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  "body": "\n## Proof\n\nWe show that both conditions say that the map $\\lambda \\id_V - T$ is not injective.\n\n1. Let $n = \\dim V$. The [[characteristic polynomial of $T$|0BZ81AJNWF]] needs $n \\ge 1$, so let $n \\ge 1$. Fix a basis $B = (b_1, \\dots, b_n)$ of $V$.\n2. Let $A = [T]_B$ be the [[matrix of $T$|0YYMY95FWS]]. By the definition and the [[invariance under change of basis|0TYMASG33Y]], $p_T(t) = p_A(t) = \\det(t I_n - A)$.\n3. Let $S = \\lambda \\id_V - T$ and $M = \\lambda I_n - A$. Since $S(b_j) = \\lambda b_j - T(b_j)$, column $j$ of $[S]_B$ is column $j$ of $M$. So $[S]_B = M$.\n4. Let $\\operatorname{ev}_\\lambda \\colon F[t] \\to F$ be the [[evaluation homomorphism|0125QVNCFQ]] at $\\lambda$. It maps each entry of $t I_n - A$ to the same entry of $M$.\n5. By the [[Leibniz formula|016PE4VZ8Q]], $p_A(t)$ is a sum of products of entries of $t I_n - A$, with signs $\\pm 1$.\n6. A ring homomorphism preserves sums, products and signs. So by step 5 and the Leibniz formula for $M$, $p_T(\\lambda) = \\operatorname{ev}_\\lambda(p_A) = \\det M$.\n7. By the definition of an [[eigenvalue|0ZND7NQ970]], $\\lambda$ is an eigenvalue of $T$ if and only if $S(v) = 0$ for some $v \\ne 0$.\n8. So $\\lambda$ is an eigenvalue if and only if the [[kernel|02YFME3RGW]] of $S$ is not $\\set{0}$. By the [[kernel criterion|0BJBYCGMWE]], this holds if and only if $S$ is not injective.\n9. Suppose $S$ is not injective. Take $v \\ne 0$ with $S(v) = 0$, and suppose that $M$ is [[invertible|0TWTS88TWN]] with inverse $N$.\n10. By the definition of the matrix of a linear map, $[S(v)]_B = M [v]_B$. So $[v]_B = N M [v]_B = N [S(v)]_B = 0$.\n11. The map $v \\mapsto [v]_B$ is injective, by the [[isomorphism with the coordinate space|059THXBRBB]]. So $v = 0$, which contradicts step 9. Hence $M$ is not invertible.\n12. By the [[determinant criterion for invertibility|03Q7RE2HXR]] over a field, $\\det M = 0$.\n13. Now suppose $S$ is injective. Since $V$ is finite-dimensional, $S$ is [[bijective|0912EEEVCT]]. Let $R \\colon V \\to V$ be its inverse map.\n14. The map $R$ is linear: for $u, w \\in V$ and $a, c \\in F$, $S(a R(u) + c R(w)) = a u + c w$. Applying $R$ gives $a R(u) + c R(w) = R(a u + c w)$.\n15. The matrix of $\\id_V$ in $B$ is $I_n$, since $\\id_V(b_j) = b_j$. By the rule for the [[matrix of a composition|05Q1B7MH0M]], $M [R]_B = [S \\circ R]_B = I_n$ and $[R]_B M = [R \\circ S]_B = I_n$.\n16. So $M$ is invertible. By the determinant criterion, $\\det M$ is a unit of $F$, so $\\det M \\ne 0$.\n17. By steps 8, 12 and 16, $\\lambda$ is an eigenvalue of $T$ if and only if $\\det M = 0$.\n18. By step 6, this holds if and only if $p_T(\\lambda) = 0$.\n\n## Idea\n\nThe scalar $\\lambda$ is an eigenvalue exactly when $\\lambda \\id_V - T$ kills a nonzero vector. In finite dimensions this happens exactly when the matrix $\\lambda I - A$ is not invertible, that is, when its determinant is $0$. That determinant is the characteristic polynomial evaluated at $\\lambda$.\n",
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