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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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Welcome to the site! You can use $ to do (la)tex markup:…. And +1 for talking about the work done before posting here :-) – Aryabhata Dec 16 '10 at 20:02
Thank you all for your help. I can sleep peacefully tonight :). – user4832 Dec 16 '10 at 20:49
up vote 4 down vote accepted

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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I think I got it. Sorry ... bad edit. T(v) = T(w1 + w2) = T(w1) + T(w2) = λw1 + λw2. Since w1 and w2 are writable from different bases we can say that T(w1) = λw1 or T(w2) = λw2 (if they're non-null, and one of them is, at least). This means that λ is an eigenvalue for T|W1 or T|W2. – user4832 Dec 16 '10 at 20:13
@Horia Coman: Basically yes. I would instead write $T(w_1)+T(w_2)=T(v) = \lambda v = \lambda (w_1+w_2)$; then note that there is one and only one way to write $T(v)$ as a sum of something in $W_1$ and something in $W_2$, and that $T(w_1)$ must be in $W_1$ and $T(w_2)$ must be in $W_2$, so then you can conclude that $T(w_1)=\lambda w_1$ and $T(w_2)=\lambda w_2$. Otherwise, fine. – Arturo Magidin Dec 16 '10 at 20:19
@Horia Coman: Once you are satisfied, be sure to "accept" the answer you found most helpful. You click on the checkmark that should appear to the left of the question, under the voting arrows. – Arturo Magidin Dec 16 '10 at 20:20
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. – user4832 Dec 16 '10 at 20:26

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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You mean $T - \lambda I$ right? – Eric Haengel Dec 16 '10 at 20:05
@Eric Hangel: Right. I have fixed it. Thanks. – user1119 Dec 16 '10 at 20:06
Going from your hint, an eigenvector in V can be decomposed as a sum of eigenvectors in W1 and W2 in a unique way (because it's a direct sum). Then, if we apply T to the whole mix, we get T(v) = T(w1 + w2) = T(w1) + T(w2) and T(v) = λv = λw1 + λw2. Since T(w1) is from W1,T(w2) is from W2, λw1 is from W1 and λw2 is from W2, then T(w1) = λw1 and T(w2) = λw2, therefore λ is an eigenvalue for T|W1 and T|W2. – user4832 Dec 16 '10 at 20:33
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. – user1119 Dec 16 '10 at 20:45

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