Exercise 6.4.16

By Schur’s Theorem $A=P^{-1}\mathit{BP}$ for some upper triangular matrix $B$ and invertible matrix $P$. Now we want to say that $f(B)=O$ first. Since the characteristic polynomial of $A$ and $B$ are the same, we have the characteristic polynomial of $A$ would be

since $B$ is upper triangular. Let $C=f(B)$ and $\left\{e_i\right\}$ the be the standard basis. We have $C{e}_1=0$ since $(B_{11}I-B)e_1=0$. Also, we have $C{e}_i=0$ since $(B_{\mathit{ii}}I-B)e_i$ is a linear combination of $e_1,e_2,\dots ,e_{i-1}$ and so this vector will vanish after multiplying the matrix

So we get that $f(B)=C=O$. Finally, we have