How to show that $(\operatorname{Im}{A})^⊥ = \ker(A^⊤)$, $\operatorname{Im}{(A^⊤)}= (\ker A)^⊥$?

Question

Let now $A ∈ L(V)$, where $V$ be an Euclidean vector space.

I need to show that:

$$(\operatorname{Im}{A})^⊥ = \ker(A^⊤)$$ $$\operatorname{Im}{(A^⊤)}= (\ker A)^⊥$$

in infinite dimension, one only has $closure(Im A^T) \subset {ker A}^\perp$ — nicolas, Nov 12 '14 at 18:48

score 1 · Accepted Answer · edited Mar 06 '18 at 18:34

1

That should do it

$$ \forall y \in V.\forall x \in V. \langle Ax, y \rangle = \langle x, A^Ty \rangle$$

edited Mar 06 '18 at 18:34

user1868607

answered Nov 11 '14 at 11:33

nicolas

How to show that $(A^⊤)^⊤ \subseteq A$? – qexi Nov 11 '14 at 11:42
@qexi, I think you meant $;\left(A^t\right)^t=A;$ ? It can't be a contention relation here as $;A;$ is not a set but a matrix. – Timbuc Nov 11 '14 at 11:47
No, I mean that $A$ is a set. Let's use $U$ for example. I don't know how to show that $(U^⊥)^⊥ = U$ & $(U^⊤)^⊤⊆U$. – qexi Nov 11 '14 at 11:50
@qexi this is another question. and one for which you might be surprised by the answer :) – nicolas Nov 11 '14 at 12:04
@qexi, but you defined $;A;$ as a matrix! – Timbuc Nov 11 '14 at 12:16
@Timbuc he refers to another question I think – nicolas Nov 11 '14 at 12:54

1 Answers1