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  "body": "\n## Proof\n\nWe reduce the statement to the Cayley–Hamilton theorem for matrices.\n\n1. If $V = 0$, then every endomorphism of $V$ is the zero map, so $p_T(T) = 0$. From now on let $n = \\dim V \\ge 1$.\n2. Choose a basis $B = (b_1, \\dots, b_n)$ of $V$. For an endomorphism $S$ of $V$, let $[S]_B$ be its [[matrix in $B$|0YYMY95FWS]]. Let $M = [T]_B$.\n3. Column $j$ of $[S]_B$ is the coordinate vector $[S(b_j)]_B$. The coordinate map is linear by the [[isomorphism with the coordinate space|059THXBRBB]]. So $[S + S']_B = [S]_B + [S']_B$ and $[c S]_B = c [S]_B$ for $c \\in F$.\n4. By the [[matrix of a composition|05Q1B7MH0M]], $[S \\circ S']_B = [S]_B [S']_B$. Also $[\\id_V]_B = I_n$, since $\\id_V(b_j) = b_j$.\n5. By step 4 and induction on $k$, $[T^k]_B = M^k$ for all $k \\ge 0$.\n6. Let $p(x) = \\sum_k c_k x^k \\in F[x]$. By the definition of the [[value at $T$|0BNEXSKNRV]] and steps 3 and 5, $[p(T)]_B = \\sum_k c_k M^k = p(M)$.\n7. By the [[definition of $p_T$|0BZ81AJNWF]], $p_T = p_M$, the characteristic polynomial of $M$.\n8. By the [[Cayley–Hamilton theorem for matrices|0F765KFC2J]], applied over the field $F$, $p_M(M) = 0$.\n9. By steps 6, 7 and 8, $[p_T(T)]_B = p_M(M) = 0$.\n10. So every column of $[p_T(T)]_B$ is zero, that is, $p_T(T)(b_j) = 0$ for each $j$.\n11. A linear map that vanishes on a basis vanishes on every linear combination of it. Hence $p_T(T) = 0$.\n\n## Idea\n\nChoosing a basis turns endomorphisms into matrices, and this translation respects sums, scalar multiples and composition. So it sends $p(T)$ to $p(M)$ for every polynomial $p$. The matrix theorem says $p_M(M) = 0$, and a map with zero matrix is zero.\n",
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