What Is the Minimal Polynomial of a Matrix?

For a polynomial

notag  phi(t) = a_kt^k + cdots + a_1t + a_0,

where a_kinmathbb{C} for all k, the matrix polynomial obtained by evaluating phi at Ainmathbb{C}^{n times n} is

notag  phi(A) = a_kA^k + cdots + a_1A + a_0 I.

(Note that the constant term is a_0 A^0 = a_0 I). The polynomial phi is monic if a_k = 1.

The characteristic polynomial of a matrix Ainmathbb{C}^{n times n} is p(t) = det(t I - A), a degree n monic polynomial whose roots are the eigenvalues of A. The Cayley–Hamilton theorem tells us that p(A) = 0, but p may not be the polynomial of lowest degree that annihilates A. The monic polynomial psi of lowest degree such that psi(A) = 0 is the minimal polynomial of A. Clearly, psi has degree at most n.

The minimal polynomial divides any polynomial phi such that phi(A) = 0, and in particular it divides the characteristic polynomial. Indeed by polynomial long division we can write phi(t) = q(t)psi(t) + r(t), where the degree of r is less than the degree of psi. Then

notag      0 = phi(A) = q(A) psi(A) + r(A) = r(A).

If rne 0 then we have a contradiction to the minimality of the degree of psi. Hence r = 0 and so psi divides phi.

The minimal polynomial is unique. For if psi_1 and psi_2 are two different monic polynomials of minimum degree m such that psi_i(A) = 0, i = 1,2, then psi_3 = psi_1 - psi_2 is a polynomial of degree less than m and psi_3(A) = psi_1(A) - psi_2(A) = 0, and we can scale psi_3 to be monic, so by the minimality of m, psi_3 = 0, or psi_1 = psi_2.

If A has distinct eigenvalues then the characteristic polynomial and the minimal polynomial are equal. When A has repeated eigenvalues the minimal polynomial can have degree less than n. An extreme case is the identity matrix, for which psi(t) = t - 1, since psi(I) = I - I = 0. On the other hand, for the Jordan block

notag  J = begin{bmatrix}      lambda & 1 & 0 \      0 & lambda & 1 \      0 & 0 & lambda   end{bmatrix}

the characteristic polynomial and the minimal polynomial are both equal to (lambda - 1)^3.

The minimal polynomial has degree less than n when in the Jordan canonical form of A an eigenvalue appears in more than one Jordan block. Indeed it is not hard to show that the minimal polynomial can be written

notag     psi(t) = displaystyleprod_{i=1}^s (t-lambda_i)^{n_i},

where lambda_1,lambda_2,dots,lambda_s are the distinct eigenvalues of A and n_i is the dimension of the largest Jordan block in which lambda_i appears. This expression is composed of linear factors (that is, n_i = 1 for all i) if and only if A is diagonalizable.

To illustrate, for the matrix

notag  A = left[begin{array}{cc|c|c|c}      lambda &  1      &         &     &   \              & lambda &         &     &   \hline              &         & lambda &     &   \hline              &         &         & mu &    \hline              &         &         &     & mu   end{array}right]

in Jordan form (where blank elements are zero), the minimal polynomial is psi(t) = (t-lambda)^2(t-mu), while the characteristic polynomial is p(t) = (t-lambda)^3(t-mu)^2.

What is the minimal polynomial of a rank-1 matrix, A = xy^* ne 0? Since A^2 = (y^*x) xy^*, we have q(A) = 0 for q(t) = t^2 - (y^*x) t = t^2 - mathrm{trace}(A) t. For any linear polynomial p(t) = t - a_0, p(A) = xy^* - a_0 I, which is nonzero since xy^* has rank 1 and I has rank n. Hence the minimal polynomial is q.

The minimal polynomial is important in the theory of matrix functions and in the theory of Krylov subspace methods. One does not normally need to compute the minimal polynomial in practice.

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