Proof that a matrix is diagonalisable iff its minimal polynomial has distinct roots

Proof

We prove the two implications separately. The converse uses Lagrange interpolation polynomials.

1. 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 of endomorphisms to matrices.
2. 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 $p$.
3. By step 2, $p(C) = 0$ if and only if $p(A) = 0$. Hence $m_C = m_A$.
4. First let $A$ be diagonalisable. 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 $f = \prod_{j=1}^{k} (t - \mu_j)$.
5. 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 $f(D) = 0$.
6. By step 5 and divisibility by the minimal polynomial, $m_D$ divides $f$.
7. 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.
8. 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 $r < k$.
9. 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 $h(\mu_{r+1}) = 0$.
10. By the factor theorem, $h = (t - \mu_{r+1}) h'$. So $g_{r+1}$ divides $m_D$, which completes the induction.
11. 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 in the integral domain $F[t]$, $a b = 1$, so $a$, $b$ are constants. Both polynomials are monic, so $a = 1$ and $m_D = f$.
12. By steps 3 and 11, $m_A = f = (t - \mu_1) \cdots (t - \mu_k)$ with pairwise distinct $\mu_j \in F$.
13. 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 $A$.
14. 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.
15. Let $h = \sum_{i=1}^{k} c_i^{-1} f_i$. For each $l$, $f_i(\lambda_l) = 0$ when $i \ne l$. So $h(\lambda_l) = c_l^{-1} c_l = 1$.
16. 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, so $h = 1$.
17. 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 $v = u_1 + \cdots + u_k$.
18. 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 of $\lambda_i$.
19. So each $u_i$ is zero or an eigenvector of $A$. By step 17, the set $S$ of all eigenvectors of $A$ spans $F^n$.
20. Every spanning set contains a basis, so $F^n$ has a basis contained in $S$. This is a basis of eigenvectors of $A$.
21. By the characterisation by eigenvectors, $A$ is diagonalisable over $F$.