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  "body": "\n## Proof\n\nWe write $t I - P^{-1} A P$ as a conjugate of $t I - A$ and use the product rule.\n\n1. Let $A$ and $P$ be $n \\times n$ matrices over $K$, with $P$ invertible. Let $Q = P^{-1}$, so $Q P = P Q = I_n$.\n2. The ring $K$ is a subring of the [[polynomial ring|0Q9EZS90GY]] $K[t]$, which is commutative. So we view $A$, $P$, $Q$ as matrices over $K[t]$. The products $QP$ and $PQ$ do not change.\n3. The scalar $t$ commutes with every matrix in products. So $Q (t I_n) P = t (Q P) = t I_n$, by the definition of the [[matrix product|076JGP6R3T]].\n4. By [[distributivity|0RXD6EZYBK]] and [[associativity|073E4XC016]], $Q (t I_n - A) P = Q (t I_n) P - Q A P = t I_n - Q A P$.\n5. By the definition of the [[characteristic polynomial|039HF58KQ5]] and step 4, $p_{QAP}(t) = \\det\\big(Q (t I_n - A) P\\big)$, a determinant over $K[t]$.\n6. By the [[product rule|04CPJ6AF00]] over $K[t]$, this equals $\\det Q \\cdot \\det(t I_n - A) \\cdot \\det P$.\n7. By commutativity of $K[t]$ and the product rule, this equals $\\det(Q P) \\det(t I_n - A) = \\det I_n \\cdot p_A(t)$.\n8. By the [[determinant of the identity|0Z4BWARC38]], $\\det I_n = 1$. So $p_{P^{-1} A P}(t) = p_A(t)$.\n9. If $B$ is [[similar|027KDW2K72]] to $A$, then $B = P^{-1} A P$ for an invertible $P$. By step 8, $p_B = p_A$.\n\n## Idea\n\nThe matrix $t I$ commutes with everything, so $t I - P^{-1} A P = P^{-1} (t I - A) P$. Taking determinants, the factors $\\det P^{-1}$ and $\\det P$ cancel.\n",
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