I'm having trouble understanding in a mathematical sense the reason why:
(1) $$Null(A) = [R(A^{T})]^\perp $$
(2) $$Null(A^T) = [R(A)]^\perp$$
What I've tried so far is picking some arbitrary vector $\vec{v}$ in $[R(A^{T})]^\perp$ . Picking some arbitrary vector $\vec{y}$ in $R(A^T)$, we then have the relation that $<\vec{v}, \vec{y}> = 0$ if (1) is true.
We can rewrite $\vec{y}$ as $A\vec{x}$ and subsitute that into the equation. This gives us:
$$<\vec{v}, A\vec{x}> = 0$$
Which can be rewritten as:
$$\vec{v}^T A\vec{x} = 0$$ $$[\vec{v}^T (A\vec{x})]^T = 0$$ since the LHS of the equation is just a 1x1 matrix.
$$ (A\vec{x})^T \vec{v} = 0$$ $$ \vec{x}^T A^T \vec{v} = 0$$
Finally, we simplify to:
$$<\vec{x}, A^T \vec{v}> = 0$$ which doesn't seem to help.
How do I show that (1) and (2) are true?
$\endgroup$ 41 Answer
$\begingroup$For the first equality, $$\begin{align} v \in N(A) &\iff Av = 0 \\ &\iff \forall w \, \langle Av,w\rangle = 0 \tag{*}\\ &\iff \forall w \, \langle v,A^Tw \rangle = 0 \\ &\iff v \in R(A^T)^\perp. \end{align}$$
The only possibly tricky step is going from (*) to the preceding line, which requires the lemma that, if $\langle x,y \rangle = 0$ for all $y$, then $x=0$.
The proof for the other equality is similar.
These equalities are special cases of a broader result: If $T:V\to W$ is a linear map and $T^*: W^*\to V^*$ its adjoint, then the image of $T^*$ annihilates the kernel of $T$, and the kernel of $T^*$ annihilates the image of $T$.
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