TITLE: Proving the intersection of distinct eigenspaces is trivial
QUESTION [7 upvotes]: Suppose $\lambda_1$ and $\lambda_2$ are different eigenvalues of $T$. Prove $E_{\lambda_1} \cap E_{\lambda_2}= \{\vec0\}$. 
I have a basic idea of what to do. Since both eigenvalues are distinct, doesn't that mean the basis for each space are linearly independent of each other so that no vector in one is in the span of the basis of the other? I'm just looking on how to formalize these ideas.

REPLY [6 votes]: Here's one way to look at it: Suppose $\vec x$ is in the intersection of the eigenspaces corresponding to the two eigenvalues $\lambda_1$ and $\lambda_2$, where $\lambda_1\ne\lambda_2$.  Then
$$
\lambda_1 \vec x = T\vec x = \lambda_2\vec x,
$$
so
$$
(\lambda_1 - \lambda_2) \vec x = \vec 0.
$$
Since $\lambda_1-\lambda_2\ne0$, the only way one can have $(\lambda_1 - \lambda_2) \vec x = \vec 0$ is if $\vec x=\vec0$.