# Geometric interpretation of complex eigenvalues

What is the geometric interpretation of complex eigenvalues? For me it is clear that real eigenvalues of a matrix $A$ are associates to eigenvectors along which the matrix $A$ contracts or expands. Complex eigenvalues are associated intuitively (but not clearly) to me to eigenvectors along which the matrix $A$ rotates the space.

• Do you know about the real Jordan form? That makes clear your intuition. en.wikipedia.org/wiki/Jordan_normal_form#Real_matrices Jul 30, 2014 at 18:00
• I know it, but I do not have it geometric interpretation clear. Jul 30, 2014 at 18:02
• I may write an answer later but here is a good geometric description: math.purdue.edu/~bkrummel/ma265_lecture5_5.pdf Jul 31, 2021 at 7:27
• @Al.G. the link is dead, do you remember the title of the note? Nov 27, 2023 at 16:55
• Sadly not, but there are other resources that explain the same. A quick googling lead me here: services.math.duke.edu/~jdr/1617f-1553/materials/11-09-web.pdf, which kinda explains what I was thinking about - a pair of complex eigenvalues means rotation by the complex value argument within a particular subspace (in contrast to real eigenvalues, which encode scaling of the axes). Nov 27, 2023 at 17:10

If $A$ is real matrix and exists its (not real) eigenvalue $\lambda=a+bi$ and eigenvector $x=v+iw$ ($a,b \in \mathbb{R}$, $v,w$-real vectors), then:
$Ax=\lambda x$, so:
$A(v+iw)=\lambda(v+iw)$
$Av+iAw=av-bw+i(aw+bv)$
So $Av=av-bw$ and $Aw=ax+bw$. You can interpret this result this way: there exist $a,b \in \mathbb{R}$ and vectors $v,w$ that $A$ transforms $v$ to $av-bw$ and $w$ to $aw+bv$.