Eigenvalues of Distinct Eigenvalues are Independent

Theorem

• If $\forall i\in [i,n],T(v_i)={\lambda }_i{v}_i$ are distinct eigenvalues an eigenvectors
• Then, the set of vectors $\{v_1,\dots ,v_n\}$ is indep

Intuition

If no two eigenvectors share the same eigenvalue, then we have independence