Kernel of linear transformation is subspace of domain
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Let $T: U \mapsto V$ be a linear transformation. Let $K = \{u \in U: T(u) = 0\}$. $K$ is called the kernel of $T$. Then $K$ is a subspace of $U$.
Proof
\[ x, y \in K \implies (T(x) = 0 \wedge T(y) = 0) \implies T(x+y) = T(x) + T(y) = 0 \implies x + y \in K \]
\[ x \in K \implies T(x) = 0 \implies T(cx) = cT(x) = 0 \implies cx \in K \]
Therefore, $K$ is a subspace of $U$.
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- Depth: 7
- Number of transitive dependencies: 13