Is annihilator of a subspace actually the space of inner products on orthogonal vectors I am reading the book Linear Algebra Done Right by Sheldon Axler and learning about inner products. There's presented (6.42) Riesz Representation Theorem which essentially shows that there's an isomorphism between $V^*$ (the dual space) and inner products on some vector $<*,v>$ (where $v \in V$).
Now see, it's known that $(Col\ A)^\perp = Row\ A^T$ ($A$ - some square matrix). And it's also known that $Null\ T' = (Range\ T)^0$ (the second is the annihilator of the range of the operator $T$). On the other hand, $T'$ has as matrix the transposed matrix of the operator $T$, that is $Row\ A^T = Null\ T'$
It is not mentioned anywhere in the book but it appears that $(Range\ T)^0$ is not just the set of all linear forms such that they equal to 0 for any vector from $Range\ T$ but they are all inner products constructed from the vectors orthogonal to $Range\ T$.
Could you please confirm my understanding on this? Thank you!
 A: Well, finally I managed to prove it to be true and it adds a rather beautiful symmetry to the theory.
Formally, my statement: Let's $\ T \in L(V)$ and $T'$ its dual map, let's also $U = \{\phi \in L(V,F):\ \phi = \langle\cdot,u\rangle\ \forall u \in (Range\ T)^\perp\}$, then $Null\ T' = (Range\ T)^0 = U$.
The first part, $Null\ T' = (Range\ T)^0$, is proved in Axler's book (3.109). Then we proceed as follows: let's $u \in (Range\ T)^\perp$ then $\phi = \langle\cdot,u\rangle$ is equal $0$ for every vector in $Range\ T$, therefore $\phi \in (Range\ T)^0$. Hence $U \subset(Range\ T)^0$. On the other hand, if $\psi \in (Range\ T)^0$, then it equals $0$ for every vector from $Range\ T$ by definition. Then, by the Riesz Representation Theorem there exists and only one $u\in V$ such that $\psi = \langle\cdot,u\rangle$. But then this $u$ is orthogonal to every vector from $Range \ T$. Thus $(Range\ T)^0 \subset U$. Which concludes the proof.
Analogous thing can be proven for $Range\ T' = \{\phi \in L(V,F):\ \phi = \langle\cdot,v\rangle\ \forall v \in (Null\ T)^\perp\} = (Null\ T')^0$.
That all gives a really beautiful relation between abstract theory of dual spaces and the matrix theory, where for the latter another thing is usually taught: for any matrix $A$ it's true that $(Col\ A)^\perp = Null\ A^T$ and $(Col\ A^T)^\perp = Null\ A$.
