Kernel of Transpose of Linear Transformation is Annihilator of Image

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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:

$\map \ker {u^\intercal}$ is the annihilator of the image of $u$

where $\map \ker {u^\intercal}$ denotes the kernel of $u^\intercal$.


Proof

From the definitions of:

the transpose $u^\intercal$
the annihilator $\paren {\map u G}^\circ$

it follows that:

$\map {u^\intercal} y = 0 \iff y \in \paren {\map u G}^\circ$

Thus:

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

$\blacksquare$


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