Let \(A\) be the \(4 \times 4\) matrix in Example 5 of Section 6.3. Find matrices \(E_1, E_2, E_3\) such that \(A = c_1E_1 + c_2E_2 + c_3E_3\), \(E_1 + E_2 + E_3 = I\), and \(E_iE_j = 0\), \(i \neq j\).
The printed exercise says “Example 6 of Section 6.3”, but that section has no Example 6. Its final example, Example 5, is the unique \(4\times4\) matrix in the section and is the intended reference.
Solution
The referenced matrix is \[ A= \begin{pmatrix} 0&1&0&1\\ 1&0&1&0\\ 0&1&0&1\\ 1&0&1&0 \end{pmatrix}, \] whose distinct characteristic values are \(0,2,-2\) [@HK71]. Take \[ c_1=0,\qquad c_2=2,\qquad c_3=-2. \] The Lagrange polynomials are \[ p_1(x)=1-\frac{x^2}{4},\qquad p_2(x)=\frac{x(x+2)}8,\qquad p_3(x)=\frac{x(x-2)}8. \] Set \(E_i=p_i(A)\). Since \[ A^2= \begin{pmatrix} 2&0&2&0\\ 0&2&0&2\\ 2&0&2&0\\ 0&2&0&2 \end{pmatrix}, \] we obtain \[ E_1=I-\frac{A^2}{4} =\frac12 \begin{pmatrix} 1&0&-1&0\\ 0&1&0&-1\\ -1&0&1&0\\ 0&-1&0&1 \end{pmatrix}, \] \[ E_2=\frac{A^2+2A}{8} =\frac14 \begin{pmatrix} 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1 \end{pmatrix}, \] and \[ E_3=\frac{A^2-2A}{8} =\frac14 \begin{pmatrix} 1&-1&1&-1\\ -1&1&-1&1\\ 1&-1&1&-1\\ -1&1&-1&1 \end{pmatrix}. \]
1. The projections resolve the identity and reconstruct \(A\).
The polynomial identities \[ p_1+p_2+p_3=1, \qquad 0p_1+2p_2-2p_3=x \] hold identically. Evaluating at \(A\) gives \[ E_1+E_2+E_3=I \] and \[ A=2E_2-2E_3=c_1E_1+c_2E_2+c_3E_3. \]
2. Distinct spectral projections annihilate one another.
Because \(A\) is diagonalizable with distinct characteristic values \(0,2,-2\), its minimal polynomial divides \[ x(x-2)(x+2). \] For \(i\ne j\), the polynomial \(p_ip_j\) vanishes at each of these three values, so it is divisible by \(x(x-2)(x+2)\). Hence \[ E_iE_j=(p_ip_j)(A)=0 \] [@HK71]. Thus the displayed matrices satisfy all the requested identities.