Proposition 42. Let M be an m n matrix. Then (32) N (M M)=N (M) (33) R(MM )=R(M)

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1 RODICA D. COSTIN. Singular Value Decomposition.1. Rectangular matrices. For rectangular matrices M the notions of eigenvalue/vector cannot be defined. However, the products MM and/or M M (which are square, even self-adoint, and even positive semi-definite matrices) carry a lot of information about M: Proposition. Let M be an m n matrix. Then () N (M M)=N (M) () R(MM )=R(M) Proof. To show (), let x N(M M); then M Mx =, so that = hm Mx, xi = hmx,mxi which implies Mx =, showing that N (M M) N (M). The converse inclusion is immediate. To show (), note that (), used for M interchanged with M implies that N (MM )=N (M ), hence N (MM )? = N (M )?, which is exactly () (recall that for any linear transformation L we have N (L )? = R(L)). Moreover, MM and M M have the same nonzero eigenvalues: Proposition. Let M be an m n matrix. The matrices MM and M M are positive semi-definite. Moreover, they have the same nonzero eigenvalues (with the same multiplicity). More precisely, let 1,..., r be the positive eigenvalues. If M Mv = v with v 1,...,v r an orthonormal set, then MM u = u for u = p 1 Mv and u 1,...,u r is an orthonormal set. Proof. MM and M M obviously self-adoint; they are positive semidefinite since hx,m Mxi = hmx,mxi and hx,mm xi = hm x,m xi. Let v 1,...,v n be an orthonormal set of eigenvectors of M M,thefirst r corresponding to nonzero eigenvalues: M Mv = v with >, for =1,...,r and M Mv = for >r. Applying M we discover that MM Mv = Mv with >, for =1,...,r and MM Mv = for >rwhichwould mean that Mv are eigenvectors to MM corresponding to the eigenvalue provided we ensure that Mv =. This is true for apple r by (). Also, all Mv 1,...,Mv r are mutually orthogonal, since hmv,mv i i = hv,m Mv i i = i i so Mv? Mv i for all i = apple r, and kmv k =. Therefore, all the nonzero eigenvalues of M M are also eigenvalues for MM, with corresponding orthonormal eigenvectors u = p 1 Mv, =1,...,r.

2 SPECTRAL PROPERTIES OF SELF-ADJOINT MATRICES The same argument can be applied replacing M by M, showing that indeed, MM and M M have the same nonzero eigenvalues and with the same multiplicity... The SVD theorem. We are going to bring any m n matrix M to a (rectangular) diagonal form by writing M = U V where is a diagonal m n matrix, and U and V are unitary (of obvious dimensions). The diagonal elements of are called the singular values of M. The SVD has a myriad applications in filtering, image reconstruction, image compression, statistics, to name ust a few. Theorem. Singular Value Decomposition Let M be an m n matrix. Then M = U V where: U is a unitary matrix whose columns are eigenvectors of MM V is a unitary matrix whose columns are eigenvectors of M M is an m n diagonal matrix More precisely: if U =[u 1,...,u r, u r+1,...,u m ] and V =[v 1,...,v r, v r+1,...,v n ] then for =1,...r the vectors u and v correspond to the eigenvalue = while all the others correspond to the eigenvalue. The diagonal matrix has = = p for =1,...,r,andall other elements are. Also, u = 1 Mv for =1,...,r. Remarks. 1. M M = V n V and MM = U m U where m,n are diagonal matrices with entries 1,..., r and everywhere else.. The singular values are preferred be listed in decreasing order 1... r for reasons coming from applications, see.. Proof of Theorem. Let v 1,...,v r and u 1,...,u r be as in Proposition ; u r+1,...,u m and v r+1,...,v n correspond to the eigenvalue. Calculating U MV = u 1. u m M [v 1,...,v n ]= u 1. u m [Mv 1,...,Mv n ]= where is a matrix with elements i = u i Mv. For > r we have M Mv =, hence by () also Mv =, hence i =, while for apple r we have u i Mv = u i (p )u = p i, showing that is the diagonal matrix stated.

3 RODICA D. COSTIN.. Examples and applications of SVD. Example 1. How does the SVD look like for a square, diagonal matrix? Say " # a1 () M = a In this case MM = " a1 a # = M M therefore = a, V = I, and u = 1 Me = a a e. By the polar decomposition of complex numbers, write a = a e i u = e i e and the SVD is " # " #" # a1 e i 1 a1 = a e i a then which is called the polar decomposition of the matrix (). In general: Proposition. Polar decomposition of square matrices. Every square matrix M can be decomposed as M = US with U unitary and S positive semidefinite. Proof. Writing the SDV of the matrix M = U V =(UV )(V V )whichis the polar decomposition since UV is a unitary matrix and V V is a selfadoint matrix with non-negative eigenvalues. Example. A rectangular diagonal matrix, say a 1 e i 1 a 1 a = e i a 1 " 1 1 # Example. A column matrix apple " 1 = 1 p p p p 1 # apple p [1] Example. An orthogonal matrix Q is its own SVD since QQ = Q Q = I hence V = I, = 1 and U = Q.

4 SPECTRAL PROPERTIES OF SELF-ADJOINT MATRICES.. The matrix of an oblique proection. Recall that a square matrix P is a proection if P = P ;thenp proects onto R = R(P ), parallel to N = N (P ). For given complementary subspaces R and N a simple formula for the matrix of P can be obtained from the singular value decomposition. Since the eigenvalues of P can only be or 1, then all = 1. The vectors v 1,...,v r in Theorem form an orthonormal basis for R(P P )= N (P P )? = N?, and u 1 = P v 1,...,u r = P v r form an orthonormal basis for R(PP )=R(P)=R. Split the matrices U, V into blocks, the first one containing the first r columns: U =[U A U ], V =[V B V ], and since has its upper-left r r diagonal sub matrix equal to the identity, the SDV of P becomes apple apple P = U V =[U A U I V ] B = U A VB Note that VB U A = I. Indeed, the elements of this matrix are (VB U A) i, = hv i, u i = 1 hv i, v i = 1 i and for proections = 1. Let y 1,...,y r be any basis of R; thenb =[y 1,...,y r ]=V B S for some invertible r r matrix S. Similarly, if x 1,...,x r is any basis of N?,thenA = [x 1,...,x r ]=U A T for some invertible r r matrix T.ThenA(B A) 1 B is the matrix of P since A(B A) 1 B = U A T (S V BU A T ) 1 S V B = U A (V BU A ) 1 V B = U A IV B = P.. Low-rank approximations, image compression. Suppose an m n matrix M is to be approximated by a matrix X of same dimensions, but lower rank k. If M = U V with singular values 1... k... r, letx = U k V where k has the same singular values 1,..., k and everywhere else. Then the sum of the squares of the singular values of M X is minimum among all matrices m n of rank k (in the sense that the Frobenius norm of M X is minimum). This low rank approximations are used in image compression, noise filtering and many other applications.. Pseudoinverse There are many ways to define a matrix which behaves, in some sense, like the inverse of a matrix which is not invertible. This section describes the Moore-Penrose pseudoinverse. Finding the best fit solution (in the least square sense) to a possibly overdetermined linear system Mx = b yields a vector x + which depends linearly on b, hence there is a matrix M + so that x = M + b;thisisthe Moore-Penrose pseudoinverse of M. Recall the construction of this solution. Step I. If Mx = b is overdetermined (i.e. has no solutions) this is because b R(M). Then find x so that kmx bk is minimum. This happens if Mx = P b where P b is the orthogonal proection of b on R(M).

5 8 RODICA D. COSTIN Step II. Now Mx = P b is solvable. The solution is not unique if N (M) is not {}, in which case, if x p is a solution, then all vectors in x p + N (M) are solutions. Choosing among them the solution of minimal length: find w N(M) with kx p + wk minimum. Since kx p + wk is the distance between x p and w N(M) itisminimumwhenw is the orthogonal proection of x p on N (M). Define x + = x p + w. ThenM + is defined by M + x = x + for all x. Example. Solve x = b for = 1 Clearly R( ) = {y R y =} hence P b = P (b 1,b,b ) T =(b 1,b, ) T. Then x = P b has the solutions x with x = b / for =1, and x, arbitrary, which has minimal norm for x, =. We obtained x + = b 1 / 1 b / = 1/ 1 1/ b 1 b + b For a general m n diagonal matrix with singular values = similar arguments show that its pseudoinverse + is an n m diagonal matrix with singular values + =1/. For a general m n matrix M with singular value decomposition M = U V, solving Mx = b is equivalent to solving y = U b where y = V x. This that the optimal solution y + = + U b, therefore (since U preserves distances) x + = V + U b.weproved Theorem. The pseudoinverse of a matrix M with singular value decomposition M = U V is M + = V + U. The pseudoinverse has many properties similar to those of an inverse. The following statements are left as exercises. 1. If M is invariable, then M 1 = M +.. MM + M = M and M + MM + = M + (though MM + and M + M are not necessarily the identity).. MM + and M + M are orthogonal proectors.. The operator + commutes with complex conugation and transposition.. ( M) + = 1 M +. If is a scalar (think M =[ ]) then + equals if = and 1/ if =.. The pseudoinverse of a vector x is x + = x kxk if x = and T if x =.

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