Exercise 5.A.25

Proof. If $u$ and $w$ are linearly dependent, then they obviously correspond to the same eigenvalues. Assume $u,w$ are linearly independent. Let $\alpha ,\beta ,\lambda$ be eigenvalues of $T$ with corresponding vectors $u,v,u+w$, respectively. Then

which implies that $(\alpha -\lambda )u=(\lambda -\beta )w$. Since $u,w$ are linearly independent, it follows that their coefficients are $0$. Thus $\lambda =\alpha =\beta$. □