Let \(p = (x - 2)(x - 3)(x - 1)\) and let \(T\) be any linear operator on \(\RR^4\) such that \(p(T) = 0\). Let \(P_1, P_2, P_3\) be the Lagrange polynomials of Exercise 3, and let \(E_i = P_i(T)\), \(i = 1, 2, 3\). Prove that \[ E_1 + E_2 + E_3 = I, \qquad E_iE_j = 0 \quad \text{if} \quad i \neq j, \] \[ E_i^2 = E_i, \quad \text{and} \quad T = 2E_1 + 3E_2 + E_3. \]
Solution
The Lagrange polynomials satisfy \[ P_i(t_j)=\delta_{ij} \qquad(t_1,t_2,t_3)=(2,3,1). \] The polynomial \[ P_1+P_2+P_3-1 \] has degree at most \(2\) and vanishes at the three distinct points \(1,2,3\), so it is identically zero. Hence \[ E_1+E_2+E_3=I. \] Similarly, \[ x-(2P_1+3P_2+P_3) \] has degree at most \(2\) and vanishes at \(1,2,3\), so \[ T=2E_1+3E_2+E_3. \]
For \(i\ne j\), the polynomial \(P_iP_j\) vanishes at each of \(1,2,3\), hence is divisible by \[ p(x)=(x-1)(x-2)(x-3). \] Therefore \(P_i(T)P_j(T)=0\), i.e. \[ E_iE_j=0. \] Likewise, \(P_i^2-P_i\) vanishes at \(1,2,3\), so it is divisible by \(p\) and \[ E_i^2-E_i=0. \] Thus each \(E_i\) is idempotent and the \(E_i\) are pairwise orthogonal, with the stated decomposition of \(I\) and \(T\).