Characteristic polynomial

In linear algebra, one associates a polynomial to every square matrix, its characteristic polynomial. This polynomial encodes several important properties of the matrix, most notably its eigenvalues, its determinant and its trace.

We start with a field K (you can think of K as the real or complex numbers) and an n-by-n matrix A over K. The characteristic polynomial of A, denoted by p_A(t), is the element of the polynomial ring K[t] defined by

p_A(t) = det(A - tI)

where I denotes the n-by-n identity matrix. This is indeed a polynomial, since determinants are defined in terms of sums of products. (Some authors define the characteristic polynomial to be det(tI - A); the difference is immaterial since the two polynomials differ at most by a sign.)

The degree of the polynomial p_A(t) is n. The most important fact about the characteristic polynomial is this: the eigenvalues of A are precisely the zeros of p_A(t). The constant coefficient p_A(0) is equal to the determinant of A, and the coefficient of t^n-1 is equal to (-1)^n-1 times the trace of A.

The Cayley-Hamilton theorem states that replacing t by A in the expression for p_A(t) yields the zero matrix: p_A(A) = 0. Simply, every matrix satisfies its own characteristic equation. As a consequence of this, one can show that the minimal polynomial of A divides the characteristic polynomial of A.

The matrix A and its transpose have the same characteristic polynomial. If A and B are similar matrices, then they also have the same characteristic polynomial. The converse however isn't true: matrices with the same characteristic polynomial need not be similar.

A is similar to a triangular matrix[?] if and only if its characteristic polynomial can be completely factored into linear factors over K. In fact, A is even similar to a matrix in Jordan normal form[?] in this case.

Common misspelling and questions (FAQ)

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