The above is our classic closest point problem and is optimized by taking θ n Q

# The above is our classic closest point problem and is

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The above is our classic closest point problem, and is optimized by taking θ n = Q T a n (since the columns of Q are orthonormal). Thus we can write the original problem ( 2 ) as minimize Q : M × r N X n =1 k a n - QQ T a n k 2 2 subject to Q T Q = I , and then take b Θ = b Q T A . Expanding the functional and using the fact that ( I - QQ T ) 2 = ( I - QQ T ), we have N X n =1 k a n - QQ T a n k 2 2 = N X n =1 a T n ( I - QQ T ) a n = N X n =1 k a n k 2 2 - a T n QQ T a n . 81 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 23:01, November 5, 2019 Subscribe to view the full document.

Since the first term does not depend on Q , our optimization program is equivalent to maximize Q : M × r N X n =1 a T n QQ T a n subject to Q T Q = I . Now recall that for any vector v , h v , v i = trace( vv T ). Thus N X n =1 a n QQ T a n = N X n =1 trace( Q T a n a T n Q ) = trace Q T N X n =1 a n a T n ! Q ! = trace Q T ( AA T ) Q . The matrix AA T has eigenvalue decomposition AA T = U Σ 2 U T , where U and Σ come from the SVD of A (we will take U to be M × M , possible adding zeros down the diagonal of Σ 2 ). Now trace Q T ( AA T ) Q = trace Q T U Σ 2 U T Q = trace W T Σ 2 W , where W = U T Q . Notice that W also has orthonormal columns, as W T W = Q T UU T Q = Q T Q = I . Thus our optimization program has become maximize W : M × r trace( W T Σ 2 W ) subject to W T W = I . 82 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 23:01, November 5, 2019 After we solve this, we can take any b Q such that c W = U T b Q . This last optimization program is equivalent to a simple linear pro- gram that is solvable by inspection. Let w 1 , . . . , w r be the columns of W . Then trace( W T Σ 2 W ) = r X p =1 w T p Σ 2 w p = r X p =1 M X m =1 | w p [ m ] | 2 σ 2 m = M X m =1 h [ m ] σ 2 m , where h [ m ] = r X p =1 | w p [ m ] | 2 . Notice that h [ m ] = r X p =1 | W [ p, m ] | 2 is a sum of the squares of a row of W . Since the sum of the squares of every column of W is one, the sum of the squares of every entry in W must be r , and so M X m =1 h [ m ] = r. It is clear that h [ m ] is non-negative, but it also true that h [ m ] 1. Here is why: since the columns of W are orthonormal, they can be considered as part of an orthonormal basis for R M . That is, there is a M × ( M - r ) matrix W 0 such that the M × M matrix W W 0 has both orthonormal columns and orthonormal rows — thus the sum of the squares of each row are equal to one. Thus the sum of the squares of the first r entries cannot be larger than this. 83 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 23:01, November 5, 2019 Subscribe to view the full document.

Thus the maximum value trace( W T Σ 2 W ) can take is given by the linear program maximize h R M M X m =1 h [ m ] σ 2 m subject to M X m =1 h [ m ] = r, 0 h [ m ] 1 . We can intuit the answer to this program. Since all of the σ 2 m and all of the h [ m ] are positive, we want to have as much weight as possible assigned to the largest singular values. Since the weights are constrained to be less than 1, this simply means we “max out” the first r terms; the solution to the program above is b h [ m ] = ( 1 , m = 1 , . . . , r 0 , m = r + 1 , . . . , M. • Fall '08
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