SVD - ECE 3250 SINGULAR-VALUE DECOMPOSITION Fall 2008 In...

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ECE 3250 SINGULAR-VALUE DECOMPOSITION Fall 2008 In what follows, I’ll be relying on a few essential facts from linear algebra. Some of them are easier to prove than others. As always, F denotes R or C . By F m × n I will mean the set of all ( m × n ) matrices with entries in F . If A F m × n , A H is the conjugate transpose, or Hermitian conjugate, of A . Note that A H F n × m . A necessarily square matrix Q is a Hermitian matrix if and only if Q H = Q . If A F m × n , the nullspace of A is the set of all x F n such that Ax = 0. Note that the nullspace of A is a subspace of F n . The range of A is the set of all y F m such that y = Ax for some x F n . The range of A is the range of the linear mapping x Ax . The range of A is a subspace of F m . Now for the facts we’ll need. Fact 1: If A F m × n , then rank of A = n - (dimension of the nullspace of A ) . One way to prove this fact is by analyzing the equation Ax = y by means of Gauss elimi- nation. Fact 2: If A F m × n , then the rank of A H A is the same as the rank of A . To see this, first observe that A and A H A have the same nullspace because A H Ax = 0 = x H A H Ax = 0 ⇐⇒ Ax 2 = 0 ⇐⇒ Ax = 0 , so the nullspace of A H A is contained in the nullspace of A . The reverse inclusion is obvi- ous. Now apply Fact 1 to conclude that the rank of A and the rank of A H A are equal. Fact 3: If Q F n × n is Hermitian, then all the eigenvalues of Q are real. To see this, let x F n be an eigenvector of Q corresponding to eigenvalue λ o . Because Q is Hermitian, x H Qx = ( x H Qx ) H = x H Q H x = x H Qx . Meanwhile, x H Qx = λ o x 2 and x H Qx = λ o x 2 , from which it follows that λ o = λ o because x = 0. Fact 4: If A F m × n , then all the eigenvalues of A H A are real and nonnegative. To see why, note first that A H A is Hermitian, so all its eigenvalues are real by Fact 3. If λ o is an eigenvalue of A H A and x a corresponding eigenvector, then 0 Ax 2 = x H A H Ax = λ o x 2 , which implies that λ o 0 because x = 0. Recall that C n is an inner-product space with inner product v, w = w H v for all v, w C n . In what follows, when I use the word “orthonormal” I’ll be referring to orthonormality with respect to that inner product. Furthermore, I’ll state all remaining results in terms of complex matrices and vectors. 1
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2 Fact 5: If Q C n × n is Hermitian, then there exists an orthonormal basis { u 1 , u 2 , . . . , u n } for C n consisting solely of eigenvectors for Q . Proving this fact is not straightforward, but you’ve all seen it before. Now for the singular-value decomposition. Let A C m × n have rank r . By Fact 2, the Hermitian matrix A H A also has rank r , so its nullspace has dimension n - r by Fact 1. A nonzero vector x is in the nullspace of A H A if and only if it is an eigenvector of A H A corresponding to eigenvalue 0. Accordingly, if we invoke Fact 5 and pick an orthonormal basis { u 1 , u 2 , . . . , u n } for C n consisting solely of eigenvectors for A H A , we can choose the u j so that { u r +1 , u r +2 , . . . , u n } is an orthonormal basis for the nullspace of A H A . If we
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