Eigenvalues and eigenvectors in combined Linear Transformations

Question

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!

Answer 1

As to 1:

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.

Answer 2

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.