alled the diagonal representation, or spectral decomposition of A. For instance, the diagonal-operator representation of the Pauli matrix Z is: (17.28) It is left as an easy exercise to determine the eigenstates and diagonal representations of the Pauli matrix Y and the Hadamard matrix H. More generally, in the diagonal representation of A, the matrix elements can be developed as follows: (17.29) which shows that all nondiagonal elements of the matrix are identically zero, while the diagonal elements are equal to the eigenvalues Ai, i.e., (21 0 ... (17.30) Quantum measurements An operator matrix that can be transformed into the diagonal form shown in Eq. (17.30) is said to de diagonalizable. For this, its n X n matrix must have n eigenstates. Clearly, if the matrix has fewer than n eigenstates, it cannot be diagonalized, as there remain some nonzero off-diagonal elements in the matrix. An operator is said to be normal if it satisfies AA+ = A+ A. A remarkable property: any normal operator is diagonalizable, and any diagonalizable operator is normal. Consequently, since they are normal, Hermitian operators are diagonalizable. Matrix trace The trace of a matrix A, noted tr(A) is the sum of all its diagonal elements Aii, according (17.31) It is easily established that for any two operators A, B and complex number a, we have the properties: tr(a A) = a tr(A) tr(A+B)=tr(A)+tr(B) (17.32) tr(AB) = tr(BA). In the diagonal representation (Eq. (17.30)), it is clear that the trace of a matrix is the sum of its eigenvalues. But the trace of a matrix does not depend on the choice of the This can be shown through the following. Since A is a linear operator, it must have at least one eigenvalue, say 2. Define P =121(21 as the projector on the corresponding eigenspace 12) (where AP = AP), and Q=I-P - the projector on the orthogonal subspace (hence P+Q=I and PQ = QP = 0). We also have p2 = P,Q2=Q, and P, Q are both Hermitian (P+ = P, hence Q+ = Q). We develop A according to the following: A=(P+Q)A=(P+Q)A(P+Q)= PAP+QAP+PAQ+QAQ PXP+