Prove that the minimal polynomial of the restricted operator divides the minimal polynomial of the operator

Question

The following is a proof from the book Linear algebra by Hoffman and Kunze(Chapter 6 Elementary canonical forms). I did not understand the highlighted portion. Could anyone please give an explanation or alternate explanation of this part?

Answer 1

The main point to use that any polynomial$~P$ that annihilates a linear operator$~\phi$ is a multiple of the minimal polynomial$~\mu_\phi$ of that operator. This is clear from Euclidean division of $P$ by $\mu_\phi$ (the remainder also annihilates $\phi$, so it must be zero).

Now to apply this to get the conclusion you want, clearly $P$ should be the minimal polynomial of$~T$, and $\phi=T_W$ the restriction of $T$ to$~W$; then $P$ is a multiple of $\mu_\phi$ is the conclusion you seek. All you need to know in order to apply the argument is that $P$ annihilates $\phi=T_W$, which just means that $P[T_W]=0$, or that the operator $P[T]$ is zero on$~W$. But by definition $P[T]=0$ everywhere, so you're done.

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This shows that there is really no need at all to choose a basis and to express $T$ by a matrix with respect to it. Such a choice was however required for the characteristic polynomial part (since it involves the determinant of a matrix), and the basis must be chosen so as to start with a basis $\mathscr B$ of the subspace$~W$, and the matrix of $T_W$ with respect to $\mathscr B$ will end up as the upper-left block $B$ of the block upper-triangular matrix$~A$ of $T$. Now with such a matrix in place, one can use it and rephrase what I said above as the fact that $P[A]=0$ (i.e., $P$ annihilates $T$) implies $P[B]=0$ (i.e., $P$ annihilates $T_W$). This is so simply because $P[B]$ appears as the same upper-left block of $P[A]$. It is this latter fact that is justified in the citation by the computation of arbitrary powers $A^k$ (and by linear combination of them). But I find the restriction argument more natural.

Answer 2

Hint:

He who can do the more can do the less, but is not the least to do the less ;o).