Question

Suppose T € L(V ) and dim range T =k. Prove that T has at most k + 1distinct eigenvalues.

Answer 1

$\text{Since dim(V ) < ∞, T has at most a finite number of distinct eigenvalues.}\\ \text{Let $λ_1, . . . , λ_m$ be the distinct eigenvalues of T, and let}:\\ v_1, . . . , v_m \text{be corresponding nonzero eigenvectors.}\\ \text{If $λ_j \neq 0,$ then $T(v_j/λ_j ) = v_j$}\\ \text{Since at most one of $λ_1, . . . , λ_m $ equals 0, this implies that at least m − 1 of the vectors $v_1, . . . , v_m$ are in range(T)}.\\ \text{ These vectors are linearly independent, which implies that $m −1 ≤ dim(rangeT) = k.$}\\ \text{Thus $m ≤ k + 1$, as desired.}$