Kernel of a polynomial which divides the minimal polynomial

Question

Question: $P_{T}$:= characteristic polynomial of T; $minP_T$:= minimal polynomial of T. Consider $P_{T}=minP_T$. If $g$ is a polynomial which divides $minP_T$, then dim(Ker($g(T)$))=deg $g$.

My Attempt: $g(T)$ can be written as a product of powers of irreducible polynomials. Suppose $g(T)=\prod_{i=1}^{k}\Phi_i^{m_i}$ where each $\Phi_i$ is irreducible. Now we know the minimal polynomial of the restriction of $T$ in Ker $\Phi_{i}^{m_i}$ i.e. $minP_{T_{Ker\Phi_{i}^{m_i}}}=\Phi_{i}^{m_i}$. Therefore dim Ker $\Phi_{i}^{m_i}$=deg $\Phi_{i}^{m_i}$. Now since each of the $\Phi_i$'s are relatively prime, their Kernels only have the zero intersection. Hence $dim(Ker(g(T))=\sum_{i=1}^k dim({Ker\Phi_{i}^{m_i}})= \sum_{i=1}^k deg \Phi_{i}^{m_i}=$deg $g$.

Question: I don't see in the question where the condition of the minimal polynomial being same as the characteristic polynomial is required. What I think is that, whenever any polynomial divides the minimal polynomial, it's Kernel has the dimension of the polynomial's degree.

I would like to know if there is any fault in my approach.

Thanks in advance

Answer 1

You had the correct idea to start the proof by induction on the number of distinct irreducibles, but made a serious mistake in not having the base case of your induction.

So let $V$ be a finite-dimensional $k$-vectorspace, and $T\in\operatorname{End}_kV$ be such that (in your notation) $\operatorname{P}_T(x)=\Phi(x)^n$, $\Phi(x)\in k[x]$ irreducible and $\operatorname{minP}_T=\operatorname{P}_T$. We need to prove that $\dim\ker\Phi^m(T)=\deg\Phi^m$ for all $0\leq m\leq n$. The obvious way to do so is to put $T$ into generalized Jordan normal form. Since $\operatorname{minP}_T=\operatorname{P}_T$, there is some $v\in V$ such that $\Phi^{m-1}(T)(v)\neq 0$. Then $$ \begin{matrix} v,&Tv,&\dots,&T^{\deg\Phi-1}v,\\ \Phi(T)v,&\Phi(T)Tv,&\dots,&\Phi(T)T^{\deg\Phi-1}v,\\ \Phi(T)^2v,&\dots,\\ &&\dots,&\Phi(T)^{n-1}T^{\deg\Phi-1}v \end{matrix} $$ is a basis of $V$ and with respect to this basis, $T$ is \begin{align*} T&=\begin{pmatrix}C\\ U & C\\ & U & C\\ &&\ddots & \ddots\\ &&&U&C \end{pmatrix},\\ C&=\begin{pmatrix} &&&\ast\\ 1&&&\ast\\ &\ddots&&\vdots\\ &&1&\ast \end{pmatrix},\quad U=\begin{pmatrix} 0&\dots&0&1\\ &&&0\\ &&&\vdots\\ &&&0 \end{pmatrix} \end{align*} where $C$ is the companion matrix of $\Phi$. The effect of $\Phi(T)$ is to push the blocks down: $$ \Phi(T)= \begin{pmatrix} 0\\ I&0\\ &I&0\\ &&I&0\\ &&&\ddots&\ddots\\ &&&&I&0 \end{pmatrix}, \Phi(T)^2= \begin{pmatrix} 0\\ 0&0\\ I&0&0\\ &I&0&0\\ &&\ddots&\ddots&\ddots\\ &&&I&0&0 \end{pmatrix}, \dots $$ So $\dim\ker\Phi^m(T)=m\deg\Phi=\deg\Phi^m$.

Answer 2

First off, you definitely need the condition that minimal and characteristic polynomials are equal, as you can seen by considering the case where $g$ is the minimal polynomial itself: the kernel of $g[T]=0$ is the entire space, whose dimension is the degree of the characteristic polynomial, so you are requiring that the degree of the minimal polynomial match that, which (due to Cayley-Hamilton) it can only do by being equal to the characteristic polynomial.

The hard part is therefore to show that this suffices to make $\dim(\ker(g[T]))=\deg(g)$ for every divisor of the minimal polynomial. Limiting to monic divisors of the minimal polynomial (which clearly is valid) let me start with facts that do not depend on minimal and characteristic polynomials being equal.

First, the map $K:g\mapsto\ker(g[T])$ from such monic divisor to subspaces is partial-order preserving, in the sense that $g\mid h\implies K(g)\subseteq K(h)$. This is quite obvious: if $h=fg$, a vector killed by $g[T]$ will also be killed by $h[T]=f[T]\circ g[T]$. Also, the map $K$ is injective, which I shall only need along a chain of divisors (though once this case is known it is pretty obvious that it is true generally): if $f$ is a strict divisor of $g$ then $K(f)$ is a strict subspace $K(g)$. The proof can be done by contradiction: supposing $K(f)=K(g)$, write the minimal polynomial as a product $gh$; then $g[T]\circ h[T]=0$ which means the image of $h[T]$ is contained in $K(g)$, but this is also $K(f)$, so $f[T]\circ h[T]=0$ which means that $fh$ is an annihilating polynomial of $T$, but is is also a strict divisor of the minimal polynomial $gh$, which is a contradiction.

So if we build up the minimal polynomial starting from $1$, multiplying on each step by a monic irreducible polynomial, we get a chain of monic divisors, which under $K$ gives a chain of (strictly) increasing subspaces.

In the chain of divisors the degree rises from $0$ to the degree of the minimal polynomial, and the dimension of the subspaces rises from $0$ to the dimension of the whole space. Now we are going to assume that these final numbers are equal (since minimal and characteristic polynomials are equal), and then prove that on each step, degree and dimension go up by the same number (therefore remain the same between them). We can prove this by showing that dimension always goes up at least by the increase of the degree, that is, by the degree of the irreducible factor we are multiplying by. So let $f,g$ be two successive products and let $d=g/f$ be that irreducible factor. Naming $V=\ker(f[T])$ and $W=\ker(g[T])$ we have $V\subset W$, a strict inclusion as we have seen, so $f[T]$ restricts to a nonzero map $W\to W$, whose image $U$ is isomorphic to $V/W$ by a first isomorphism theorem; we wish to show that $\dim(U)\geq\deg(d)$ (and we know $\dim(U)\neq0$). Now $d[T]$ vanishes on $U$, since writing any $u\in U$ as $u=f[T](w)$ for some $w\in W$, one has $d[T](u)=d[T](f[T](w))=g[T](w)=0$, as $w\in W=\ker(g[T])$. So $d$ is an annihilating polynomial of the restriction $T|_U$, and with $d$ irreducible and $\dim(U)$ nonzero that forces $d$ to be its minimal polynomial. Then $\deg(d)\leq\dim(U)$ follows from the Cayley-Hamilton theorem, and we have obtained our goal.