Generalized Eigenspace Decomposition

Definition A polynomial with coefficients in $\mathbb{F}$ splits over $\mathbb{F}$ if it is a constant times a product of linear factors $t - c$ with every $c \in \mathbb{F}$. The minimal polynomial of $T$ splits over $\mathbb{F}$ when

with distinct $\lambda_1, \dots, \lambda_r \in \mathbb{F}$ and exponents $s_j \ge 1$. The roots of $\mu_T$ in $\mathbb{F}$ are exactly the eigenvalues of $T$, so the $\lambda_j$ are then all the eigenvalues of $T$. Over $\mathbb{C}$ every polynomial splits, by the fundamental theorem of algebra, so the hypothesis holds for every complex operator. Over $\mathbb{R}$ it holds exactly when every complex root of $\mu_T$ is real. The factors $a_j = (t - \lambda_j)^{s_j}$ are pairwise coprime, even when some exponents exceed $1$, because no two of them have a common root.

Definition Let $\mu_T = a_1 a_2 \cdots a_r$ with monic polynomials $a_j$ of positive degree that are pairwise coprime, and let $b_j$ be the product of all the factors except $a_j$. Each $b_j$ is coprime to $a_j$, so Bézout’s identity gives polynomials $u_j, v_j$ with $u_j a_j + v_j b_j = 1$. Put $h_j = v_j b_j$ and $P_j = h_j(T)$, and let $K_j = \ker a_j(T)$. Then

the image of $P_j$ is $K_j$, each $K_j$ is invariant under $T$, and

with $P_j$ the projection onto $K_j$ along the sum of the other pieces. When the $a_j$ are the powers of the distinct monic irreducible factors of $\mu_T$, this is the primary decomposition of $V$ under $T$. When $\mu_T$ splits, those factors are $a_j = (t - \lambda_j)^{s_j}$.

For the proof, note that $h_j = 1 - u_j a_j$ leaves remainder $1$ on division by $a_j$, while $h_k$ for $k \ne j$ is a multiple of $a_j$, because $a_j$ is a factor of $b_k$. So every $a_j$ divides $1 - (h_1 + \cdots + h_r)$. Pairwise coprime polynomials that each divide $f$ have a product that divides $f$. For two of them, $a$ and $b$ with $ua + vb = 1$, write $f = ag$. Then $g = uag + vbg$, and $b$ divides both terms because $ag = f$, so $b$ divides $g$ and $ab$ divides $f$. Induction handles more factors. Hence $\mu_T$ divides $1 - \sum h_k$, and evaluating at $T$ gives $\sum P_k = I$. For $j \ne k$ the product $b_j b_k$ contains every $a_i$, so $\mu_T$ divides $h_j h_k$ and $P_j P_k = 0$. Multiplying $\sum P_k = I$ by $P_j$ gives $P_j^2 = P_j$. Since $a_j b_j = \mu_T$, we get $a_j(T) P_j = v_j(T) \mu_T(T) = 0$, so the image of $P_j$ lies in $K_j$, and on $K_j$ the operator $P_j = I - u_j(T) a_j(T)$ is the identity. Every $x \in V$ equals $P_1 x + \cdots + P_r x$ with $P_k x \in K_k$, so the pieces span $V$. Finally, if $w_1 + \cdots + w_r = 0$ with $w_k \in K_k$, then $P_i$ fixes $w_i$ and kills every other $w_k$, since $a_k$ is a factor of $b_i$. Applying $P_i$ gives $w_i = 0$, so the sum is direct. Each $K_j$ is invariant because $a_j(T)$ commutes with $T$.

Definition Suppose $\mu_T$ splits over $\mathbb{F}$ as above. Then for each $j$

where $G_{\lambda}(T) = \ker (T - \lambda I)^n$ is the generalized eigenspace, and

Each $G_{\lambda_j}(T)$ is invariant under $T$, and the projection onto it along the others is the polynomial projector $P_j$. The operator need not be diagonalizable. The splitting hypothesis cannot be dropped. If $V$ is the sum of the generalized eigenspaces for some eigenvalues $\lambda_1, \dots, \lambda_r$ in $\mathbb{F}$, then $\prod_j (t - \lambda_j)^n$ annihilates $T$, so $\mu_T$ divides it and splits.

For the proof write $K_j = \ker (T - \lambda_j I)^{s_j}$, so that $V = K_1 \oplus \cdots \oplus K_r$ by the primary decomposition. Since $\mu_T$ divides $\chi_T$, which has degree $n$, we have $s_j \le n$ and $K_j \subseteq G_{\lambda_j}(T)$. For $k \ne j$ the operator $T - \lambda_j I$ is injective on $K_k$. Indeed, if $w \in K_k$ and $Tw = \lambda_j w$, then $0 = (T - \lambda_k I)^{s_k} w = (\lambda_j - \lambda_k)^{s_k} w$, so $w = 0$. Now let $x \in G_{\lambda_j}(T)$ and write $x = x_1 + \cdots + x_r$ with $x_k = P_k x \in K_k$. Then

and each term lies in $K_k$ by invariance. The sum is direct, so every term is zero. For $k \ne j$ the power $(T - \lambda_j I)^n$ is injective on $K_k$, so $x_k = 0$. Hence $x = x_j \in K_j$.

Definition A linear operator $N$ on $V$ is nilpotent if $N^k = 0$ for some positive integer $k$, and the smallest such $k$ is its index of nilpotency. On a space of dimension $d \ge 1$ a nilpotent operator satisfies $N^d = 0$, because the kernels of its powers stabilize by step $d$ and must eventually be the whole space. Its only eigenvalue is $0$. Zero is an eigenvalue because an injective $N$ would have injective powers, yet $N^k = 0$ is not injective. No other scalar is an eigenvalue, since $Nw = \alpha w$ with $w \ne 0$ gives $0 = N^k w = \alpha^k w$, so $\alpha = 0$. When $\mu_T$ splits as above, the restriction $N_j$ of $T - \lambda_j I$ to $G_{\lambda_j}(T)$ is nilpotent with index exactly $s_j$, so on that subspace

The power $N_j^{s_j}$ is zero because $G_{\lambda_j}(T) = \ker (T - \lambda_j I)^{s_j}$. If $N_j^{r} = 0$ for some $r < s_j$, let $q$ be $\mu_T$ with the factor $(t - \lambda_j)^{s_j}$ replaced by $(t - \lambda_j)^{r}$. Then $q(T)$ kills $G_{\lambda_j}(T)$ by assumption and every other $G_{\lambda_k}(T)$ through its own factor. These subspaces span $V$, so $q(T) = 0$ with $\deg q < \deg \mu_T$, which contradicts minimality. So the restriction of $T$ has minimal polynomial $(t - \lambda_j)^{s_j}$, and $s_j$ is the largest rank of a generalized eigenvector for $\lambda_j$.

Definition Suppose $\mu_T$ splits over $\mathbb{F}$ as above, and let $m_j$ be the algebraic multiplicity of $\lambda_j$, its multiplicity as a root of $\chi_T$. Then $\dim G_{\lambda_j}(T) = m_j$ and

In particular $\chi_T$ splits too, the dimensions $m_1 + \cdots + m_r$ add up to $n$, and the geometric multiplicity $\dim \ker (T - \lambda_j I)$ is at most $m_j$, because the eigenspace lies inside $G_{\lambda_j}(T)$.

For the proof, join bases of the generalized eigenspaces into a basis of $V$. The sum is direct and each piece is invariant, so the matrix of $T$ is block diagonal, with a block $A_j$ of size $d_j = \dim G_{\lambda_j}(T)$ for each restriction, and $\chi_T = \prod_j \chi_{A_j}$. Each block has the form $A_j = \lambda_j I + M_j$ with $M_j$ nilpotent. Every complex root of $\chi_{M_j}$ is an eigenvalue of $M_j$ viewed as a complex matrix, which is still nilpotent, so that root is $0$. A monic polynomial of degree $d_j$ whose only complex root is $0$ equals $t^{d_j}$, hence

Multiplying over $j$ and comparing the exponents of the distinct factors $t - \lambda_j$ gives $m_j = d_j$. No Jordan basis is needed for this.