Exercise 8.D.7

Proof. Let ${\lambda }_1,\dots ,{\lambda }_m$ denote the distinct zeros of $p$ and $q$. Then

and

for some positive integers $d_1,\dots ,d_m,h_1,\dots ,h_m$, where $h_j\ge d_j$ for each $j$ (because $q$ is a polynomial multiple of $p$). Let $A$ equal the following block diagonal matrix

where each $A_j$ is the $h_j$-by-$h_j$ matrix defiend by

where the $1$’s appear up to the $d_j$-th column and the $0$’s fill the rest. Note that $A$ is a $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->q$-by-$\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->q$ matrix. Define $T\in L({ℂ}^{\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->q})$ such that the matrix of $T$ with respect to the standard basis is $A$. Then each ${\lambda }_j$ is an eigenvalue of $T$, because $A$ is upper triangular and ${\lambda }_j$ appears on the diagonal of $A$. Moreover, the multiplicity of ${\lambda }_j$ as an eigenvalue of $T$ is $h_j$, by Exercise 11 in section 8B. Hence, the characteristic polynomial of $T$ is $q$.

It is easy to see that ${(A_j-{\lambda }_{j}I)}^{d_j-1}\ne 0$ but ${(A_j-{\lambda }_{j}I)}^{d_j}=0$ (we can use a reasoning similar to that of Exercise 3 to show this). This, together with Exercise 9 from section 8B, shows that the $j$-th block is nonzero in ${(A-{\lambda }_{j}I)}^{d_j-1}$ and is zero in ${(A-{\lambda }_{j}I)}^{d_j}$ (pay attention to this fact). It follows that

Hence $p(T)=0$. We claim now $p$ is the minimal polynomial of $T$. To see this, suppose that it is not. Then we can subtract $1$ from one of the exponents in the equation above and it will still hold. Thus, we can write

for some $j\in \{1,\dots ,m\}$ and some polynomial $p^{\prime }$, with $p^{\prime }({\lambda }_j)\ne 0$. Let $v\in G({\lambda }_j,T)$ such that ${(T-{\lambda }_j)}^{d_j-1}v\ne 0$ (this $v$ exists due to the fact we pinpointed before). We have

but this is contradiction because ${(T-{\lambda }_j)}^{d_j-1}v\in G({\lambda }_j,T)$ and ${\lambda }_j$ is not an zero of $p^{\prime }$. □