Let \(A\) be an \(n \times n\) matrix with real entries. Let \(T\) be the linear operator on \(\RR^n\) 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. Use the result of Exercise 20 to prove the following: If the only subspaces invariant under \(T\) are \(\RR^n\) and the zero subspace, then \(U\) is diagonalizable.
Solution
1. Every nonzero real vector is cyclic for \(T\).
Let \(0\ne v\in\mathbb R^n\). Its cyclic subspace \[ Z(v;T)=\operatorname{span}\{v,Tv,T^2v,\ldots\} \] is nonzero and \(T\)-invariant. By hypothesis the only such subspaces are \(0\) and \(\mathbb R^n\). Therefore \[ Z(v;T)=\mathbb R^n. \] So every nonzero vector is cyclic.
2. The real characteristic polynomial is irreducible and has degree at most two.
By Exercise 20, step 1 implies that the characteristic polynomial \(\chi_A\in\mathbb R[x]\) is irreducible [@HK71]. Every irreducible polynomial over \(\mathbb R\) has degree one or two: a polynomial of odd degree has a real root, while after removing real linear factors any nonreal complex root occurs with its conjugate and gives a real quadratic factor. Hence \[ n=\deg\chi_A\le2. \]
3. Over \(\mathbb C\), the characteristic polynomial has distinct roots.
If \(n=1\), the complex operator \(U\) is already represented by a \(1\times1\) matrix and is diagonal.
If \(n=2\), irreducibility over \(\mathbb R\) means \(\chi_A\) has no real root. Its two complex roots are therefore a nonreal conjugate pair \[ \lambda,\overline\lambda, \] which are distinct. Thus \(U\) has two distinct characteristic values on the two-dimensional complex vector space \(\mathbb C^2\). Eigenvectors belonging to distinct characteristic values are linearly independent, so they form a basis of \(\mathbb C^2\) [@HK71]. Hence \(U\) is diagonalizable.
Thus in all cases \[ \boxed{U\text{ is diagonalizable over }\mathbb C.} \]