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  "body": "\n## Proof\n\nNecessity is a computation. For sufficiency we split eigenvectors of $A$ into eigenvectors of $B$ with polynomials in $B$.\n\n1. Suppose $P$ is invertible and $P^{-1} A P = D_1$ and $P^{-1} B P = D_2$ are [[diagonal|0Q9PGXD9E4]]. The product of two diagonal matrices is diagonal with the products of the diagonal entries, so $D_1 D_2 = D_2 D_1$.\n2. Then $A B = P D_1 P^{-1} P D_2 P^{-1} = P D_1 D_2 P^{-1} = P D_2 D_1 P^{-1} = B A$. So $A B = B A$ is necessary.\n3. Conversely, let $A B = B A$. If $n = 0$, then $P = I_0$ works, so let $n \\ge 1$. Since $B$ is [[diagonalisable|0Y17PEKF24]], its minimal polynomial is $m_B = (t - \\mu_1) \\cdots (t - \\mu_l)$ with pairwise distinct $\\mu_j \\in F$, by the [[minimal polynomial criterion|0MPAJRB23T]]. Here $l \\ge 1$, since the constant $1$ does not annihilate $B$.\n4. For each $j$, let $f_j = \\prod_{i \\ne j} (t - \\mu_i)$ and $c_j = f_j(\\mu_j) \\ne 0$. Let $h = \\sum_j c_j^{-1} f_j$. Then $h(\\mu_i) = 1$ for each $i$, since $f_j(\\mu_i) = 0$ for $j \\ne i$.\n5. The polynomial $h - 1$ has degree at most $l - 1$ and $l$ roots. A nonzero polynomial has [[at most as many roots as its degree|0KJVQV3E0F]], so $h = 1$. Hence $\\sum_j c_j^{-1} f_j(B) = I_n$, using the [[value at $B$|0BNEXSKNRV]].\n6. Since $A$ commutes with $B$, it commutes with each power of $B$, hence with $g(B)$ for every polynomial $g$.\n7. Let $w$ be an [[eigenvector|0B5H4208Z9]] of $A$, with $A w = \\lambda w$. Let $u_j = c_j^{-1} f_j(B) w$. By step 5, $w = u_1 + \\cdots + u_l$.\n8. By step 6, $A u_j = c_j^{-1} f_j(B) A w = \\lambda u_j$.\n9. Since $(t - \\mu_j) f_j = m_B$, we get $(B - \\mu_j I_n) u_j = c_j^{-1} m_B(B) w = 0$. So $B u_j = \\mu_j u_j$.\n10. By steps 8 and 9, each nonzero $u_j$ is an eigenvector of both $A$ and $B$. Call such a vector a common eigenvector, and let $S$ be the set of common eigenvectors.\n11. By step 7, every eigenvector of $A$ is a sum of elements of $S$.\n12. Since $A$ is diagonalisable, $F^n$ has a basis of eigenvectors of $A$, by the [[characterisation by eigenvectors|056SW1E0AC]]. With step 11, $S$ spans $F^n$.\n13. [[Every spanning set contains a basis|06MRNP4BNT]]. So there is a basis $v_1, \\dots, v_n$ of $F^n$ contained in $S$.\n14. Let $P$ be the matrix with columns $v_1, \\dots, v_n$. Each $v_i$ is an eigenvector of $A$ and of $B$. By the [[formula for $P^{-1} A P$|056SW1E0AC]], both $P^{-1} A P$ and $P^{-1} B P$ are diagonal.\n\n## Idea\n\nCommuting matrices preserve each other's eigenspaces. Concretely, the interpolation polynomials in $B$ that add up to $I_n$ split each eigenvector of $A$ into pieces. Each piece is an eigenvector of $B$ and still an eigenvector of $A$. So common eigenvectors span the space, and a basis of them diagonalises both matrices.\n",
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