# If the only sub-spaces invariant under $Τ$ are $R^n$ and the zero subspace, then $U$ is diagonalizable.

Let $A$ be an $n \times n$ matrix with real entries. Let $Τ$ be the linear operator on $R^n$ which is represented by $A$ in the standard ordered basis, and let $U$ be the linear operator on $C^n$ which is represented by $A$ in the standard ordered basis.

Prove that every $0\neq v \in V$ is a cyclic vector if and only if the characteristic polynomial of $T$ is irreducible over $F$.

Use the above result to prove the following: If the only sub-spaces invariant under $Τ$ are $R^n$ and the zero subspace, then $U$ is diagonalizable.

I have proved that every $0\neq v \in V$ is a cyclic vector if and only if the characteristic polynomial of $T$ is irreducible over $F$.

But unable to prove the second part.

For any irreducible factor of the minimal polynomial of $T$, you get a non-trivial invariant subspace of $\mathbb{R}^n$. Therefore if $0$ and $\mathbb{R}^n$ are the only invariant subspaces, the minimal polynomial of $T$, $m_T(x)$, must be an irreducible quadratic over $\mathbb{R}$.
But $U$ also satisfies $m_T(x)$. So $m_U(x) | m_T(x)$. However, $m_T(x) = (x-\mu)(x-\overline{\mu})$ for some $\mu \in \mathbb{C}\backslash \mathbb{R}$. So $m_U(x)$ is either a linear function or a quadratic with distinct roots. In either case $m_U(x)$ factors into distinct linear factors which is equivalent to $U$ being diagonalizable.