Is orthogonal subspace to Eigen subspace another eigensubspace?

Question

Is Orthogonal subspace to eigen subspace another eigen subspace?

Say A is matrix. $\lambda$ is an eigen value and $U=E_{\lambda} $ is its corresponding eigen subspace. $U^{\perp}$ is another eigen subspace?

$Au = \lambda u , u \in E_{\lambda}\\ $

Now say $\vec v$ is another vector in orthogonal subspace to U. $v^Tu=0 \\ u^TAv = v^TAu = v^T(\lambda u) = 0$

It means $A \vec v$ is also perpendicular to $\vec u$.

It means $A \vec v$ lies in V(orthogonal subspace of $E_{\lambda}$). Would that mean that $A \vec v = \lambda' v$ ???? making V another eigen subspace for $\lambda'$. If V happened to be spanned by single vector then V becomes eigen subspace right? It can happen that V may be spanned by more than 1 vector. In that case Av though lies in V (orthogonal subspace), Av may not be equal to scaled version of $\vec v$(Only in scaled version of v vector right, we call that eigen subspace!?)

Answer 1

No, in addition to the example GReyes gave of a $3\times 3$ matrix with only one eigenvalue, if you have a matrix with more than one eigenvalue then $U^{\perp}$ will contain all of the other eigenspaces and so will not usually itself be an eigenspace.

In fact for $U^{\perp}$ to be an eigenspace, the matrix must have precisely two eigenvalues, and be diagonalizable (ie there must exist a basis for $\mathbb{R}^{n}$ consisting entirely of eigenvectors).

Answer 2

This is not true in general. Just consider a nontrivial rotation $\mathbb{R}^3$. The axis of rotation is an eigenspace with eigenvalue $1$, but the orthogonal complement is not a (real) eigenspace (there are no other real eigenvalues).