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:
- $\map \ker {u^\intercal} = \paren {\map u G}^\circ$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.12$