decomposing elements of the Lie group $SO(n)$ by means of diagonalization

Question

I am reading the proof of the corollary 0.2 of this link https://www.math.tamu.edu/~rojas/son.pdf and several questions arise. I would like to know what the $\Lambda$ matrix is ​​like, I don't know very well how this matrix is ​​composed graphically, could someone show me what this matrix looks like for a specific dimension?

Answer 1

$\Lambda$ is a diagonal matrix, which means that even though it has $n^2$ entries, the vast majority of the entries equal zero. Only the $n$ entries on the diagonal are non-zero. The entries of $\Lambda$ on the diagonal are not limited to being real numbers - they may also be complex, i.e. of the form $a+bi$ for some numbers $a$ and $b$, and $i=\sqrt{-1}$ is the imaginary unit. Even further, when plotted at $(a,b)$ on the plane these complex numbers lie on the unit circle, which is equivalent to saying that $a^2+b^2=1$. Thus, when the non-zero entries of $\Lambda$ are plotted in the complex plane, they become $n$ points on the unit circle (circle of radius $1$ centered at the origin).

Geometrically, the values on the diagonal of $\Lambda$ represent its eigenvalues. Since they are complex numbers on the unit circle, they can be written as $e^{i\theta}$, where $\theta$ represents an angle (in radians). In fact, the non-real eigenvalues come in pairs $e^{\pm i\theta}$, corresponding to a 2-dimensional eigenspace, and the vectors in that eigenspace are rotated by an angle of $\pm\theta$.