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I am trying to understand the proof of Theorem 2 from Paul R. Halmos's Finite-Dimensional Vector Spaces ($\S{58}$ Jordan form, pp 113-114) which is part of the proof of generalized eigenspace decomposition. Here is the theorem and proof:

the theorem

and proof

The only part I don't understand is in the end where it is concluded that

each of the subspaces $M_j$ is disjoint from the span of all others.

It feels like I'm missing something obvious, but I can't figure out what this follows directly from.

Duong Ngo
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Bara Like
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  • I re-cut my proof here: https://math.stackexchange.com/questions/3878092/prove-that-the-intersection-of-2-generalised-eigenspaces-is-the-zero-space/ to address your question. Also you should be able to figure out how to make a small modification to the Bezout identity argument to answer your question. – user8675309 Dec 19 '24 at 20:14
  • In my answer here: https://math.stackexchange.com/a/2121062/407833 I provided a proof for the more general statement "If $p_1,\ldots,p_n$ are mutually coprime polynomials, and $v_1,\ldots,v_n$ are vectors $\neq 0$ satisfying $p_i(A)\cdot v_i = 0;;\forall i\in{1,\ldots,n},$ then the vectors $v_1,\ldots,v_n$ are linearly independent." If you apply this to the polynomials $p_i(x)=\left(x−\lambda_i\right)^{m_i},$ then you get the desired result. – Reinhard Meier Dec 27 '24 at 23:31

1 Answers1

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The statement

each of the subspaces $M_{j}$ is disjoint from the span of all others.

is not trivial.

The following explanation is from Sheldon Axler's Linear Algebra Done Right (4th edition), Theorem 8.11 and Theorem 8.12.

Assume $\lambda_{i}, \lambda_{j}$ be eigenvalues of $A$. Let $v$ be a generalized eigenvector in $\mathfrak{M}_{i} \cap \mathfrak{M}_{j}$ (from the definition of generalized eigenvector, $v$ is nonzero).

Let $q_{j}$ be the nilpotent index of $A - \lambda_{j}I$ (restricted to $\mathfrak{M}_{j}$). Let $m$ be the smallest positive integer such that ${(A - \lambda_{i}I)}^{m}v = 0$.

\begin{align*} 0 & = {(A - \lambda_{j}I)}^{q_{j}}v \\ & = {(A - \lambda_{i}I + (\lambda_{i} - \lambda_{j})I)}^{q_{j}}v \\ & = \sum^{q_{j}}_{k=0}\dbinom{q_{j}}{k}{(\lambda_{i} - \lambda_{j})}^{q_{j} - k}{(A - \lambda_{i}I)}^{k}v \\ & = {(\lambda_{i} - \lambda_{j})}^{q_{j}}v + \sum^{q_{j}}_{k=1}\dbinom{q_{j}}{k}{(\lambda_{i} - \lambda_{j})}^{q_{j} - k}{(A - \lambda_{i}I)}^{k}v \\ \end{align*}

Apply ${(A - \lambda_{i}I)}^{m - 1}$ to both sides, we obtain \begin{align*} 0 = {(\lambda_{i} - \lambda_{j})}^{q_{j}}{(A - \lambda_{i}I)}^{m - 1}v \end{align*}

Since ${(A - \lambda_{i}I)}^{m - 1}v \ne 0$, we conclude that $\lambda_{i} = \lambda_{j}$. Hence a nonzero vector cannot be a generalized eigenvector corresponding to distinct eigenvalues. It remains to show that generalized eigenvectors corresponding to different eigenvalues are linearly independent.

Suppose on the contrary that there is a minimal positive integer $m$ such that there $m$ linearly dependent generalized eigenvectors $v_{1}, \ldots, v_{m}$ of distinct eigenvalues $\lambda_{1}, \ldots, \lambda_{m}$. For each $k$, let $q_{k}$ be the nilpotent index of $A - \lambda_{k}I$ (restricted to $\mathfrak{M}_{k}$). There exist nonzero scalars $a_{1}, \ldots, a_{m}$ (mind the minimality of $m$) such that $$a_{1}v_{1} + \cdots + a_{m}v_{m} = 0.$$

Apply ${(A - \lambda_{m}I)}^{q_{m}}$ to both sides. ${(A - \lambda_{m}I)}^{q_{m}}v_{m} = 0$ and ${(A - \lambda_{m}I)}^{q_{m}}v_{k} \ne 0$ for $k \ne m$. For each $k \ne m$, ${(A - \lambda_{m}I)}^{q_{m}}v_{k}$ is a generalized eigenvector of the eigenvalue $\lambda_{k}$ since ${(A - \lambda_{k}I)}^{q_{k}}{(A - \lambda_{m}I)}^{q_{m}}v_{k} = {(A - \lambda_{m}I)}^{q_{m}}{(A - \lambda_{k}I)}^{q_{k}}v_{k} = 0$. On the other hand $$ a_{1}{(A - \lambda_{m}I)}^{q_{m}}v_{1} + \cdots + a_{m-1}{(A - \lambda_{m}I)}^{q_{m}}v_{m-1} = 0. $$

so ${(A - \lambda_{m}I)}^{q_{m}}v_{1}, \cdots, {(A - \lambda_{m}I)}^{q_{m}}v_{m-1}$ is a shorter list of linearly dependent generalized eigenvectors corresponding to different eigenvalues, which contradicts the minimality of $m$.

Thus generalized eigenvectors of distinct eigenvalues are linearly independent, which implies the statement.

Duong Ngo
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    Doesn't this only prove that the generalized eigenspaces are pairwise disjoint? How do I know there isn't some linear combination of vectors from some generalized eigenspaces, that is a generalized eigenvector from another? – Bara Like Dec 19 '24 at 10:28
  • You are right, I will update my answer. – Duong Ngo Dec 19 '24 at 10:48
  • @BaraLike I updated the answer and made it community wiki. It should be ok now. – Duong Ngo Dec 19 '24 at 11:20