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HomeMatlabWhat Is the Minimal Polynomial of a Matrix? – Nick Higham

What Is the Minimal Polynomial of a Matrix? – Nick Higham


For a polynomial

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

the place 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.

(Word that the fixed time period is a_0 A^0 = a_0 I). The polynomial phi is monic if a_k = 1.

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

The minimal polynomial divides any polynomial phi such that phi(A) = 0, and particularly it divides the attribute polynomial. Certainly by polynomial lengthy division we will write phi(t) = q(t)psi(t) + r(t), the place the diploma of r is lower than the diploma of psi. Then

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

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

The minimal polynomial is exclusive. For if psi_1 and psi_2 are two totally different monic polynomials of minimal diploma m such that psi_i(A) = 0, i = 1,2, then psi_3 = psi_1 - psi_2 is a polynomial of diploma lower than m and psi_3(A) = psi_1(A) - psi_2(A) = 0, and we will 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 attribute polynomial and the minimal polynomial are equal. When A has repeated eigenvalues the minimal polynomial can have diploma lower than n. An excessive case is the identification matrix, for which psi(t) = t - 1, since psi(I) = I - I = 0. However, for the Jordan block

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

the attribute polynomial and the minimal polynomial are each equal to (lambda - 1)^3.

The minimal polynomial has diploma lower than n when within the Jordan canonical type of A an eigenvalue seems in multiple Jordan block. Certainly it’s not laborious to indicate that the minimal polynomial could be written

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

the place lambda_1,lambda_2,dots,lambda_s are the distinct eigenvalues of A and n_i is the dimension of the biggest Jordan block by which lambda_i seems. This expression consists of linear elements (that’s, n_i = 1 for all i) if and provided that A is diagonalizable.

For example, for the matrix

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

in Jordan type (the place clean components are zero), the minimal polynomial is psi(t) = (t-lambda)^2(t-mu), whereas the attribute polynomial is p(t) = (t-lambda)^3(t-mu)^2.

What’s the minimal polynomial of a rank-1 matrix, A = xy^* ne 0? Since A^2 = (y^*x) xy^*, we have now 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. Therefore the minimal polynomial is q.

The minimal polynomial is necessary within the concept of matrix features and within the concept of Krylov subspace strategies. One doesn’t usually have to compute the minimal polynomial in observe.

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