I am supposed to show that if we have a $3\times 3$ matrix with $3$ distinct eigenvalues then the non-cyclic vectors are a union of $3$ distinct, $2$-dimensional subspaces. I am not really sure how to tackle this problem.
My try:
Let $A$ be the matrix in question. Then $A=PBP^{-1}$ where $B$ is the diagonal matrix with the distinct eigenvectors. My first instinct was to check if $B$ has any noncyclic vectors, $v,Bv,B^2v$ gives this matrix: \begin{bmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{bmatrix} where $a,b,c$ are distinct eigenvalues, this matrix is invertible so all vectors (where no entry is $0$) are cyclic. Thus only vectors with some entry $0$ are non cyclic. clearly all such vectors are non-cyclic. Thus I found my $3$ distinct subspaces whose union gives the non cyclic vectors. However, my problem is justifying why noncyclic vectors of $B$ correspond to noncyclic vectors of $A$. Afterall, $A$ might not be diagonal only similar to $B$