 Solution

The columns of $P$ are the coordinates of the vectors ${\mathbf{f}}_j$, so they form a basis exactly whenFor a linear transformation $\alpha$, define its matrix representation in the standard basis byThe change-of-basis matrix satisfies $[\mathbf{v}{]}_{\mathbf{e}}=P[\mathbf{v}{]}_{\mathbf{f}}$. Thereforeand multiplication by $P^{-1}$ gives the similar matrix relation

The characteristic polynomial of the displayed $A$ isEigenvectors $(2,-1,2{)}^T$, $(1,0,0{)}^T$, and $(0,-1,3{)}^T$ giveIndeed, $\det P=1$ andso $P^{-1}AP=B$. Finally, the diagonalization of a matrix gives