eigenvalue (original) (raw)

(5) det⁡(B)=0, i.e. det⁡(λ⁢I-A)=0.

But if V is of infinite dimension, (5) has no meaning and the conditions (2) and (4) are not equivalent to (1). A scalar λ satisfying (2) (called a spectral value ofA) need not be an eigenvalue. Consider for example the complex vector space V of all sequences (xn)n=1∞ of complex numbers with the obvious operations, and the map A:V→V given by

Zero is a spectral value of A, but clearly not an eigenvalue.

Now suppose again that V is of finite dimension, say n. The function

is a polynomial of degree n over k in thevariable λ, called the characteristic polynomial of the endomorphism A. (Note that some writers define the characteristic polynomial as det⁡(A-λ⁢I) rather than det⁡(λ⁢I-A), but the two have the same zeros.)

If k is ℂ or any other algebraically closed field, or if k=ℝand n is odd, then χ has at least one zero, meaning that Ahas at least one eigenvalue. In no case does A have more than neigenvalues.

Although we didn’t need to do so here, one can compute the coefficients of χ by introducing a basis of V and the corresponding matrix forB. Unfortunately, computing n×n determinants and finding roots of polynomials of degree n are computationally messy procedures for even moderately large n, so for most practical purposes variations on this naive scheme are needed. See the eigenvalue problem for more information.

If k=ℂ but the coefficients of χ are real (and in particular ifV has a basis for which the matrix of A has only real entries), then the non-real eigenvalues of A appear in conjugate pairs. For example, if n=2 and, for some basis, A has the matrix

then χ⁢(λ)=λ2+1, with the two zeros ±i.