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  "body": "\n## Proof\n\nWe prove the two implications separately. The converse uses Lagrange interpolation polynomials.\n\n1. For a square matrix $C$, its annihilating polynomials are those of the map $x \\mapsto C x$. Indeed, $p(C) x$ is the value of $p$ at this map applied to $x$. So we may apply results on [[minimal polynomials|0V58FPM3EQ]] of endomorphisms to matrices.\n2. Let $P$ be invertible and $C = P^{-1} A P$. By induction on $k$, $C^k = P^{-1} A^k P$. So $p(C) = P^{-1} p(A) P$ for each $p \\in F[t]$, by the definition of the [[value of $p$|0BNEXSKNRV]].\n3. By step 2, $p(C) = 0$ if and only if $p(A) = 0$. Hence $m_C = m_A$.\n4. First let $A$ be [[diagonalisable|0Y17PEKF24]]. Choose $P$ with $P^{-1} A P = D = \\operatorname{diag}(d_1, \\dots, d_n)$. Let $\\mu_1, \\dots, \\mu_k$ be the distinct values among $d_1, \\dots, d_n$, and let $f = \\prod_{j=1}^{k} (t - \\mu_j)$.\n5. Powers and linear combinations of diagonal matrices are computed entry by entry. So $f(D) = \\operatorname{diag}(f(d_1), \\dots, f(d_n))$, and each $f(d_i) = 0$. Hence $f(D) = 0$.\n6. By step 5 and [[divisibility by the minimal polynomial|0VTAD25A3G]], $m_D$ divides $f$.\n7. Each $\\mu_j$ equals some $d_i$, and $D e_i = d_i e_i$ for the standard basis vector $e_i$. So $\\mu_j$ is an eigenvalue of $D$, and $m_D(\\mu_j) = 0$ by [[the roots of the minimal polynomial|0QF4ZEFG52]].\n8. We show by induction on $r \\le k$ that $g_r = \\prod_{j=1}^{r} (t - \\mu_j)$ divides $m_D$. For $r = 0$, $g_0 = 1$. Suppose $m_D = g_r h$ with $r \u003c k$.\n9. Evaluating at $\\mu_{r+1}$ gives $0 = g_r(\\mu_{r+1}) h(\\mu_{r+1})$. The factor $g_r(\\mu_{r+1}) = \\prod_{j \\le r} (\\mu_{r+1} - \\mu_j)$ is nonzero, so $h(\\mu_{r+1}) = 0$.\n10. By the [[factor theorem|0K3MWA7DMC]], $h = (t - \\mu_{r+1}) h'$. So $g_{r+1}$ divides $m_D$, which completes the induction.\n11. By steps 6 and 10, $f$ and $m_D$ divide each other. Write $f = m_D a$ and $m_D = f b$. Then $f = f a b$. By [[cancellation|0GJM24EXP9]] in the integral domain $F[t]$, $a b = 1$, so $a$, $b$ are constants. Both polynomials are monic, so $a = 1$ and $m_D = f$.\n12. By steps 3 and 11, $m_A = f = (t - \\mu_1) \\cdots (t - \\mu_k)$ with pairwise distinct $\\mu_j \\in F$.\n13. Conversely, let $m_A = (t - \\lambda_1) \\cdots (t - \\lambda_k)$ with pairwise distinct $\\lambda_i \\in F$. If $n = 0$, then $A$ is the empty matrix, which is diagonal. So let $n \\ge 1$. Then $k \\ge 1$, since the constant $1$ does not annihilate $A$.\n14. For each $i$, let $f_i = \\prod_{j \\ne i} (t - \\lambda_j)$ and $c_i = f_i(\\lambda_i)$. Then $c_i \\ne 0$, because the $\\lambda_j$ are distinct.\n15. Let $h = \\sum_{i=1}^{k} c_i^{-1} f_i$. For each $l$, $f_i(\\lambda_l) = 0$ when $i \\ne l$. So $h(\\lambda_l) = c_l^{-1} c_l = 1$.\n16. The polynomial $h - 1$ has degree at most $k - 1$ and the $k$ roots $\\lambda_1, \\dots, \\lambda_k$. A nonzero polynomial has [[at most as many roots as its degree|0KJVQV3E0F]], so $h = 1$.\n17. By steps 2 and 16, $I_n = \\sum_i c_i^{-1} f_i(A)$. Let $v \\in F^n$ and $u_i = c_i^{-1} f_i(A) v$. Then $v = u_1 + \\cdots + u_k$.\n18. Since $(t - \\lambda_i) f_i = m_A$, we get $(A - \\lambda_i I_n) u_i = c_i^{-1} m_A(A) v = 0$. So $u_i$ lies in the [[eigenspace|0AYEPBQMAW]] of $\\lambda_i$.\n19. So each $u_i$ is zero or an [[eigenvector|0B5H4208Z9]] of $A$. By step 17, the set $S$ of all eigenvectors of $A$ spans $F^n$.\n20. [[Every spanning set contains a basis|06MRNP4BNT]], so $F^n$ has a basis contained in $S$. This is a basis of eigenvectors of $A$.\n21. By the [[characterisation by eigenvectors|056SW1E0AC]], $A$ is diagonalisable over $F$.\n\n## Idea\n\nA diagonal matrix is killed by the product of $t - \\mu$ over its distinct diagonal entries. Each of these factors is needed, since each $\\mu$ is a root of the minimal polynomial. Conversely, when $m_A$ has distinct roots, the interpolation polynomials $f_i / f_i(\\lambda_i)$ add up to $1$. Applied to a vector, they split it into eigenvectors.\n",
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