Let \(V\) be finite-dimensional over a field \(F\), and let \(L:V\to V\) have characteristic polynomial \[ \chi=\chi_1\chi_2, \] where \(\chi_1,\chi_2\in F[x]\) are relatively prime. Prove that \[ V=V_1\oplus V_2 \] for subspaces \(V_1,V_2\) satisfying \[ \chi_i(L)V_i=0, \qquad i=1,2. \]
Solution
Set \[ V_1\coloneqq\ker\chi_1(L), \qquad V_2\coloneqq\ker\chi_2(L). \]
1. There exist polynomials \(a,b\in F[x]\) such that\[ a\chi_1+b\chi_2=1. \]
The polynomials \(\chi_1\) and \(\chi_2\) are relatively prime, so Bézout’s identity in the Euclidean domain \(F[x]\) gives such \(a\) and \(b\).
2. Every \(v\in V\) can be written as a sum \(v=v_1+v_2\) with\(v_i\in V_i\).
For \(v\in V\), define \[ v_1\coloneqq b(L)\chi_2(L)v, \qquad v_2\coloneqq a(L)\chi_1(L)v. \] Evaluating the identity from step 1 at \(L\) gives \[ I_V=a(L)\chi_1(L)+b(L)\chi_2(L), \] so \(v=v_1+v_2\).
By the Cayley–Hamilton theorem, \[ \chi(L)=\chi_1(L)\chi_2(L)=0. \] Since all polynomials in \(L\) commute, \[ \chi_1(L)v_1 =b(L)\chi_1(L)\chi_2(L)v =0, \] and similarly \[ \chi_2(L)v_2 =a(L)\chi_2(L)\chi_1(L)v =0. \] Hence \(v_1\in V_1\) and \(v_2\in V_2\).
3. One has \(V_1\cap V_2=\{0\}\).
Let \(w\in V_1\cap V_2\). Then \[ \chi_1(L)w=0 \qquad\text{and}\qquad \chi_2(L)w=0. \] Applying the operator identity from step 2 gives \[ w =a(L)\chi_1(L)w+b(L)\chi_2(L)w =0. \] Thus the intersection is trivial.
4. The required decomposition is\[ \boxed{V=V_1\oplus V_2}, \]and \(\chi_i(L)V_i=0\) for \(i=1,2\).
Step 2 gives \(V=V_1+V_2\), while step 3 shows that the sum is direct. By the definition of \(V_i\), the operator \(\chi_i(L)\) vanishes on \(V_i\).
5. Q.E.D.
Step 4 is exactly the required conclusion.