# Eigenvalues and eigenvectors in combined Linear Transformations

I am new to linear algebra, and cannot work out the following question, despite the fact that I have been thinking about it for a long while.

Let $$V$$ be a linear space of n dimensions over R, and let $$S,T:V \to V$$ be linear transformations.

True or False?

1. If $$v$$ is an eigenvector of $$S$$ and of $$T$$, then $$v$$ is also an eigenvector of $$S + T$$.
2. If $$λ_1$$ is an eigenvalue of $$S$$ and $$λ_2$$ is an eigenvalue of $$T$$, then $$λ_1$$ + $$λ_2$$ is a eigenvalue of $$S + T$$.

I am not just looking for the right answer, but also for the reasoning behind it…

Thank you!

If $$v$$ is an eigenvector of $$S$$ then $$Sv=\lambda v$$ for some scalar $$\lambda$$.
If $$v$$ is an eigenvector of $$T$$ then $$Tv=\mu v$$ for some scalar $$\mu$$.
Then $$(S+T)(v)=Sv + Tv = \lambda v + \mu v = (\lambda +\mu)v$$. So $$v$$ is indeed an eigenvector of $$S+T$$ with eigenvalue $$\lambda+\mu$$.
As to 2: this reasoning won't work as we will have (in general) different eigenvectors $$v$$ and $$w$$, say, and we cannot really say anything about $$Tv$$ or $$Sw$$ etc. So there a counterexample will most likely exist. SO in a pure true/false setting go for false.
Hints: for the first, consider the definition of eigenvalues. We have $$Sv=\lambda_1v$$ and $$Tv=\lambda_2v$$ for some $$\lambda_1, \lambda_2$$. What can we then say about $$(S+T)(v)$$? For the second, the same line of reasoning won't work. Try messing around with some common linear transformations and their eigenvalues and you'll quickly find a counter-example.