# Eigenvalue question

1. May 22, 2009

### jeff1evesque

Hello,

I was reading something in my text/wikipedia, and they both said that "...the eigenvalues of a matrix are the zeros of its characteristic polynomial." Do they mean that λ in the characteristic polynomial causes det (A - λI) = 0 (in particular A = λI)?

JL

Last edited: May 22, 2009
2. May 23, 2009

### dx

The characteristic polynomial is p(x) = det(A - xI). The zeroes of the characteristic polynomial are the x values that satisfy det(A-xI) = 0.

3. May 23, 2009

### HallsofIvy

Staff Emeritus
Yes, that's exactly what they mean.

If the determinant of $A- \lambda I$ is not 0, then $A- \lambda I$ has an inverse and so the equation $(A- \lambda I)v= 0$ has a unique solution $(A- \lambda I)^{-1}(A- \lambda I)v= v= (A- \lambda I)^{-1}0= 0$ which contradicts the definition of "eigenvalue" which is that the equation $Av= \lambda v$, equivalent to $(A- \lambda I)v= 0$, has "non-trivial" (non-zero) solutions.