Characteristic polynomial
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In linear algebra, one associates a polynomial to every square matrix, its characteristic polynomial or secular equation. This polynomial encodes several important properties of the matrix, most notably its eigenvalues, its determinant and its trace.
Motivation
Given a square matrix A, we want to find a polynomial whose roots are precisely the eigenvalues of A. For a diagonal matrix A, the characteristic polynomial is easy to define: if the diagonal entries are a, b, c the characteristic polynomial will be
- (t − a)(t − b)(t − c)...
For a general matrix A, one can proceed as follows. If λ is an eigenvalue of A, then there is an eigenvector v≠0 such that
- A v = λv,
- (λI − A)v = 0
Formal definition
We start with a field K (you can think of K as the real or complex numbers) and an n×n matrix A over K. The characteristic polynomial of A, denoted by pA(t), is the polynomial defined by
- pA(t) = det(t I − A)
Example
Suppose we want to compute the characteristic polynomial of the matrix
- [A=\begin2 & 1\\-1& 0\end.]
- [t I-A = \begint-2&-1\\1&t\end]
- [(t-2)t - 1(-1) = t^2-2t+1.\,\!]
Properties
The polynomial pA(t) is monic (its leading coefficient is 1) and its degree is n. The most important fact about the characteristic polynomial was already mentioned in the motivational paragraph: the eigenvalues of A are precisely the roots of pA(t). The constant coefficient pA(0) is equal to (−1)n times the determinant of A, and the coefficient of t n − 1 is equal to the negative of tr(A), the matrix trace of A. For a 2×2 matrix A, the characteristic polynomial is nicely expressed then as
- t 2 − tr(A)t + det(A).
The Cayley-Hamilton theorem states that replacing t by A in the expression for pA(t) yields the zero matrix: pA(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.
Two similar matrices have the same characteristic polynomial. The converse however is not true in general: two matrices with the same characteristic polynomial need not be similar.
The matrix A and its transpose have the same characteristic polynomial. 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.
Characteristic polynomial of a product of two matrices
If A and B are two square n×n matrices then characterictic polymonials of AB and BA coincide:
- [p_(t)=p_(t).\,]
- [ p_(t) = t^ p_(t).\,]
- [BA = A^ (AB) A.\,]
See also
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