[proofplan] The proof is obtained by expanding the definitions. If $0$ is an eigenvalue, then there is a nonzero vector $v \in V$ such that $T(v)=0v$, and this equation says exactly that $v$ lies in the kernel of $T$. Conversely, a nonzero vector in the kernel satisfies $T(v)=0=0v$, so it is an eigenvector with eigenvalue $0$. [/proofplan]

[step:Show that a zero eigenvector gives a nonzero kernel vector]Assume that $0 \in k$ is an eigenvalue of $T$. By the definition of eigenvalue, there exists a vector $v \in V$ such that $v \ne 0$ and \begin{align*} T(v)=0v. \end{align*} Since scalar multiplication by $0 \in k$ gives the zero vector of $V$, this means \begin{align*} T(v)=0. \end{align*} By the definition of the kernel of $T$, the equality $T(v)=0$ implies $v \in \ker T$. Since $v \ne 0$, the kernel contains a nonzero vector, and therefore \begin{align*} \ker T \ne \{0\}. \end{align*}[/step]

[guided]Assume that $0 \in k$ is an eigenvalue of $T$. The definition of eigenvalue includes the nonzero-vector condition: there must exist a vector $v \in V$ with $v \ne 0$ such that \begin{align*} T(v)=0v. \end{align*} Here $0v$ denotes scalar multiplication of $v$ by the zero element $0 \in k$. In any [vector space](/page/Vector%20Space), scalar multiplication by $0$ gives the zero vector of $V$, so the displayed equation becomes \begin{align*} T(v)=0. \end{align*} The kernel of $T$ is the set of vectors sent to the zero vector: \begin{align*} \ker T=\{w \in V : T(w)=0\}. \end{align*} Thus $T(v)=0$ implies $v \in \ker T$. Because the eigenvector $v$ is nonzero, $\ker T$ contains a nonzero vector. Hence \begin{align*} \ker T \ne \{0\}. \end{align*}[/guided]

[step:Show that a nonzero kernel vector is a zero eigenvector] Assume that \begin{align*} \ker T \ne \{0\}. \end{align*} Then there exists $v \in \ker T$ with $v \ne 0$. By the definition of the kernel, \begin{align*} T(v)=0. \end{align*} Since $0v=0$, we have \begin{align*} T(v)=0v. \end{align*} Thus there exists a nonzero vector $v \in V$ satisfying $T(v)=0v$, so $0$ is an eigenvalue of $T$. [/step]