Review : Eigenvalues & Eigenvectors — Question 3

Question 3 – Solution

Strategy. Complex eigenvectors encode a real invariant plane; they need not describe a real line fixed in direction.

Step 1: Find and verify the complex pairs. The polynomial is $(\lambda-1)^2+4$, so $$\boxed{\lambda_\pm=1\pm 2i,\qquad E_{1+2i}=\operatorname{span}_{\mathbb C}\{(1,-i)^T\},\quad E_{1-2i}=\operatorname{span}_{\mathbb C}\{(1,i)^T\}.}$$ Indeed $C(1,-i)^T=(1+2i,2-i)^T=(1+2i)(1,-i)^T$. Conjugation verifies the other pair. The distinct eigenvalues give independent complex eigenvectors.

Step 2: Exclude a real eigenvector. If a real nonzero $x$ satisfied $Cx=\lambda x$, a nonzero coordinate of $x$ would make $\lambda$ the ratio of two real numbers. Thus $\lambda$ would be real, but the characteristic polynomial has no real root. No such vector exists.

Step 3: Extract the real plane relations. For $(1,-i)^T=p+iq$, take $p=(1,0)^T$, $q=(0,-1)^T$. Equating real and imaginary parts of $C(p+iq)=(1+2i)(p+iq)$ gives $$\boxed{Cp=p-2q,\qquad Cq=2p+q.}$$ Direct multiplication verifies both. These independent real vectors span the plane even though neither is a real eigenvector.

Step 4: Recover the rotation and scale. Let $\theta\in(0,\pi/2)$ satisfy $\cos\theta=1/\sqrt 5$ and $\sin\theta=2/\sqrt 5$; equivalently $\theta=\arctan(2)$. Then $$C=\sqrt 5\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix},\qquad \boxed{C^m e_1=5^{m/2}(\cos m\theta,\sin m\theta)^T.}$$ Its length is $5^{m/2}$. The figure shows $e_1$ and its image $(1,2)^T$, with unit and image circles drawn on equal scales.