Generalized Eigenspace Decomposition

We use the following named lemmas, each proved from scratch, and the technique of primary decomposition via coprime polynomial factors of the characteristic (equivalently minimal) polynomial, combined with Cayley–Hamilton.

Step 0: Setup via Cayley–Hamilton. By the Cayley-Hamilton Theorem, the characteristic polynomial $\chi_T(z) = \det(zI-T)$ satisfies $\chi_T(T) = 0$. Since $\mathbb{C}$ is algebraically closed, $\chi_T$ factors completely:

where $\lambda_1,\dots,\lambda_m$ are the distinct roots (eigenvalues) and $d_i \ge 1$ their multiplicities. Since $\chi_T(T)=0$, we have

Lemma A (kernels of coprime polynomials in $T$ split the space). Suppose $p_1(z),\dots,p_m(z)$ are polynomials, pairwise coprime (no common root, since we work over $\mathbb C$ this means no shared linear factor), such that $p_1(T)p_2(T)\cdots p_m(T) = 0$. Let $W_i = \ker p_i(T)$. Then $V = W_1 \oplus \cdots \oplus W_m$. *Proof of Lemma A. Apply this with $p_i(z) = (z-\lambda_i)^{d_i}$; these are pairwise coprime since the $\lambda_i$ are distinct. We prove the general lemma by induction on $m$. Base case $m=1$: trivial, $p_1(T) = 0$ forces $W_1 = \ker p_1(T) = V$. Case $m=2$:* Since $p_1,p_2$ are coprime, by Bézout's identity for polynomials over the field $\mathbb C$, there exist polynomials $a(z), b(z)$ with

Substituting $T$ (all these are polynomials in the single operator $T$, so they commute):

For $v \in V$, write

Since $p_1(T)p_2(T) = 0$ (as $p_1p_2$ divides the given product, or directly from $(\ast)$ when $m=2$), we get $p_2(T) v_2 = p_2(T)a(T)p_1(T) v= a(T)\big(p_1(T)p_2(T)\big) v = 0$, so $v_2 \in \ker p_2(T) = W_2$. Similarly $v_1 = b(T)p_2(T)v \in \ker p_1(T)=W_1$ because $p_1(T)v_1 = b(T) p_1(T)p_2(T) v = 0$. This shows $V = W_1 + W_2$. For the sum to be direct, suppose $w_1 + w_2 = 0$ with $w_i \in W_i$. Applying $a(T)p_1(T)$ to $0 = w_1+w_2$: $a(T)p_1(T)w_1 + a(T)p_1(T) w_2 = 0$. But $a(T)p_1(T)w_2 = a(T)p_1(T) w_2$; using $a(T)p_1(T) = I - b(T)p_2(T)$, we get $a(T)p_1(T) w_2 = w_2 - b(T)p_2(T) w_2 = w_2 - 0 = w_2$ since $p_2(T) w_2 = 0$. Meanwhile $a(T)p_1(T) w_1 = 0$ because $p_1(T) w_1 = 0$. Hence the equation becomes $0 + w_2 = 0$, i.e. $w_2 = 0$, and then $w_1 = 0$ too. So $W_1 \cap W_2 = \{0\}$, and $V = W_1 \oplus W_2$. Inductive step: Suppose the lemma holds for $m-1$ pairwise coprime polynomials. Given $p_1,\dots,p_m$ pairwise coprime with $\prod p_i(T) = 0$, set $q(z) = \prod_{i=2}^m p_i(z)$. Since $p_1$ is coprime to each $p_i$ ($i\ge 2$), and coprimality of $p_1$ with each factor of $q$ implies coprimality of $p_1$ with $q$ (a standard property of coprime polynomials: if $\gcd(p_1,p_i)=1$ for all $i$, then $\gcd(p_1, \prod p_i) = 1$, provable by Bézout multiplicatively), $p_1$ and $q$ are coprime. Also $p_1(T) q(T) = \prod_i p_i(T) = 0$. By the $m=2$ case, $V = \ker p_1(T) \oplus \ker q(T) = W_1 \oplus \ker q(T)$. Now restrict attention to $U := \ker q(T)$. We claim $U = \ker p_2(T) \oplus \cdots \oplus \ker p_m(T)$ inside $U$, using the inductive hypothesis applied to $T|_U$ (note $U$ is $T$-invariant since $T$ commutes with $q(T)$, so $q(T)Tu = Tq(T)u = 0$ for $u \in U$). We have $q(T|_U) = 0$ by definition of $U$, matching the hypothesis of the inductive claim for $p_2,\ldots,p_m$ pairwise coprime. Hence $U = \ker(p_2(T)|_U)\oplus\cdots\oplus\ker(p_m(T)|_U)$. Since for $i \ge 2$, $\ker(p_i(T)|_U) = \ker p_i(T) \cap U = \ker p_i(T)$ — the last equality because $\ker p_i(T) \subseteq U=\ker q(T)$ automatically (if $p_i(T) v = 0$ then $q(T) v = 0$ too, since $p_i \mid q$) — we conclude

Combining, $V = W_1 \oplus U = W_1 \oplus W_2 \oplus \cdots \oplus W_m$. This completes the induction and proves Lemma A. $\blacksquare$

Applying Lemma A. Take $p_i(z) = (z-\lambda_i)^{d_i}$ in $(\ast)$. These are pairwise coprime (distinct linear factors, over $\mathbb C$ coprimality of powers of distinct linear polynomials is immediate: any common root would have to be a root of both $z - \lambda_i$ and $z-\lambda_j$, impossible for $\lambda_i \ne \lambda_j$). Lemma A gives

Step 1: Identify $\ker(T-\lambda_i I)^{d_i}$ with $G(\lambda_i,T)$. By definition $G(\lambda_i,T) = \bigcup_k \ker(T-\lambda_iI)^k$, and this chain stabilizes at $k=n=\dim V$ at the latest (shown in the definition above). We must show it in fact stabilizes by $k = d_i$, matching the exponent from Cayley–Hamilton, and that no more room is needed. First, $\ker(T-\lambda_i I)^{d_i} \subseteq G(\lambda_i,T)$ trivially, by definition of $G(\lambda_i,T)$ as a union including the $k=d_i$ term. Conversely, we show $G(\lambda_i,T) \subseteq \ker(T-\lambda_i I)^{d_i}$, i.e. the chain of kernels for $\lambda_i$ has already stabilized by step $d_i$. Suppose $v \in \ker(T-\lambda_iI)^k$ for some $k > d_i$; we show $v \in \ker(T-\lambda_i I)^{d_i}$. Write $v = \sum_j w_j$ with $w_j \in G(\lambda_j,T) = \ker(T-\lambda_j I)^{d_j}$ using the direct sum decomposition just established (this uses the decomposition as a set-theoretic fact, not yet assuming the dimension count, so there is no circularity). Apply $(T-\lambda_i I)^k$:

For $j = i$: since $w_i \in \ker(T-\lambda_i I)^{d_i}$ and $k > d_i \ge d_i$, certainly $(T-\lambda_i I)^k w_i = 0$ as well (apply extra copies of $T - \lambda_i I$ to the already-zero vector $(T-\lambda_i I)^{d_i} w_i$). For $j \neq i$: we claim $(T - \lambda_i I)$ restricted to $\ker(T-\lambda_j I)^{d_j}$ is invertible on that subspace. Indeed, on $\ker(T-\lambda_j I)^{d_j}$, the operator $T - \lambda_j I$ is nilpotent (its $d_j$-th power is zero there by definition of the kernel), so $T-\lambda_i I = (T - \lambda_j I) + (\lambda_j - \lambda_i) I$ is (nilpotent) $+$ (nonzero scalar times identity, since $\lambda_i \ne \lambda_j$), which is invertible: if $N$ is nilpotent with $N^r=0$ and $c \ne 0$, then $N + cI$ is invertible with inverse $\frac1c\sum_{t=0}^{r-1}(-N/c)^t$, as one checks by direct multiplication (a telescoping geometric-series identity, valid since $N$ and $I$ commute). Hence $T-\lambda_i I$ is invertible on $\ker(T-\lambda_jI)^{d_j}$, so any power $(T-\lambda_iI)^k$ is also invertible there (composition of invertible maps). So from $0 = \sum_j (T-\lambda_iI)^kw_j = (T-\lambda_i I)^k w_i + \sum_{j\ne i}(T-\lambda_i I)^k w_j$, and the first term is already $0$, we get $\sum_{j \ne i} (T-\lambda_i I)^k w_j = 0$. Since $(T - \lambda_iI)^k$ is injective on each $\ker(T-\lambda_jI)^{d_j}$ for $j \ne i$ (invertibility just shown), and these subspaces intersect trivially by directness of the sum, injectivity on each summand plus the vanishing of the total sum forces each term $(T-\lambda_i I)^k w_j$, $j\ne i$, to vanish individually: more precisely, apply the same argument used for uniqueness of direct-sum decompositions — the vectors $(T-\lambda_iI)^kw_j$ for $j \neq i$ lie in the respective $\ker(T-\lambda_j I)^{d_j}$ (since these are $T$-invariant, being kernels of a polynomial in $T$, hence invariant under the commuting operator $(T-\lambda_iI)^k$), so their sum being $0$ with the subspaces in direct sum forces each to be $0$. By injectivity of $(T-\lambda_iI)^k$ on $\ker(T-\lambda_jI)^{d_j}$, this gives $w_j = 0$ for all $j \ne i$. Therefore $v = w_i \in \ker(T-\lambda_iI)^{d_i}$, as required. This proves $G(\lambda_i, T) = \ker(T-\lambda_iI)^{d_i}$, establishing part 3's claim that the generalized eigenspace equals this specific kernel, and combined with Step 0's decomposition proves part 2.

Step 2: Dimension count. From $V = \bigoplus_i \ker(T-\lambda_iI)^{d_i} = \bigoplus_i G(\lambda_i,T)$ we get

We must show $\dim G(\lambda_i,T) = d_i$ exactly (not merely that the dimensions sum correctly). Fix $i$ and let $U_i = G(\lambda_i,T) = \ker(T-\lambda_iI)^{d_i}$. Since $U_i$ is $T$-invariant (kernel of a polynomial in $T$ commutes with $T$: if $(T-\lambda_iI)^{d_i}u=0$ then $(T-\lambda_i I)^{d_i}(Tu) = T(T-\lambda_iI)^{d_i}u = 0$), we may consider $S := T|_{U_i}$. On $U_i$, $S - \lambda_i I$ is nilpotent by construction (its $d_i$-th power is zero). Choose a basis of $V$ adapted to the direct sum $V = U_1\oplus\cdots\oplus U_m$ by concatenating bases of each $U_j$; in this basis $T$ is block diagonal with blocks $T|_{U_j}$. The characteristic polynomial multiplies over a block-diagonal matrix:

Since $T|_{U_i} - \lambda_i I$ is nilpotent on $U_i$ (dimension $\dim U_i$), its only eigenvalue is $0$ (a nilpotent operator's characteristic polynomial is $z^{\dim U_i}$, since all eigenvalues of a nilpotent operator are $0$ — if $Nv=\mu v$ with $v\ne 0$ then $N^r v = \mu^r v$, and $N^r=0$ forces $\mu^r=0$ hence $\mu=0$, and since we're over $\mathbb C$ the char. poly of $N$ splits and can only have root $0$). Hence $\chi_{T|_{U_i}-\lambda_iI}(z) = z^{\dim U_i}$, and shifting back,

Substituting into the product formula:

But we already know from Step 0 that $\chi_T(z) = \prod_j (z-\lambda_j)^{d_j}$, with the $\lambda_j$ distinct. Unique factorization of polynomials over $\mathbb{C}$ (or over any field) into irreducible (here linear) factors forces the exponents to match termwise:

This proves part 3.

Step 3: Invariance and structure of the restriction. We already noted $G(\lambda_i,T)=U_i$ is $T$-invariant, proving part 1. Define $N_i := (T-\lambda_iI)|_{U_i}$, an operator on $U_i$. By construction $N_i^{d_i} = 0$ on $U_i$ (since $U_i = \ker(T-\lambda_iI)^{d_i}$), so $N_i$ is nilpotent. And $T|_{U_i} = \lambda_i I|_{U_i} + N_i$ by rearranging the definition of $N_i$. This proves part 4, and completes the proof of the theorem. $\blacksquare$