Recipe

Find an eigenvalue and eigenvector for a real matrix, over C

Use when: a real matrix's characteristic polynomial has a non-real root, or you're asked to work over \mathbb C even when every root happens to be real

Goal: a complex eigenvalue \lambda and a genuine eigenvector for it, verified by direct substitution

Taught in Every operator on a complex space has an eigenvalue

1. 1

   Compute \det(A-\lambda I) and set it to zero exactly as over \mathbb R, then solve without discarding any non-real root.

   Example

   For T rotating \mathbb R^2 by 90°, matrix \begin{pmatrix}0&-1\\1&0\end{pmatrix}: \det(A-\lambda I)=\lambda^2+1=0, so \lambda=\pm i.

   • the polynomial has only real roots — every eigenvalue is already real, and nothing here changes.
   • the polynomial has non-real roots — treat each one as a genuine eigenvalue over \mathbb C, not a sign of failure.

   Why: this section's theorem guarantees a root always exists over \mathbb C, so nothing gets discarded the way a real-numbers-only course would discard a negative discriminant.

2. 2

   Row-reduce exactly as usual, letting the non-real \lambda you found participate in the arithmetic like any other number.

   Example

   With \lambda=i: (A-iI)\mathbf v=\mathbf 0 gives \begin{pmatrix}-i&-1\\1&-i\end{pmatrix}\mathbf v=\mathbf 0. Row 1 gives -iv_1-v_2=0, so v_2=-iv_1; taking v_1=1 gives \mathbf v=(1,-i).

   Why: an eigenvector is defined exactly the same way over \mathbb C as over \mathbb R, a nonzero solution of (A-\lambda I)\mathbf v=\mathbf 0, with no separate complex definition needed.

3. 3

   Compute A\mathbf v directly and confirm it equals \lambda\mathbf v exactly, carrying out the complex multiplication in full.

   Example

   A(1,-i)=(0\cdot1+(-1)(-i),\ 1\cdot1+0\cdot(-i))=(i,1), and i(1,-i)=(i,-i^2)=(i,1): the two sides match.

   Why: a sign or conjugation slip during complex row reduction is easy to make, and this check catches it before the eigenvector is used anywhere downstream.

4. 4

   A real matrix's non-real eigenvalues always come in conjugate pairs \lambda,\bar\lambda, with conjugate eigenvectors, so finding one member hands you the other for free.

   Example

   For the same T, \lambda=i pairs with \lambda=-i, and (1,-i) pairs with its own conjugate (1,i), an eigenvector for -i with no further row reduction.

   Why: the characteristic polynomial has real coefficients whenever A is real, and a real polynomial's non-real roots are always mirrored across the real axis.