Question Let $T \in L(U)$ and $\lambda_1,...\lambda_k$ be some distinct eigenvalues of $T$ such that $T$ is reducible over the eigenspaces $E_{\lambda_1},...E_{\lambda_k}$. Show that $\lambda_1,...\lambda_k$ are all the eigenvalues of $T$. In other words, if $\lambda$ is any eigenvalue of $T$, then $\lambda$ is among $\lambda_1,...,\lambda_k$.
I think the key to this proof is that $T$ is reducible over the eigenspaces. However, $T$ is not given a dimension, and there is no geometric multiplicity given, so I'm a little lost.
Since $T$ is reducible over the eigenspaces. That means that $\dim(U)=\dim(\Sigma E_{\lambda_1} +...+E_{\lambda_k})$, I think. Also it means that $T$ can have a matrix $A$ that is a box triangular matrix, but I'm not quite sure what that would imply. Please help