Exercise 6.4.12

Since the characteristic polynomial splits, we may apply Schur’s Theorem and get an orthonormal basis $\beta$ such that ${[T]}_{\beta }$ is upper triangular. Denote the basis by

We already know that $v_1$ is an eigenvector. Pick $t$ to be the maximum integer such that $v_1,v_2,\dots ,v_t$ are all eigenvectors with respect to eigenvalues ${\lambda }_i$. If $t=n$ then we’ve done. If not, we will find some contradiction. We say that ${[T]}_{\beta }=\left\{A_{i,j}\right\}$. Thus we know that

Since the basis is orthonormal, we know that

by Theorem 6.15(c). This means that $v_{t+1}$ is also an eigenvector. This is a contradiction. So $\beta$ is an orthonormal basis. By Theorem 6.17 we know that $T$ is self-adjoint.