Let \(T\) be the linear operator on \(\RR^2\), the matrix of which in the standard ordered basis is \[ A = \begin{bmatrix} 1 & -1 \\ 2 & 2 \end{bmatrix}. \]
Prove that the only subspaces of \(\RR^2\) invariant under \(T\) are \(\RR^2\) and the zero subspace.
If \(U\) is the linear operator on \(C^2\), the matrix of which in the standard ordered basis is \(A\), show that \(U\) has 1-dimensional invariant subspaces.
Solution
The characteristic polynomial of \[ A=\begin{bmatrix}1&-1\\2&2\end{bmatrix} \] is \[ p(x)=\det(xI-A)=x^2-3x+4. \] Its discriminant is \(9-16=-7\), so it has no real root.
For (a), every proper nonzero subspace of \(\mathbb R^2\) is one-dimensional. If such a line \(W=\mathbb Rv\) were \(T\)-invariant, then \(Tv\in W\), so \(Tv=cv\) for some real \(c\). This would make \(c\) a real characteristic value, impossible. Hence the only invariant subspaces are \(0\) and \(\mathbb R^2\).
For (b), over \(\mathbb C\) the polynomial factors as \[ p(x)=\left(x-\frac{3+i\sqrt7}{2}\right) \left(x-\frac{3-i\sqrt7}{2}\right). \] Thus \(U\) has complex eigenvectors for these two eigenvalues, and the span of any such eigenvector is a one-dimensional invariant subspace.