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I'm working my way through Linear Algebra Done Right. To help with one proof, I want to prove the following:

Given $\mathbf{V}$, a vector space and $T$, a linear operator on it, then:

If $\mathbf{W}_1$ and $\mathbf{W}_2$ are subspaces of $\mathbf{V}$ such that:

  1. $\mathbf{V}$ is a direct sum of $\mathbf{W}_1$ and $\mathbf{W}_2$.

  2. $\mathbf{W}_1$ and $\mathbf{W}_2$ are invariant under $T$.

  3. The restriction of $T$ to $\mathbf{W}_1$ has at most $k$ eigenvalues.

  4. The restriction of $T$ to $\mathbf{W}_2$ has at most $p$ eigenvalues.

Then $T$ has at most $k+p$ eigenvalues.

I've done a sketch of a proof using determinants, but it was based on old knowledge about the properties of determinants with regards to eigenvalues, so it may not be correct. The book doesn't emphasize using them though, and maybe there's a proof of this without using determinants.

I've tried a proof by contradiction, trying to find something weird by assuming that T can have more than $k+p$ eigenvalues, but I haven't been able to find anything.

Any help would be appreciated.

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    Thank you all for your help. I can sleep peacefully tonight :).2010-12-16

2 Answers 2

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Hint: Suppose $\lambda$ is an eigenvalue. Then there exists an eigenvector $\mathbf{v}$ corresponding to $\lambda$. Write $\mathbf{v}=\mathbf{w}_1+\mathbf{w}_2$; since $\mathbf{v}$ is nonzero, at least one of $\mathbf{w}_1$ and $\mathbf{w}_2$ is nonzero.

Now, evaluate $T(\mathbf{v})$; since $\mathbf{V}$ is a direct sum, and each $\mathbf{W}_i$ is invariant, what can you say about $T(\mathbf{w}_1)$ and $T(\mathbf{w}_2)$?

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    Your version is sharper. For mine you'd have to look at the associated matrices and do operations with them to reach the same conclusion.2010-12-16
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Hint: $ker(T) = ker(T|W_1)\ \oplus\ ker(T|W_2)$.

Here $ker$ stands for kernel, $T|W_i$ stands for $T$ restricted to $W_i$, etc.. Actually you have to do this with the transformation $T -\lambda I$ instead of $T$.

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    Yes, with just one caveat: One of the $\lambda w_i$ might be zero. So $\lambda$ might be an eigenvalue for $T$ restricted to just one of these subspaces.2010-12-16