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

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

2023-04-26 03:48:31

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.

(Notice 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 might 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 specifically it divides the attribute polynomial. Certainly by polynomial lengthy division we are able to 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 now we have 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 are able to 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 id matrix, for which psi(t) = t - 1, since psi(I) = I - I = 0. Alternatively, 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 a couple of Jordan block. Certainly it isn’t exhausting to indicate that the minimal polynomial might be written

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

See Also

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 wherein lambda_i seems. This expression consists of linear components (that’s, n_i = 1 for all i) if and provided that A is diagonalizable.

As an example, for the matrix

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

in Jordan kind (the place clean parts 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^*, now 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. Therefore the minimal polynomial is q.

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

Source Link

What's Your Reaction?
Excited
0
Happy
0
In Love
0
Not Sure
0
Silly
0
View Comments (0)

Leave a Reply

Your email address will not be published.

2022 Blinking Robots.
WordPress by Doejo

Scroll To Top