset F^\perp $ and hence $ (A^\perp)^\perp\subset (F^\perp)^\perp $