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

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

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.

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