How do we find eigenvalues without spinning a vector around by hand? Rewrite A\vec{v} = \lambda\vec{v} as (A - \lambda I)\vec{v} = \vec{0}. For a non-zero \vec{v} to satisfy this, the matrix A - \lambda I must collapse space — its determinant must be zero:

This is the characteristic equation. For a 2×2 matrix it expands into a neat quadratic in \lambda — using the trace T = a + d and the determinant D = ad - bc:

Roots are eigenvalues

Plotted against \lambda, the characteristic polynomial is a parabola, and the eigenvalues are exactly where it crosses zero. Slide the trace and determinant and watch the roots move. Push the curve until it just kisses the axis (a repeated root) or lifts off it entirely (complex eigenvalues — the matrix is a pure rotation).

A handy shortcut

The quadratic hands you two tidy facts for free: the eigenvalues add up to the trace (\lambda_1 + \lambda_2 = T) and multiply to the determinant (\lambda_1\lambda_2 = D). So a quick glance at a 2×2 matrix often gives its eigenvalues by inspection. With the eigenvalues in hand, the eigenvectors come next.