Let A∈M2(R) have coordinates (z,w). Show that A has real eigenvalues if and only if ∣w∣≥∣Imz∣, and give the eigenvalues in that case.
Hint
The eigenvalues are the roots of t2−(trA)t+detA. Compute the discriminant using trA=2Rez and detA=∣z∣2−∣w∣2.
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Solution
Idea. A real 2×2 matrix has real eigenvalues iff its characteristic polynomial has nonnegative discriminant. In the (z,w) coordinates, this discriminant is a difference of two squares.
Step 1: trace and determinant.trA=a+d=2Rez and detA=∣z∣2−∣w∣2 (previous problems).
Step 2: discriminant. The characteristic polynomial t2−(trA)t+detA has discriminant Δ=(trA)2−4detA=4(Rez)2−4∣z∣2+4∣w∣2=4(∣w∣2−(Imz)2), since ∣z∣2=(Rez)2+(Imz)2.
Step 3: conclusion.A has real eigenvalues iff Δ≥0 iff ∣w∣≥∣Imz∣. In that case the eigenvalues are λ±=2trA±Δ=Rez±∣w∣2−(Imz)2. If ∣w∣<∣Imz∣ the eigenvalues are the non-real pair Rez±i(Imz)2−∣w∣2.
Translation to entries.Imz=2c−b and ∣w∣2=4(a−d)2+(b+c)2, so ∣w∣2−(Imz)2=4(a−d)2+(b+c)2−(c−b)2=4(a−d)2+4bc, which is the familiar discriminant (a−d)2+4bc≥0.
Check. Rotation R(θ): z=eiθ, w=0, so ∣w∣=0<∣sinθ∣=∣Imz∣ unless θ∈πZ; real eigenvalues only for R(0)=I and R(π)=−I. Reflection diag(1,−1): z=0, w=1, eigenvalues 0±1=±1. Numerically, the formula Rez±∣w∣2−(Imz)2 matches `numpy` eigenvalues for 200 random matrices.
Remark. The set of matrices with a real eigenvalue is {∣w∣≥∣Imz∣}: it is closed, with interior {∣w∣>∣Imz∣} and boundary {∣w∣=∣Imz∣} (matrices with a repeated real eigenvalue). This recovers the description of the interior in terms of tr2>4det.
Lesson. When a discriminant condition appears, look for coordinates in which it factors or becomes a comparison of two moduli.