Let \(T\) be the diagonalizable linear operator on \(\RR^3\) which we discussed in Example 3 of Section 6.2. Use the Lagrange polynomials to write the representing matrix \(A\) in the form \(A = E_1 + 2E_2\), \(E_1 + E_2 = I\), \(E_1E_2 = 0\).
Solution
In the referenced example, the distinct characteristic values of \(T\) are \(1\) and \(2\) [@HK71]. The corresponding Lagrange polynomials are \[ p_1(x)=\frac{x-2}{1-2}=2-x, \qquad p_2(x)=\frac{x-1}{2-1}=x-1. \] Therefore define \[ \boxed{E_1=p_1(A)=2I-A, \qquad E_2=p_2(A)=A-I.} \]
1. These matrices resolve the identity and recover \(A\).
Directly, \[ E_1+E_2=(2I-A)+(A-I)=I, \] and \[ E_1+2E_2=(2I-A)+2(A-I)=A. \]
2. The two spectral projections annihilate one another.
Since the only characteristic values are \(1\) and \(2\) and \(A\) is diagonalizable, its minimal polynomial is \[ m_A(x)=(x-1)(x-2). \] Hence \[ (A-I)(A-2I)=0 \] [@HK71]. Therefore \[ E_1E_2=(2I-A)(A-I)=-(A-2I)(A-I)=0. \] Thus \[ \boxed{A=E_1+2E_2,\qquad E_1+E_2=I,\qquad E_1E_2=0}, \] as required.