When are the characteristic and the minimal polynomials of a linear map equal?

Posted: May 26, 2010 in Elementary Algebra; Problems & Solutions, Linear Algebra
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Problem 1. Let V be a vector space over some field F and suppose that \dim_F V = n. Let T: V \longrightarrow V be a linear transformation. Prove that the minimal and the characteristic polynomials of T are equal if and only if there exists some v \in V such that v,T(v), \cdots , T^{n-1}(v) are linearly independent over F.

Solution. Let p(x) be the minimal polynomial of T.

Proof of \Longleftarrow. We cannot have \deg p(x) < n because then, since p(T) = 0, we’d  have p(T)(v)=0 which contradicts our assumption that v, T(v), \cdots , T^{n-1}(v) are linearly independent. Thus \deg p(x)=n.

First proof of \Longrightarrow. For every v \in V let I_v= \{f(x) \in F[x]: \ f(T)(v) = 0 \}. See that I_v is an ideal of F[x] and thus, since I_v= \langle f_v(x) \rangle, for some f_v(x) \in F[x]. Clearly p(x) \in I_v, for all v \in V and so f_v(x) \mid p(x). But the number of divisors of p(x) is finite, which means that the number of distinct elements of the set \{I_v: \ v \in V \} is finite. Let I_{v_1}, \cdots , I_{v_m} be those distinct elements and put

V_i = \{v \in V: \ f_{v_i}(T)(v)=0 \}.

See that V_i is a subspace of V, for all 1 \leq i \leq k, and V=\bigcup_{i=1}^m V_i. Therefore V=V_k, for some 1 \leq k \leq m. Thus f_{v_k}(T)(v)=0, for all v \in V, i.e. f_{v_k}(T)=0. That implies p(x) \mid f_{v_k}(x), by the minimality of p(x), and so f_{v_k}(x)=p(x) because we already know that f_{v_k}(x) \mid p(x) and that p(x) is a monic polynomial. Finally, suppose that

c_0v_k + c_1T(v_k) + \cdots + c_{n-1}T^{n-1}(v_k)=0,

for some c_j \in F, and let g(x)=c_0 + c_1x + \cdots + c_{n-1}x^{n-1} \in F[x]. Then g(T)(v_k)=0, which gives us

g(x) \in I_{v_k} = \langle f_{v_k}(x) \rangle = \langle p(x) \rangle,

that is p(x) \mid g(x). Therefore g(x)=0, because \deg p(x)=n > \deg g(x), and hence c_j = 0, for all j.

Second proof of \Longrightarrow.  Give V a structure of an F[x] module by defining the multiplication by f(x)v = f(A)(v). Then see that V becomes a finitely generated torsion F[x] module and thus, by the fundamental theorem for finitely generated modules over PIDs, there exists some v \in V such that \text{ann}_{F[x]}(v)=\text{ann}_{F[x]}(V). It’s easy now to show that v, T(v), \cdots , T^{n-1}(v) are linearly independent over F. \Box

An important result of Problem 1 is the following:

Problem 2. Let V be a vector space over F and \dim_F V = n. Let T: V \longrightarrow V be a linear transformation. If T^j, \ 0 \leq j \leq n-1, are linearly independent over F, then there exists some v \in V such that T^j(v), \ 0 \leq j \leq n-1, are linearly independent over F.

Solution. Since the operators T^j, \ 0 \leq j \leq n-1, are linearly independent, the degree of the minimal polynomial of T is n and thus the characteristic and the minimal polynomials of T are equal. Now the result follows by Problem 1. \Box

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