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  "body": "\n## Proof\n\nWe compute $T(b'_j)$ in the basis $C$ in two ways, and then show that $Q$ is invertible.\n\n1. Let $n = \\dim V$ and $m = \\dim W$. Write $B = (b_1, \\dots, b_n)$, $B' = (b'_1, \\dots, b'_n)$, $C = (c_1, \\dots, c_m)$ and $C' = (c'_1, \\dots, c'_m)$.\n2. By the definition of the [[change of basis matrix|0MKRHZB3WS]], $P = (p_{kj})$ and $Q = (q_{ik})$ satisfy $b'_j = \\sum_{k=1}^n p_{kj} b_k$ and $c'_k = \\sum_{i=1}^m q_{ik} c_i$.\n3. Let $M = [T]_{C,B} = (\\mu_{ik})$ and $M' = [T]_{C',B'} = (\\mu'_{kj})$. By the definition of the [[matrix of a linear map|0YQSGNYE3K]], $T(b_k) = \\sum_{i} \\mu_{ik} c_i$ and $T(b'_j) = \\sum_{k} \\mu'_{kj} c'_k$.\n4. Fix $j$. By step 3 and then step 2, $T(b'_j) = \\sum_{k=1}^m \\mu'_{kj} \\sum_{i=1}^m q_{ik} c_i = \\sum_{i=1}^m (Q M')_{ij}\\, c_i$, using the [[matrix product|032Q9FB8MT]].\n5. By step 2 and linearity, $T(b'_j) = \\sum_{k=1}^n p_{kj} T(b_k) = \\sum_{k=1}^n p_{kj} \\sum_{i=1}^m \\mu_{ik} c_i = \\sum_{i=1}^m (M P)_{ij}\\, c_i$.\n6. The [[coordinates|04CYQHN6MC]] of $T(b'_j)$ in the [[basis|0KCQ9Y3QKA]] $C$ are unique. Comparing steps 4 and 5 gives $(QM')_{ij} = (MP)_{ij}$ for all $i, j$. So $QM' = MP$.\n7. Now we show that $Q$ is invertible. Let $R = (r_{ki})$ be the change of basis matrix from $C'$ to $C$, so that $c_i = \\sum_{k=1}^m r_{ki} c'_k$.\n8. By steps 2 and 7, $c'_l = \\sum_{i=1}^m q_{il} c_i = \\sum_{k=1}^m \\left(\\sum_{i=1}^m r_{ki} q_{il}\\right) c'_k = \\sum_{k=1}^m (RQ)_{kl}\\, c'_k$.\n9. Also $c'_l = \\sum_k \\delta_{kl} c'_k$, where $\\delta_{kl}$ is $1$ if $k = l$ and $0$ otherwise. By uniqueness of coordinates in $C'$, $RQ = I_m$.\n10. By [[one-sided inverses of square matrices|0753Y2X5K6]], $Q$ is [[invertible|0EB9N0S33C]] with $Q^{-1} = R$.\n11. Multiply $QM' = MP$ on the left by $Q^{-1}$. By [[associativity|073E4XC016]], $M' = (Q^{-1} Q) M' = Q^{-1}(QM') = Q^{-1}(MP) = Q^{-1} M P$.\n12. So $[T]_{C',B'} = Q^{-1} [T]_{C,B} P$.\n13. For an endomorphism $T$ of $V$ with $C = B$ and $C' = B'$, $Q$ is the change of basis matrix from $B$ to $B'$. So $Q = P$, and $[T]_{B'} = P^{-1} [T]_B P$.\n\n## Idea\n\nExpand $T(b'_j)$ in the old output basis $C$ in two ways. Going through the new output basis $C'$ gives column $j$ of $QM'$. Going through the old input basis $B$ gives column $j$ of $MP$. Equating them and inverting $Q$ gives the formula.\n",
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