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  "body": "\n## Proof\n\nWe show that every basis gives the same value of $\\det [T]_B$. Then the definition of $\\det T$ gives this value for each basis.\n\n1. Let $n = \\dim V$. The [[determinant|0C3CFHDN6R]] is defined for $n \\times n$ matrices with $n \\ge 1$, so $n \\ge 1$ here.\n2. Let $B$ and $B'$ be two bases of $V$. Let $[T]_B$ and $[T]_{B'}$ be the [[matrices of $T$|0YYMY95FWS]] in these bases.\n3. Let $P$ be the [[change of basis matrix|0KWXYZPJKN]] from $B$ to $B'$. By the [[change of basis formula|0MQXVZEBMT]], $P$ is invertible and $[T]_{B'} = P^{-1} [T]_B P$.\n4. By the [[product rule|04CPJ6AF00]], $\\det [T]_{B'} = \\det(P^{-1}) \\det [T]_B \\det P$.\n5. The ring $F$ is commutative, so the right side equals $\\det(P^{-1}) \\det P \\cdot \\det [T]_B$.\n6. By the product rule, the definition of the [[inverse|0TWTS88TWN]] and the [[determinant of the identity|0Z4BWARC38]], $\\det(P^{-1}) \\det P = \\det(P^{-1} P) = \\det I_n = 1$.\n7. By steps 4 to 6, $\\det [T]_{B'} = \\det [T]_B$ for any two bases $B$ and $B'$.\n8. The [[determinant of $T$|03F5KP94K0]] is defined as $\\det [T]_{B'}$ for a chosen basis $B'$. By step 7, $\\det T = \\det [T]_B$ for the given basis $B$.\n\n## Idea\n\nChanging the basis replaces the matrix $[T]_B$ by a similar matrix $P^{-1} [T]_B P$. The determinant is multiplicative, so the factors $\\det P^{-1}$ and $\\det P$ cancel.\n",
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