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Let $V$ be a finite-dimensional complex vector space and let $U$ be a subspace of $V$ invariant under the linear operator $T$:

$$\forall u\in U: Tu \in U.$$

Must $U$ also be the subspace associated with a generalized eigenvector of $T$ (which I call a “generalized eigenspace” in the post title)? I.e., must we have:

$$\exists \lambda:(T-\lambda I)^{\dim V} u=0\; ?$$


I always refer to Sheldon Axler’s “Linear Algebra Done Right” for questions like this, but from what I can tell his theorems don’t answer this. The converse is true, which is the content of his Theorem 8.23 (in Ed. 2), but I’m not sure about the other direction.

For context, I’m trying to decide whether $3\times 3$ matrices of the form

$$\pmatrix{1&a&b\\0&1&c\\0&0&1}$$

can have any 2-dimensional invariant subspaces. It is quick to show that its only eigenspace is the one spanned by $(1,0,0)$ and that its only generalized eigenspace is all of $\mathbb R^3$ with eigenvalue $1$. But does this imply that 2-dimensional invariant subspaces can’t exist? It seems this would only necessarily follow if the answer to this post’s question is “yes.” Other opinions welcome.

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  • $\begingroup$ Re: Answer that was retracted. What was wrong with it? $\endgroup$ – WillG Dec 19 '19 at 19:24
  • $\begingroup$ No. Direct sums of generalized eigenspaces with different eigenvalues are also invariant. $\endgroup$ – Conifold Dec 20 '19 at 0:51
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The subspace of $\mathbb R^3$ spanned by $(1,0,0)$ and $(0,1,0)$ is $A$-invariant. And this shows that an invariant subspace doesn't have to be a generalized eigenspace.

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  • $\begingroup$ As the other answer (now retracted) pointed out, this 2D subspace is in fact equal to $\text{null}(T-1\cdot I)^2$. I suppose I should have asked if all invariant subspaces must be of this form. $\endgroup$ – WillG Dec 19 '19 at 20:45
  • $\begingroup$ Accepting this answer and posting a follow-up here: math.stackexchange.com/questions/3482218/… $\endgroup$ – WillG Dec 19 '19 at 20:58

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