Let \(T\) be the linear operator on \(\RR^2\), the matrix of which in the standard ordered basis is \[ \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}. \] Let \(W_1\) be the subspace of \(\RR^2\) spanned by the vector \(\epsilon_1 = (1, 0)\).
Prove that \(W_1\) is invariant under \(T\).
Prove that there is no subspace \(W_2\) which is invariant under \(T\) and which is complementary to \(W_1\): \[ \RR^2 = W_1 \oplus W_2. \] (Compare with Exercise 1 of Section 6.5.)
Solution
Let \[ A=\begin{pmatrix}2&1\\0&2\end{pmatrix}, \] so \(T(x,y)=(2x+y,2y)\).
1. The line \(W_1=\operatorname{span}(1,0)\) is \(T\)-invariant.
For every scalar \(a\), \[ T(a,0)=(2a,0)=2(a,0)\in W_1. \] Thus \(T(W_1)\subseteq W_1\).
2. No complementary line can be invariant.
Any one-dimensional complement \(W_2\) of \(W_1\) is spanned by a vector \[ v=(x,y) \] with \(y\ne0\); otherwise \(v\in W_1\). If \(W_2\) were \(T\)-invariant, then because it is one-dimensional there would be a scalar \(\lambda\) with \[ Tv=\lambda v. \] Thus \(v\) would be an eigenvector of \(A\). The characteristic polynomial is \((t-2)^2\), so necessarily \(\lambda=2\). But \[ (A-2I)v =\begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}y\\0\end{pmatrix}, \] which vanishes only when \(y=0\). Hence every eigenvector lies in \(W_1\). Therefore no line complementary to \(W_1\) is invariant under \(T\) [@HK71].