Prove that every linear transformation \[ T:\mathbb R^3\to\mathbb R^3 \] has
- a one-dimensional invariant subspace;
- a two-dimensional invariant subspace.
Solution
1. The characteristic polynomial of \(T\) has a real root.
After choosing a basis, the characteristic polynomial is a real polynomial of degree \(3\). Every real polynomial of odd degree has at least one real root.
2. The transformation \(T\) has a one-dimensional invariant subspace.
1. Then\[ \det(T-\lambda I)=0, \]so there is a nonzero vector \(u\) with\[ Tu=\lambda u. \]Therefore the line\[ \operatorname{span}\{u\} \]is \(T\)-invariant.
3. The transpose \(T^t\) has a nonzero real eigenvector.
The characteristic polynomials of \(T\) and \(T^t\) are equal because \[ \det(tI-T^t) = \det\bigl((tI-T)^t\bigr) = \det(tI-T). \] Hence step 1 also supplies a real eigenvalue \(\mu\) of \(T^t\), with a nonzero vector \(v\) satisfying \[ T^tv=\mu v. \]
4. The plane\[ P\coloneqq v^\perp \]is \(T\)-invariant.
Since \(v\neq0\) in \(\RR^3\), its orthogonal complement has dimension \(2\). If \(x\in P\), then \[ \langle x,v\rangle=0. \] Using step 3, \[ \begin{aligned} \langle Tx,v\rangle &= \langle x,T^tv\rangle\\ &= \langle x,\mu v\rangle\\ &= \mu\langle x,v\rangle\\ &=0. \end{aligned} \] Thus \(Tx\in v^\perp=P\), proving invariance.
5. The transformation \(T\) has both a one-dimensional and atwo-dimensional invariant subspace.
Step 2 supplies the invariant line, and step 4 supplies the invariant plane.
6. Q.E.D.
Step 5 proves both requested assertions.