Exercise 8.D.8

Proof. Suppose thre does not exist a direct sum decomposition of $V$ into two proper subspaces invariant under $T$. Therefore, the block diagonal matrix of $T$ with respect to some Jordan basis for $T$ only has one block (the span of the basis vectors that correspond to each block is invariant under $T$). Therefore $T$ only has one eigenvalue, call it $\lambda$. Exercise 3 now implies that ${(z-\lambda )}^{\dim <!--\; FUNCTION\; APPLICATION\; -->V}$ is minimal polynomial of $T$.

For the other direction, we will prove the contrapositive. Suppose we can decompose $V$ into two proper subspaces invariant under $T$. If $T$ has more than one eigenvalue the result is obvious. Assume $T$ has only one eigenvalue, call it $\lambda$. Then $V=G(\lambda ,T)$ and there are proper subspaces $U_1,U_2$ of $G(\lambda ,T)$ invariant under $T$ such that $G(\lambda ,T)=U_1\oplus U_2$ with $1\le \dim <!--\; FUNCTION\; APPLICATION\; -->U_1,\dim <!--\; FUNCTION\; APPLICATION\; -->U_2<\dim <!--\; FUNCTION\; APPLICATION\; -->V$. We have that ${(z-\lambda )}^{\dim <!--\; FUNCTION\; APPLICATION\; -->U_1}$ and ${(z-\lambda )}^{\dim <!--\; FUNCTION\; APPLICATION\; -->U_2}$ are the characteristic polynomials of $T{\text{|}}_{U_1}$ and $T{\text{|}}_{U_2}$. Thus ${(T-\lambda )}^{\max <!--\; FUNCTION\; APPLICATION\; -->\{\dim <!--\; FUNCTION\; APPLICATION\; -->U_1,\dim <!--\; FUNCTION\; APPLICATION\; -->U_2\}}=0$ and so the minimal polynomial of $T$ cannot be $z^{\dim <!--\; FUNCTION\; APPLICATION\; -->V}$. □