Say we have an $n \times n$ matrix A, which is both orthogonal and symmetric, I figured out that its eigenvalues are $\lambda=1$ and $\lambda=-1$, however, there are two things I can't figure out:
- What can we say about the Eigenspaces?
- How to interpret the linear transformation $T(\vec{x})=A\vec{x}$ geometrically in the cases $n=2$ and $n=3$?
Best Answer
You have the decomposition $\mathbb{R}^n = V \oplus W$, where $V$ is the eigenspace associated to the eigenvalue $1$, and $W$ the eigenspace associated to the eigenvalue $-1$. Moreover, $V$ and $W$ are orthogonal.
So for any vector $x \in \mathbb{R}^n$, write $x=v+w$, with $v \in V$ and $w \in W$. You have $T(x)=v-w$ : you can see this vector as the symmetric of the vector $x$ w.r.t. the subspace $V$ (because $V$ is fixed and its orthogonal is "reversed"). Therefore, $T$ is like a reflection, except that the fixed subspace is not necessarily an hyperplane.