Let \(V\) be a vector space over a field \(F\), and let \(L: V \to V\) be a linear operator. Suppose that \(1\) is not an eigenvalue of \(L\) (that is, \(1 \notin \operatorname{spec}(L)\)). Prove that \(x = 0\) is the unique fixed point of \(L\) (i.e. \(L(x) = x \implies x = 0\)). Moreover, if \(V\) is finite-dimensional, show that \(I - L\) is an invertible linear operator on \(V\).
Solution
1. \(\ker(I-L)=0\); that is, \(0\) is the only fixed point of \(L\).
If \(L(x)=x\), then \((I-L)x=0\). A nonzero such \(x\) would be an eigenvector of \(L\) with eigenvalue \(1\), contrary to hypothesis.
2. If \(\dim V<\infty\), then \(I-L\) is invertible.
By step 1, \(I-L\) is injective, and rank-nullity gives \(\dim\operatorname{im}(I-L)=\dim V\), so \(I-L\) is also surjective.