Exercise 7.B.2

Proof. Since $T$ is self-adjoint, there exists a basis consisting of eigenvectors of $T$, by either 7.24 or 7.29 (depending on the choice of $𝔽$). Because $x^2-5x+6=(x-2)(x-3)$, we have

Thus, applying it to any vector in the basis will give $0$ (for the eigenvectors corresponding to $2$ just switch the order to $(T-3I)(T-2I)$, which is valid by 5.20). □