 Solution

The characteristic polynomial factors as $\det (\lambda I-A)=(\lambda -1)({\lambda }^2-1)$. Thus $\boxed{\lambda =1\text{ twice},\lambda =-1\text{ once}}$ in algebraic multiplicity. An orthonormal basis of corresponding eigenvectors isDirect multiplication verifies $A{v}_1=v_1$, $A{v}_2=v_2$, $A{v}_3=-v_3$. Their mutual Hermitian inner products vanish and each has norm one, which proves linear independence. For the repeated eigenvalue, Gaussian elimination givesHence $\boxed{\mathrm{rank}(A-I)=1,\dim \ker (A-I)=2}$. The second equality follows from the rank-nullity theorem and agrees with the two-dimensional eigenspace. A repeated eigenvalue alone would not imply a two-dimensional eigenspace for an arbitrary matrix; here the Hermitian matrix is diagonalizable.