Polynomial Projectors and Diagonalizability

Definition Let $a, b \in \mathbb{F}[t]$, not both zero. A common divisor of $a$ and $b$ is a polynomial that divides both. The greatest common divisor $\gcd(a, b)$ is the monic common divisor that is divisible by every common divisor, and it is also the monic common divisor of largest degree. The Euclidean algorithm computes it when $b \ne 0$. Set $r_0 = a$ and $r_1 = b$, and as long as $r_k \ne 0$ divide

with $r_{k+1} = 0$ or $\deg r_{k+1} < \deg r_k$. The degrees of the nonzero remainders strictly decrease, so after finitely many steps $r_{N+1} = 0$ while $r_N \ne 0$. Because $r_{k+1} = r_{k-1} - q_k r_k$, a polynomial divides both $r_{k-1}$ and $r_k$ exactly when it divides both $r_k$ and $r_{k+1}$. Following the chain to the pair $r_N, 0$ shows that the common divisors of $a$ and $b$ are exactly the divisors of $r_N$. So $\gcd(a, b)$ exists and equals $r_N$ divided by its leading coefficient. The algorithm uses only arithmetic with the coefficients, so for real polynomials it gives the same greatest common divisor over $\mathbb{R}$ as over $\mathbb{C}$.

Definition Polynomials $a, b \in \mathbb{F}[t]$, not both zero, are coprime if $\gcd(a, b) = 1$, that is, if their only common divisors are the nonzero constants. Over $\mathbb{C}$ two nonzero polynomials are coprime exactly when they have no common root. A common root $\lambda$ gives the common factor $t - \lambda$, and a nonconstant greatest common divisor has a complex root by the fundamental theorem of algebra. Since the greatest common divisor does not change from $\mathbb{R}$ to $\mathbb{C}$, two real polynomials are coprime exactly when they have no common complex root. Comparing roots also shows that if $f_1, \dots, f_s$ are pairwise coprime, then $f_1$ is coprime to the product $f_2 \cdots f_s$.

Definition Let $a, b \in \mathbb{F}[t]$, not both zero, and $g = \gcd(a, b)$. Bézout’s identity says that there are $u, v \in \mathbb{F}[t]$ with

To find them, run the Euclidean algorithm. Each remainder $r_{k+1} = r_{k-1} - q_k r_k$ is a combination of the two before it, so substituting backward expresses the last nonzero remainder $r_N$ as a combination of $r_0 = a$ and $r_1 = b$, and dividing by its leading coefficient gives $g$. In particular $a$ and $b$ are coprime if and only if $ua + vb = 1$ for some $u, v \in \mathbb{F}[t]$. The converse direction holds because a common divisor of $a$ and $b$ divides $ua + vb = 1$ and so is constant. The pair $u, v$ is not unique. For coprime $a, b$ and every $h \in \mathbb{F}[t]$, the polynomials $u + hb$ and $v - ha$ form a valid pair as well.

Definition Let $a, b \in \mathbb{F}[t]$ be coprime with $a(T)\,b(T) = 0$, which holds for example when $ab = \mu_T$ or $ab = \chi_T$. Choose $u, v$ with $ua + vb = 1$ and set

Then $P + Q = I$, $P^2 = P$, $Q^2 = Q$ and $PQ = QP = 0$, both commute with $T$, and

where both kernels are invariant under $T$. So $P$ is the projection onto $\ker a(T)$ along $\ker b(T)$, and it does not depend on the choice of $u$ and $v$. Evaluating $ua + vb = 1$ at $T$ gives $P + Q = I$. Next $a(T)P = v(T)\,a(T)\,b(T) = 0$, so $\operatorname{im} P \subseteq \ker a(T)$, and for $x \in \ker a(T)$ we get $Qx = u(T)\,a(T)\,x = 0$, hence $Px = x$. So $P$ maps onto $\ker a(T)$ and fixes it, which gives $P^2 = P$. The same argument works for $Q$. If $z$ lies in both kernels, then $z = v(T)\,b(T)\,z + u(T)\,a(T)\,z = 0$, so the sum is direct, and $x = Px + Qx$ shows that it is all of $V$. The product $PQ$ contains the factor $a(T)\,b(T)$, so it is $0$. A kernel $\ker f(T)$ is invariant because $f(T)\,Tx = T f(T)\,x$. For pairwise coprime $f_1, \dots, f_s$ with $(f_1 \cdots f_s)(T) = 0$, splitting off one factor at a time gives

Definition The operator $T$ is diagonalizable over $\mathbb{F}$, meaning that $V$ has a basis of eigenvectors of $T$, if and only if

for distinct $\lambda_1, \dots, \lambda_s \in \mathbb{F}$. Then the $\lambda_j$ are the distinct eigenvalues, $V = E_{\lambda_1} \oplus \cdots \oplus E_{\lambda_s}$ with $E_\lambda = \ker(T - \lambda I)$, and the projection onto $E_{\lambda_j}$ along the other eigenspaces is

To prove that the factorization implies diagonalizability, let $L_j(t) = \prod_{k \ne j} (t - \lambda_k)/(\lambda_j - \lambda_k)$, so that $L_j(\lambda_i)$ is $1$ for $i = j$ and $0$ otherwise. The polynomial $L_1 + \cdots + L_s - 1$ has degree at most $s - 1$ and the $s$ distinct roots $\lambda_1, \dots, \lambda_s$, so it is zero, and evaluating at $T$ gives $P_1 + \cdots + P_s = I$. Moreover $(t - \lambda_j)L_j(t)$ is a constant multiple of $\mu_T$, so $(T - \lambda_j I)P_j = 0$ and $\operatorname{im} P_j \subseteq E_{\lambda_j}$. Every $v = P_1 v + \cdots + P_s v$ is therefore a sum of vectors from the eigenspaces. Eigenspaces for distinct eigenvalues form a direct sum, so joining bases of the $E_{\lambda_j}$ gives an eigenbasis. The same splitting also follows from the decomposition into kernels in the previous card, because distinct linear factors $t - \lambda_j$ are pairwise coprime. Conversely, if $V$ has an eigenbasis and $\lambda_1, \dots, \lambda_s$ are its distinct eigenvalues, the polynomial $p(t) = (t - \lambda_1) \cdots (t - \lambda_s)$ satisfies $p(T)v = p(\lambda)v = 0$ for each basis vector $v$ with eigenvalue $\lambda$, so $p(T) = 0$ and $\mu_T \mid p$. Every $\lambda_j$ is a root of $\mu_T$, so $\mu_T = p$. Equivalently, $T$ is diagonalizable over $\mathbb{F}$ exactly when some product of distinct linear factors over $\mathbb{F}$ annihilates it.

Definition Let $T_1, \dots, T_k \in \operatorname{Hom}(V, V)$ be diagonalizable over $\mathbb{F}$ and pairwise commuting, $T_i T_j = T_j T_i$. Then $V$ has a basis whose vectors are eigenvectors of every $T_i$. In matrix terms, commuting diagonalizable $A_1, \dots, A_k \in \mathbb{F}^{n \times n}$ have one invertible $S \in \mathbb{F}^{n \times n}$ with every $S^{-1}A_i S$ diagonal. Conversely, operators with a common eigenbasis commute, because diagonal matrices commute. The proof is by induction on $k$, and $k = 1$ is the definition. Let $E$ be an eigenspace of $T_1$ for the eigenvalue $\lambda$. For $v \in E$,

so $T_j E \subseteq E$. The polynomial $\mu_{T_j}$ annihilates the restriction $T_j|_E$, so the minimal polynomial of the restriction divides $\mu_{T_j}$ and is again a product of distinct linear factors over $\mathbb{F}$. By the distinct-root criterion every $T_j|_E$ is diagonalizable, and the restrictions of $T_2, \dots, T_k$ still commute. By induction $E$ has a basis of common eigenvectors of $T_2, \dots, T_k$, and they are eigenvectors of $T_1$ too. Joining these bases over the eigenspaces of $T_1$, whose direct sum is $V$, gives the common eigenbasis. The basis need not be orthonormal. That stronger conclusion holds for a commuting family of normal operators on a finite-dimensional complex inner product space, which has an orthonormal basis of common eigenvectors.