Let \(F\) be the field of real numbers, \[ A = \begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, \qquad p = (x - 2)(x - 3)(x - 1). \]
Show that \(p(A) = 0\).
Let \(P_1, P_2, P_3\) be the Lagrange polynomials for \(t_1 = 2\), \(t_2 = 3\), \(t_3 = 1\). Compute \(E_i = P_i(A)\), \(i = 1, 2, 3\).
Show that \(E_1 + E_2 + E_3 = I\), \(E_iE_j = 0\) if \(i \neq j\), \(E_i^2 = E_i\).
Show that \(A = 2E_1 + 3E_2 + E_3\).
Solution
Since \(A\) is diagonal with diagonal entries \(2,2,3,1\), \[ p(A)=\operatorname{diag}(p(2),p(2),p(3),p(1))=0. \]
The Lagrange polynomials are \[ P_1(x)=-(x-3)(x-1), \] \[ P_2(x)=\frac{(x-2)(x-1)}2, \qquad P_3(x)=\frac{(x-2)(x-3)}2. \] Therefore \[ E_1=P_1(A)=\operatorname{diag}(1,1,0,0), \] \[ E_2=P_2(A)=\operatorname{diag}(0,0,1,0), \qquad E_3=P_3(A)=\operatorname{diag}(0,0,0,1). \] It follows immediately that \[ E_1+E_2+E_3=I, \qquad E_iE_j=0\ (i\ne j), \qquad E_i^2=E_i. \] Finally, \[ 2E_1+3E_2+E_3 =\operatorname{diag}(2,2,3,1)=A. \]