Eigenvalues Basics

Solve λ^2 - ( tr)λ + = 0 — trace and determinant do the work.

The idea

An eigenvalue of a square matrix $A$ is a number $\lambda$ such that $Av = \lambda v$ for some nonzero vector $v$. Along the line of such a $v$ the matrix does not rotate at all: it only stretches or shrinks by the factor $\lambda$, so $Av$ stays on $v s own line — while for a typical vector $u$ the image $Au$ turns off the line through $u$, as the figure shows. These directions are where the action of a matrix is simplest to read, and the eigenvalues are the scaling factors it applies there.

Rewriting $Av = \lambda v$ as $(A - \lambda I)v = 0$ turns the definition into a computation. The matrix $A - \lambda I$ must send a nonzero vector to zero, and a square matrix does so exactly when its determinant is zero. The eigenvalues of $A$ are therefore the solutions of $\det(A - \lambda I) = 0.$

For a $2 \times 2$ matrix this equation is a quadratic. Writing $\operatorname{tr} A$ for the trace of $A$, the sum of its diagonal entries, it reads $\lambda^{2} - (\operatorname{tr} A)\,\lambda + \det A = 0.$ The roots of this quadratic sum to $\operatorname{tr} A$ and multiply to $\det A$, so the trace and determinant give a fast way to find the eigenvalues of a $2 \times 2$ matrix, and a check on any answer.

Ways to work on it

• Walkthrough. Find both eigenvalues of a 2 × 2 matrix, step by step.
• Practice. Three steps. Random 2 × 2 with small integer eigenvalues.
• Hardest. Two steps. A random 2 × 2 whose eigenvalues may be negative.

Not sure where to start? Take the ten-question placement test.