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  "body": "\n## Proof\n\nThe [[trace of $T$|0KZDNGA561]] is defined as $\\tr [T]_{B_0}$ for some basis $B_0$ of $V$. We show that every basis gives the same number.\n\n1. Let $B$ and $B'$ be two bases of $V$. Let $M = [T]_B$ and $M' = [T]_{B'}$ be the [[matrices of $T$|0YQSGNYE3K]], and let $n = \\dim V$.\n2. Let $P$ be the [[change of basis matrix|0MKRHZB3WS]] from $B$ to $B'$. By the [[change of basis formula|0MQXVZEBMT]], $P$ is [[invertible|0EB9N0S33C]] and $M' = P^{-1} M P$.\n3. By [[associativity|073E4XC016]], $M' = P^{-1} (MP)$.\n4. By the [[symmetry of the trace of a product|0Y4NF1DRKD]], applied to the $n \\times n$ matrices $P^{-1}$ and $MP$, we get $\\tr\\left(P^{-1}(MP)\\right) = \\tr\\left((MP)P^{-1}\\right)$.\n5. By associativity again, $(MP)P^{-1} = M(PP^{-1}) = M I_n = M$.\n6. By steps 3 to 5, $\\tr M' = \\tr M$, with the [[trace of a matrix|0XC1RDBY7A]]. So $\\tr [T]_{B'} = \\tr [T]_B$ for any two bases $B$ and $B'$.\n7. In particular, taking $B' = B_0$, we get $\\tr T = \\tr [T]_{B_0} = \\tr [T]_B$ for the given basis $B$.\n\n## Idea\n\nChanging the basis replaces the matrix $M$ by $P^{-1} M P$. The trace does not see the order of the factors in a product, so the factors $P^{-1}$ and $P$ cancel inside the trace. This makes the trace a property of the map and not of the chosen basis.\n",
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        "locator": "Trace of a linear operator",
        "url": "https://en.wikipedia.org/wiki/Trace_(linear_algebra)"
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