# Proving the intersection of distinct eigenspaces is trivial

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.

• You are making things too complicated. Proceed from the definition. Suppose a vector $v$ belongs to both eigenspaces of $T$. What does this say about $\lambda_1 v$ and $\lambda_2 v$? Dec 3, 2015 at 4:56
• Oh that $T(\vec v) = \lambda_1 \vec v\ = \lambda_2 \vec v$ which would imply the eigenvalues are equal by cancellation?
– Jay3
Dec 3, 2015 at 5:10
• So suppose v is an element of the intersection, then T(v) = Lambda 1v = lambda 2v. Since this isn't possible as the eigenvalues aren't equal v must equal the zero vector. So The intersection is a subset of the trivial subspace and conversely, since the intersection of two subspaces is a subspace, the trivial subspace is a subset of intersection of the Eigenspaces. So the intersection equals the trivial subspace. Is that kind of how the proof should look?
– Jay3
Dec 3, 2015 at 5:15
• Yes, you need to be a little careful how you argue "equal by cancellation". We cannot cancel $v$ from both sides directly, but Michael Hardy shows how to use the assumption the eigenvalues are unequal to get there. We can multiply by $(\lambda_1 - \lambda_2)^{-1}$ after grouping terms on one side. Dec 3, 2015 at 12:16

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 $\vec x=\vec0$.