[proofplan] We prove the result by unpacking the definitions of kernel, range, and invariant subspace. First we record that $\ker(T)$ and $\operatorname{Range}(T)$ are subspaces of $V$, using linearity of $T$. Then we show that applying $T$ to an element killed by $T$ again produces an element killed by $T$, and that applying $T$ to an element already of the form $T(x)$ again produces an element of the form $T(z)$. [/proofplan] [step:Verify that the kernel and range are subspaces of $V$] Because $T \in \operatorname{End}_F(V)$, the map \begin{align*} T: V \to V \end{align*} is $F$-linear. Define \begin{align*} \ker(T) = \{v \in V : T(v)=0\} \end{align*} and \begin{align*} \operatorname{Range}(T) = \{T(v) : v \in V\}. \end{align*} The kernel is a subspace of $V$: if $u,v \in \ker(T)$ and $a,b \in F$, then linearity gives \begin{align*} T(au+bv)=aT(u)+bT(v)=a0+b0=0, \end{align*} so $au+bv \in \ker(T)$. The range is a subspace of $V$: if $y_1,y_2 \in \operatorname{Range}(T)$ and $a,b \in F$, choose $x_1,x_2 \in V$ such that $y_1=T(x_1)$ and $y_2=T(x_2)$. Then \begin{align*} ay_1+by_2=aT(x_1)+bT(x_2)=T(ax_1+bx_2), \end{align*} so $ay_1+by_2 \in \operatorname{Range}(T)$. [/step] [step:Show that applying $T$ preserves the kernel] Let $v \in \ker(T)$. By definition of the kernel, $T(v)=0$. Since $T$ is linear, $T(0)=0$. Therefore \begin{align*} T(T(v))=T(0)=0. \end{align*} Thus $T(v) \in \ker(T)$. Since $v \in \ker(T)$ was arbitrary, \begin{align*} T(\ker(T)) \subseteq \ker(T). \end{align*} [guided] We need to prove that $\ker(T)$ is invariant under $T$. By definition, this means that whenever an element starts inside $\ker(T)$, its image under $T$ remains inside $\ker(T)$. Take an arbitrary element $v \in \ker(T)$. The defining property of the kernel is \begin{align*} T(v)=0. \end{align*} To prove that $T(v)$ is again in the kernel, we must check the defining condition for membership in $\ker(T)$, namely that applying $T$ to $T(v)$ gives $0$. Since $T$ is linear, it sends the zero vector of $V$ to itself: \begin{align*} T(0)=0. \end{align*} Hence \begin{align*} T(T(v))=T(0)=0. \end{align*} This is exactly the condition $T(v) \in \ker(T)$. Because the choice of $v \in \ker(T)$ was arbitrary, we conclude \begin{align*} T(\ker(T)) \subseteq \ker(T). \end{align*} [/guided] [/step] [step:Show that applying $T$ preserves the range] Let $y \in \operatorname{Range}(T)$. By definition of the range, there exists $x \in V$ such that \begin{align*} y=T(x). \end{align*} Applying $T$ gives \begin{align*} T(y)=T(T(x)). \end{align*} Since $T(x)\in V$ and $\operatorname{Range}(T)=\{T(z):z\in V\}$, the vector $T(T(x))$ belongs to $\operatorname{Range}(T)$. Hence $T(y)\in \operatorname{Range}(T)$. Since $y \in \operatorname{Range}(T)$ was arbitrary, \begin{align*} T(\operatorname{Range}(T)) \subseteq \operatorname{Range}(T). \end{align*} [/step] [step:Conclude that both subspaces are invariant under $T$] By the previous steps, $\ker(T)$ and $\operatorname{Range}(T)$ are subspaces of $V$, and they satisfy \begin{align*} T(\ker(T)) \subseteq \ker(T) \end{align*} and \begin{align*} T(\operatorname{Range}(T)) \subseteq \operatorname{Range}(T). \end{align*} Therefore $\ker(T)$ and $\operatorname{Range}(T)$ are $T$-invariant subspaces of $V$. [/step]