The Eigenvalue Problem and the Characteristic Polynomial

The Geometric Intuition

Imagine a linear operator \(A\) acting on a vector space. Most vectors change both their length and their direction when \(A\) is applied. However, there are special directions in the space where the vector is merely stretched or compressed, without rotating. These special directions are the eigenvectors, and the amount of stretching is the eigenvalue. If you stand in an eigenvector direction, the operator \(A\) looks like a simple scalar multiplication. The characteristic polynomial is simply the algebraic condition that asks: "For which scalars \(\lambda\) does the transformation \(A - \lambda I\) fail to be invertible?" If \(A - \lambda I\) is not invertible, it has a non-trivial null space, meaning there exists a non-zero vector that maps to zero, which is precisely the definition of an eigenvector.

From Geometry to Algebra

To find these special scalars, we start with the defining equation \(A\mathbf{v} = \lambda\mathbf{v}\), where \(\mathbf{v} \neq \mathbf{0}\). We can rearrange this to group the terms involving \(\mathbf{v}\):

For a non-trivial solution \(\mathbf{v}\) to exist, the matrix \(A - \lambda I\) must be singular. A square matrix is singular if and only if its determinant is zero. This leads directly to the characteristic equation:

We define the characteristic polynomial \(p_A(\lambda) = \det(A - \lambda I)\). The roots of this polynomial are the eigenvalues. Note that we use \(A - \lambda I\) rather than \(\lambda I - A\); both give exactly the same roots in every dimension, because \(\det(\lambda I - A) = (-1)^n \det(A - \lambda I)\); they differ only in the sign of the leading coefficient, which is \((-1)^n\) for \(\det(A - \lambda I)\).

Basis Independence and Invariance

A critical question arises: if we change our coordinate system, do the eigenvalues change? The answer is no. The eigenvalues are intrinsic properties of the linear operator \(A\), not of the specific matrix representation. If \(S\) is an invertible change-of-basis matrix, the new matrix representation is \(A' = S^{-1}AS\). The characteristic polynomial of the new matrix is:

Because \(\lambda I = S^{-1}(\lambda I)S\), we can write \(S^{-1}AS - \lambda I = S^{-1}(A - \lambda I)S\), so

using the product law and \(\det(S^{-1})\det(S) = 1\). Thus the roots, and therefore the eigenvalues, are invariant under similarity transformations. This is why we can speak of "the" eigenvalues of an operator without specifying a basis.

Computing the Polynomial

For an \(n \times n\) matrix, \(p_A(\lambda)\) is a polynomial of degree \(n\). By the Fundamental Theorem of Algebra, it has exactly \(n\) roots in the complex field, counting multiplicity. For a \(2 \times 2\) matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), the calculation is straightforward:

Expanding this gives \(\lambda^2 - (a+d)\lambda + (ad-bc)\). The coefficient of \(\lambda\) is \(-\text{tr}(A)\) and the constant term is \(\det(A)\). For larger matrices, one typically expands along a row or column, or uses row reduction to simplify the determinant before expanding. The key is to treat \(\lambda\) as a variable and the matrix entries as constants.