Consider an operator with one block [[2,1],[0,2]] and a separate scalar block 5. The eigenspace for 2 is only one-dimensional, but the first two coordinate directions belong to ker(T−2I)². On the final direction, T−2I is multiplication by 3 and no positive power kills it. The larger kernel captures the full two-dimensional component associated with eigenvalue 2, including its non-eigenvector direction.
Predict
Predict whether generalized eigenspaces for different eigenvalues can overlap in a nonzero vector. Repeated powers can kill more vectors than a first power, but the different linear factors remain coprime. This algebraic separation will force their intersection to be zero. The required exponents need not be identical; using dimension n as a uniform exponent is convenient, while the minimal polynomial often supplies smaller exact exponents.
Experiment
For the three-dimensional block example, write the kernel dimensions of (T−2I)ᵏ for k=0,1,2,3 as 0,1,2,2. For λ=5 the corresponding list is 0,1,1,1. Identify the actual basis vectors in each kernel. Then replace the size-two block by a size-three block and predict the new kernel growth before calculating. This growth measures chain length, not just the number of eigenvectors.
Understand
If polynomials p and q are coprime, Bézout gives a(t)p(t)+b(t)q(t)=1. When p(T)q(T)=0, each v decomposes as b(T)q(T)v+a(T)p(T)v. The first summand is in ker p(T), the second in ker q(T), and a vector killed by both must be zero by the same identity. This proves a direct-sum decomposition. Apply it repeatedly to pairwise coprime powers of irreducible factors of m_T to obtain primary components. If m_T splits as ∏(t−λᵢ)^rᵢ, these components are ker(T−λᵢI)^rᵢ, called generalized eigenspaces. Each is invariant because T commutes with its defining polynomial. On that component T=λᵢI+Nᵢ with Nᵢ nilpotent.
Apply
Why may one use ker(T−λI)ⁿ in dimension n? The increasing sequence of kernels can strictly grow at most n times. Once two successive kernels agree, the sequence stabilizes: apply the equality to (T−λI)v to prove the next equality, then induct. The primary decomposition shows the stabilized kernel is exactly the λ-component. On every other primary component T−λI is invertible, by a Bézout relation between t−λ and the corresponding factor. Explain this invertibility rather than assuming that an absence of visible eigenvectors is enough in arbitrary infinite-dimensional settings.
Canonical components do not require canonical projectors to be guessed. In the coprime decomposition, the polynomial operators b(T)q(T) and a(T)p(T) act as identity on their assigned kernels and zero on the other. Their sum is I. Different Bézout coefficients give the same resulting projections because every vector already has a unique decomposition into the two kernels. Verify this assertion separately on each component, then use their spanning property.
Continue
Over a field where the characteristic polynomial does not split, irreducible primary factors still exist, but they need not correspond to scalar eigenvalues in that field. A real rotation provides the basic example. The next lesson chooses chain bases for nilpotent pieces and obtains Jordan form only under the split-field hypothesis. Primary decomposition is canonical as a decomposition into invariant subspaces; the individual basis vectors inside those pieces are not canonical.
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A is block diagonal with a 3 by 3 Jordan block for 2 and a 2 by 2 Jordan block for 5. Find dim ker(A−2I)⁵.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
N has nilpotent Jordan blocks of sizes 3 and 2. Find dim ker N².
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Suppose p and q are coprime and p(T)q(T)=0. Prove V=ker p(T)⊕ker q(T).
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Why can a generalized λ-eigenvector not also be a nonzero generalized μ-eigenvector when λ≠μ?
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What replaces real generalized eigenspaces for a real rotation with polynomial t²+1?
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Prove ker Nʳ=ker Nʳ⁺¹ implies ker Nʳ⁺¹=ker Nʳ⁺².
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Reflect and continue
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