Finding the eigenvalues

You can see eigenvectors in the picture. To find them, you need one clever reframing — and then it's a quadratic equation, which you've been solving since algebra.

The reframing

Av = λv says "A acts like plain scaling on v." Move everything to one side:

$(A - \lambda I)v = 0$

Read it as a transformation: the matrix (A − λI) sends the nonzero vector v to zero — it collapses v's direction. But a transformation that collapses anything has determinant zero (module 2's fatal case). So the eigenvalues are exactly the λ that make:

$\det(A - \lambda I) = 0$

That's the characteristic equation, and for a 2×2 it expands to a quadratic with famous coefficients:

$\lambda^2 - \operatorname{tr}(A)\,\lambda + \det(A) = 0$

where tr(A) = a + d is the trace (the diagonal sum). Trace and determinant — the two numbers you already know how to read — are the quadratic's coefficients: the eigenvalues sum to the trace and multiply to the determinant.

Worked: A = [2 1 / 1 2]

tr = 4, det = 3, so λ² − 4λ + 3 = 0 → (λ − 1)(λ − 3) = 0 → λ = 3 and λ = 1. Check: 3 + 1 = 4 ✓ and 3 · 1 = 3 ✓.

For the eigenvectors, plug each λ back into (A − λI)v = 0. With λ = 3: (A − 3I) = [−1 1 / 1 −1], and [−1 1 / 1 −1]·(1, 1) = (0, 0) ✓ — so (1, 1) is the λ = 3 direction, exactly the diagonal the picture showed.

The discriminant tells the story

The quadratic's discriminant, tr² − 4det, sorts matrices into the cases from last lesson:

The visualizer computes all of this per matrix — use it to check your hand work.

Why this matters

One determinant, one quadratic, done. But hold onto the shape of the argument — "the interesting values of λ are where a determinant dies" — because it returns in the final lesson wearing a calculus costume.