Proof of Theorem KLTS

We can apply the three-part test of Theorem TSS. First $\lt{T}{\zerovector_U}=\zerovector_V$ by Theorem LTTZZ, so $\zerovector_U\in\krn{T}$ and we know that the kernel is nonempty.

Suppose we assume that $\vect{x},\,\vect{y}\in\krn{T}$. Is $\vect{x}+\vect{y}\in\krn{T}$? \begin{align*} \lt{T}{\vect{x}+\vect{y}}&=\lt{T}{\vect{x}}+\lt{T}{\vect{y}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\\ &=\zerovector+\zerovector&&\vect{x},\,\vect{y}\in\krn{T}\\ &=\zerovector&&\knowl{./knowls/property.Z.knowl}{\text{Property Z}} \end{align*}

This qualifies $\vect{x}+\vect{y}$ for membership in $\krn{T}$. So we have additive closure.

Suppose we assume that $\alpha\in\complex{\null}$ and $\vect{x}\in\krn{T}$. Is $\alpha\vect{x}\in\krn{T}$? \begin{align*} \lt{T}{\alpha\vect{x}}&=\alpha\lt{T}{\vect{x}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\\ &=\alpha\zerovector&&\vect{x}\in\krn{T}\\ &=\zerovector&&\knowl{./knowls/theorem.ZVSM.knowl}{\text{Theorem ZVSM}} \end{align*}

This qualifies $\alpha\vect{x}$ for membership in $\krn{T}$. So we have scalar closure and Theorem TSS tells us that $\krn{T}$ is a subspace of $U$.