Image of Transpose of Linear Transformation is Annihilator of Kernel

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Theorem

Let $G$ and $H$ be $n$-dimensional vector spaces over a field.

Let $\map \LL {G, H}$ be the set of all linear transformations from $G$ to $H$.

Let $u \in \map \LL {G, H}$.

Let $u^\intercal$ be the transpose of $u$.

Then:

The image of $u^\intercal$ is the annihilator of $\map \ker u$.

where $\map \ker u$ denotes the kernel of $u$.


Proof

Let $x \in \map \ker u$.

Let $H^*$ be the algebraic dual of $H$.

Let $\innerprod x t$ be the evaluation linear transformation.


Then:

$\forall y \in H^*: \innerprod x {\map {u^\intercal} y} = \innerprod {\map u x} y = \innerprod 0 y = 0$

So:

$\map {u^\intercal} {H^*} \subseteq \paren {\map \ker u}^\circ$


From Rank Plus Nullity Theorem and Results Concerning Annihilator of Vector Subspace:

\(\ds \map \dim {\map {u^\intercal} {H^*} }\) \(=\) \(\ds n - \map \dim {\map \ker {u^\intercal} }\)
\(\ds \) \(=\) \(\ds n - \map \dim {\paren {\map u G}^\circ}\)
\(\ds \) \(=\) \(\ds \map \dim {\map u G}\)
\(\ds \) \(=\) \(\ds n - \map \dim {\map \ker u}\)
\(\ds \) \(=\) \(\ds \map \dim {\paren {\map \ker u}^\circ}\)


Hence:

$\map {u^\intercal} {H^*} = \paren {\map \ker u}^\circ$

$\blacksquare$


Sources

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