# Visualize the geometric interpretation of the matrix power of a real matrix with complex eigenvectors

I can understand the geometric meaning of $$A^n$$ (here $$A \in R^{n \times n}$$) when the eigenvalues of $$A$$ are all real. Basically, you scale up the each eigenvector $$v_i$$ along its direction by $$\lambda_i$$.

But what happens if the eigenvalues are complex? In that case, the eigenvectors will have complex elements too. I think I can sort to guess, $$\lambda_i$$ here does a bit of the rotation. But how do I define the direction of an eigenvector, with complex elements?

• You only can interpret directions in $\mathbb R^n$ for vectors with real entries. Jun 20, 2021 at 1:40

Perhaps a good example where to see what complex number has to do with real entries' matrices are of the kind $$\left(\begin{array}{cc}a&-b\\b&a\end{array}\right),$$ viewed as linear transformations $$\mathbb R^2\to\mathbb R^2$$.

Its characteristic polynomial is $$\lambda^2-2a\lambda+a^2+b^2$$ whose zeroes are $$\lambda_{\pm}=\frac{2a\pm\sqrt{-4b^2}}{2}$$ that is $$\lambda_+=a+ib,$$ $$\lambda_-=a-ib.$$ But the Eigen-Vectors would be $$v_+=ie_1+e_2$$ and $$v_-=-ie_1+e_2$$ which do not belong to $$\mathbb R^2$$.

However for the matrix above $$\left(\begin{array}{cc}a&-b\\b&a\end{array}\right)=r \left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right),$$ with $$r=\sqrt{a^2+b^2}$$ and $$\theta=\arctan\frac{b}{a}$$ and which allows the interpretation (for the transformation) of being an expansion by the $$r$$ factor and a rotation by an angle $$\theta$$ of the plane $$\mathbb R^2$$ which lets nothing invariant unless $$b=0$$.

• Thanks for the detailed answer. But I am specifically looking for a geometric interpretation of the complex eigenvectors. LIke for real eigenvectors, I know a matrix power is scaling along the direction of that eigenvector. But how do I interpret the complex eigenvectors? Jun 20, 2021 at 2:03
• They are directions on complex vector spaces that do the same things as solving non trivially the equation $Av=\lambda v$ Jun 20, 2021 at 2:08