submodule of monomial representation Let $T$ be a linear transformation acting on a vector space $V$, and suppose that
there is a basis of $V$ such that $T$ looks like the product of a diagonal matrix 
and a permutation matrix. I have heard this referred to as a monomial
representation.
If $W$ is a $k[T]$-submodule of $V$, then must $W$ also be representation of this type?
 A: If $k=\mathbb{R}$, consider $T=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}$ and let $W$ be the subspace orthogonal to $(1,1,1)$, which is a two-dimensional subspace upon which $T$ acts by rotation by $120$ degrees.  If the matrix of $T$ restricted to $W$ were monomial, the first column in any basis would have to be a multiple of $(0,1)$, so the second basis vector would be $120$ degrees from the first, and then the second column couldn't be a multiple of $(1,0)$.  With this example, lack of algebraic closure was key, and switching to $k=\mathbb{C}$ the subspace decomposes into two $1$-dimensional representations of $k[T]$ since $\langle T\rangle\cong\mathbb{Z}/3\mathbb{Z}$ is a finite abelian group.
If $k$ is algebraically closed with characteristic $0$, then $T^{\dim V}$ is diagonal, hence $T$ is diagonalizable (that is, semisimple), so every invariant subspace has a basis with respect to which $T$ is diagonal.
If $k$ has characteristic $3$, then same matrix from before has Jordan normal form $PJP^{-1}$ with
$$P=\begin{pmatrix}1&2&1\\1&1&0\\1&0&0\end{pmatrix},
J=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}.$$
Thus, there is a two-dimensional invariant subspace with a basis such that the matrix of $T$ restricted to the subspace is $\begin{pmatrix}1&1\\0&1\end{pmatrix}$.  There is no basis with respect to which this matrix is monomial; one reason is that $2\times 2$ non-diagonal invertible monomial matrices have characteristic polynomials which do not have double roots.
