Recipe

Solve a characteristic equation with no real roots

Use when: you've formed A-\lambda I for a 2\times2 matrix and the resulting quadratic's discriminant, (\operatorname{tr}A)^2-4\det A, comes out negative

Goal: the complex conjugate pair of eigenvalues, plus the correct conclusion that A has no real eigenvector

Taught in When the characteristic polynomial has no real roots

1. 1

   Build A-\lambda I and expand its determinant into a quadratic in \lambda, exactly as for any 2\times2 matrix.

   Example

   For A = \begin{bmatrix} 0 & -2 \\ 1 & 2 \end{bmatrix}: p_A(\lambda) = \det\begin{bmatrix} -\lambda & -2 \\ 1 & 2-\lambda \end{bmatrix} = (-\lambda)(2-\lambda) - (-2)(1) = \lambda^2 - 2\lambda + 2

   Why: this is the same characteristic polynomial every earlier section built — whether its roots turn out real or not isn't known until the next step.

2. 2

   Compute (\operatorname{tr}A)^2 - 4\det A (the discriminant of the quadratic) before doing any more algebra.

   Example

   \operatorname{tr}A = 0+2=2 and \det A = (0)(2)-(-2)(1)=2, so the discriminant is 2^2-4(2) = 4-8=-4, which is negative — the roots will be complex.

   • discriminant \ge 0 — the roots are real; use the ordinary quadratic formula (or the find-eigenvalues recipe) instead of this one.
   • discriminant < 0 — the roots are a complex conjugate pair; continue to Step 3.

   Why: checking the sign first avoids stalling on a negative square root partway through the quadratic formula, and it's the exact algebraic signature of "no real eigenvector exists" this section derives.

3. 3

   Plug into \lambda = \dfrac{\operatorname{tr}A \pm \sqrt{(\operatorname{tr}A)^2-4\det A}}{2}, then rewrite the negative square root as i\sqrt{\,|\text{discriminant}|\,}.

   Example

   \lambda = \frac{2 \pm \sqrt{-4}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i

   Why: the quadratic formula still works over the complex numbers; the only change from the real case is trading \sqrt{-1} for i instead of declaring "no solution."

4. 4

   State the two eigenvalues as a conjugate pair a\pm bi, and note that since neither is real, A has no real eigenvector for either one.

   Example

   \lambda = 1+i and \lambda = 1-i are genuine eigenvalues of A, but since A is a real matrix with no real \lambda, there is no real nonzero \mathbf v solving A\mathbf v = \lambda\mathbf v for either root.

   Why: a real matrix's characteristic polynomial always has real coefficients, and a real polynomial's non-real roots are guaranteed to come in exactly this conjugate-pair form — never a single isolated complex root.