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Assume that we are working over a complex space $W$ of dimension $n$. When would an operator on this space have the same characteristic and minimal polynomial?

I think the easy case is when the operator has $n$ distinct eigenvalues, but what about if it is diagonalizable? Is that sufficient, or can there be cases (with repeated eigvals) when char poly doesn't equal min poly? What are the general conditions when the equality holds? Is it possible to define them without use of determinant? (I am working by Axler and he doesn't like it.)

Thanks.

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One example: the operator that maps everything to $0$ has minimal polynomial $t$, but characteristic polynomial $(-1)^n t^n$. –  Henning Makholm Nov 12 '11 at 22:23
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3 Answers 3

up vote 11 down vote accepted

Theorem. Let $T$ be an operator on the finite dimensional complex vector space $\mathbf{W}$. The characteristic polynomial of $T$ equals the minimal polynomial of $T$ if and only if the dimension of each eigenspace of $T$ is $1$.

Proof. Let the characteristic and minimal polynomial be, respectively, $\chi(t)$ and $\mu(t)$, with $$\begin{align*} \chi(t) &= (t-\lambda_1)^{a_1}\cdots (t-\lambda_k)^{a_k}\\ \mu(t) &= (t-\lambda_1)^{b_1}\cdots (t-\lambda_k)^{b_k}, \end{align*}$$ where $1\leq b_i\leq a_i$ for each $i$. Then $b_i$ is the size of the largest Jordan block associated to $\lambda_i$ in the Jordan canonical form of $T$, and the sum of the sizes of the Jordan blocks associated to $\lambda_i$ is equal to $a_i$. Hence, $b_i=a_i$ if and only if $T$ has a unique Jordan block associated to $\lambda_i$. Since the dimension of $E_{\lambda_i}$ is equal to the number of Jordan blocks associated to $\lambda_i$ in the Jordan canonical form of $T$, it follows that $b_i=a_i$ if and only if $\dim(E_{\lambda_i})=1$. QED

In particular, if the matrix has $n$ distinct eigenvalues, then each eigenvalue has a one-dimensional eigenspace.

Also in particular,

Corollary. Let $T$ be a diagonalizable operator on a finite dimensional vector space $\mathbf{W}$. The characteristic polynomial of $T$ equals the minimal polynomial of $T$ if and only if the number of distinct eigenvalues of $T$ is $\dim(\mathbf{W})$.

Using the Rational Canonical Form instead, we obtain:

Theorem. Let $W$ be a finite dimensional vector space over the field $\mathbf{F}$, and $T$ an operator on $W$. Let $\chi(t)$ be the characteristic polynomial of $T$, and assume that the factorization of $\chi(t)$ into irreducibles over $\mathbf{F}$ is $$\chi(t) = \phi_1(t)^{a_1}\cdots \phi_k(t)^{a_k}.$$ Then the minimal polynomial of $T$ equals the characteristic polynomial of $T$ if and only if $\dim(\mathrm{ker}(\phi_i(T)) = \deg(\phi_i(t))$ for $i=1,\ldots,k$.

Proof. Proceed as above, using the Rational Canonical forms instead. The exponent $b_i$ of $\phi_i(t)$ in the minimal polynomial gives the largest power of $\phi_i(t)$ that has a companion block in the Rational canonical form, and $\frac{1}{d_i}\dim(\mathrm{ker}(\phi_i(T)))$ (where $d_i=\deg(\phi_i)$) is the number of companion blocks. QED

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Which Rational Canonical Form are you using? The form I found in Wikipedia (badly described) and MathWorld is related to invariant factors, and doesn't use any factorization into irreducibles (which would vary with the field used). It is this notion I used in my answer. –  Marc van Leeuwen Nov 17 '11 at 13:52
    
The version I generally use is described in Friedberg, Insel, and Spence. You factor the characteristic polynomial into irreducibles, $\phi_1^{a_1}\cdots\phi_k^{a_t}$; the minimal polynomial is then $\phi_1^{b_1}\cdots\phi_k^{b_k}$, and the generalized eigenspaces are $\mathrm{Ker}(\phi_i(T)^{a_i})$. You then decompose each of those. –  Arturo Magidin Nov 17 '11 at 14:02
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The following equivalent criteria, valid for an arbitrary field, are short to state. Whether or not any one of the conditions is easy to test computationally may depend on the situation, though 2. is in priciple always doable.

Proposition. The following are equivalent for a linear operator on a vector space of nonzero finite dimension.

  1. The minimal polynomial is equal to the characteristic polynomial.
  2. The list of invariant factors has length (at most) one.
  3. The Rational Canonical Form has (at most) a single block.
  4. The operator has a matrix similar to a companion matrix (or it is a $0\times0$ matrix).
  5. There exists a (so-called cyclic) vector whose images by the operator span the whole space.

Point 1. and 2. are equivalent because the minimal polynomial is the largest invariant factor (or $1$ if there are none) and the characteristic polynomial is the product of all invariant factors. The invariant factors are in bijection with the blocks of the Rational Canonical Form, giving the equivalence of 2. and 3. These blocks are companion matrices, so 3. implies 4., and by the uniqueness of the RCF 4. also implies 3 (every companion matrix is its own RCF). Finally 4. implies 5. (take the first basis vector as cyclic vector) and 5. implies 4. by taking a basis consisting of $n$ successive images (counting from $0$) of the cyclic vector.

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Here is a generalization to principal ideal domains.

Let $A$ be a principal ideal domain, $p$ an irreducible element of $A$, and $M$ a finitely generated $A$-module annihilated by some power of $p$.

Then there is a unique nondecreasing tuple $(n_1,\dots,n_k)$ of positive integers such that $M$ is isomorphic to the direct sum of the $A/(p^{n_i})$.

The characteristic ideal of $M$ is $(p^s)$, where $s$ is the sum of the $n_i$; and the annihilator of $M$ is $(p^{n_k})$. Let $\phi$ be the endomorphism $x\mapsto px$ of $M$.

The following conditions are clearly equivalent:

  • $k=1$,

  • $s=n_k$,

  • $\text{Ker }\phi\simeq A/(p)$,

  • $\text{Coker }\phi\simeq A/(p)$.

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