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Notion
Decomposition Theorems
Spectral Decomposition
Any normal operator $M$﻿ on a vector space $V$﻿ is diagonal with respect to some orthonormal basis for $V$﻿.
Conversely, any diagonalizable operator is normal.
Proof
$M = (P+Q)M(P+Q) = PMP+QMP+PMQ+QMQ = PMP + QMQ$
Now we have
$QM = QM(P+Q) = QMQ \text{ and } QM^\dagger = QM^\dagger Q\\ QMQQM^\dagger Q = QMM^\dagger Q = QM^\dagger MQ = QM^\dagger QQMQ$
Thus, if $M$﻿ is normal then $M = PMP + QMQ$﻿ where $PMP = \lambda P^2 = \lambda P$﻿ and thus is diagonalizable wrt orthonormal basis for $P$﻿. Similarly, $QMQ$﻿ is also diagonalizable wrt some orthonormal basis for $Q$﻿.
Thus, $M$﻿ is diagonalizable for orthonormal basis of the entire vector space.
$M = \Sigma \lambda_i \vert i\rangle \langle i \vert = \Sigma \lambda_i P_i$
Polar Value Decomposition
Let $A$﻿ be a matrix on vector space $V$﻿. Then there exists unitary $U$﻿ and positive operators $J$﻿ and $K$﻿ such that, $A= UJ=KU$﻿ where the unique positive operators shall satisfy the equations $J \equiv \sqrt{A^\dagger A}$﻿ and $K \equiv \sqrt{AA^\dagger}$﻿.
Moreover, if $A$﻿ is invertible then $U$﻿ is unique.
Singular Value Decomposition
Let $A$﻿ be a square matrix. Then there exist unitary matrices $U$﻿ and $V$﻿, and a diagonal matrix $D$﻿ with non-negative entries such that $A = UDV$﻿.
The diagonal elements of $D$﻿ are called the singular values of $A$﻿.
Proof
From polar value decomposition we have, $A = SJ = STDT^\dagger = (ST)D(T^\dagger) = UDV$﻿.