Decomposition Theorems
Decomposition Theorems
Spectral Decomposition
Any normal operator MM on a vector space VV is diagonal with respect to some orthonormal basis for VV.
Conversely, any diagonalizable operator is normal.
Now we have
QM=QM(P+Q)=QMQ and QM=QMQQMQQMQ=QMMQ=QMMQ=QMQQMQQM = QM(P+Q) = QMQ \text{ and } QM^\dagger = QM^\dagger Q\\ QMQQM^\dagger Q = QMM^\dagger Q = QM^\dagger MQ = QM^\dagger QQMQ
Thus, if MM is normal then M=PMP+QMQM = PMP + QMQ where PMP=λP2=λPPMP = \lambda P^2 = \lambda P and thus is diagonalizable wrt orthonormal basis for PP. Similarly, QMQQMQ is also diagonalizable wrt some orthonormal basis for QQ.
Thus, MM is diagonalizable for orthonormal basis of the entire vector space.
M=Σλiii=ΣλiPiM = \Sigma \lambda_i \vert i\rangle \langle i \vert = \Sigma \lambda_i P_i
Polar Value Decomposition
Let AA be a matrix on vector space VV. Then there exists unitary UU and positive operators JJ and KK such that, A=UJ=KUA= UJ=KU where the unique positive operators shall satisfy the equations JAAJ \equiv \sqrt{A^\dagger A} and KAAK \equiv \sqrt{AA^\dagger}.
Moreover, if AA is invertible then UU is unique.
Singular Value Decomposition
Let AA be a square matrix. Then there exist unitary matrices UU and VV, and a diagonal matrix DD with non-negative entries such that A=UDVA = UDV.
The diagonal elements of DD are called the singular values of AA.
From polar value decomposition we have, A=SJ=STDT=(ST)D(T)=UDVA = SJ = STDT^\dagger = (ST)D(T^\dagger) = UDV.