The Unapologetic Mathematician

Mathematics for the interested outsider

Generalized Eigenspaces are Disjoint

We know that eigenspaces of distinct eigenvalues are disjoint. That is, no vector is an eigenvector of two different eigenvalues. This extends to generalized eigenvectors as well. If we have a nonzero v\in\mathrm{Ker}\left((T-\lambda_11_V)^d\right) then we know for sure that v\notin\mathrm{Ker}\left((T-\lambda_21_V)^d\right) for \lambda_2\neq\lambda_1.

To see this, let’s say that we have a v that’s in two generalized eigenspaces, and start hitting it with T-\lambda_11_V over and over again. Eventually, it disappears, but before it does, it’s a nonzero eigenvector with eigenvalue \lambda_1. I say that this (not generalized) eigenvector is also in the generalized eigenspace of \lambda_2. Indeed, each time we hit v with T-\lambda_11_V we write it as the sum T(v)-\lambda_1v. Since v\in\mathrm{Ker}\left((T-\lambda_21_V)^d\right), the same is true of the scalar multiple \lambda_1v\in\mathrm{Ker}\left((T-\lambda_21_V)^d\right). And since (T-\lambda_2)^d is a polynomial in T, its kernel is invariant. Thus we have T(v)\in\mathrm{Ker}\left((T-\lambda_21_V)^d\right) as well. And thus \mathrm{Ker}\left((T-\lambda_21_V)^d\right) is invariant under the transformation T-\lambda_11_V, and we can assume (without loss of generality) that v is an actual eigenvector with eigenvalue \lambda_1, as well as a generalized eigenvector with eigenvalue \lambda_2.

But now let’s start hitting v with T-\lambda_21_V. Applying it once we find T(v)-\lambda_2v=(\lambda_1-\lambda_2)v. So no matter how many times we apply the transformation, we just keep multiplying by \lambda_1-\lambda_2, and we’ve assumed that this is nonzero. So v cannot possibly be in the generalized eigenspace of \lambda_2. And so generalized eigenspaces corresponding to distinct eigenvalues can only intersect trivially.

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February 25, 2009 - Posted by | Algebra, Linear Algebra


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