ECE 174 Fall 2017 Supplemental Solutions to …

1 ECE 174 fall 2017 supplemental solutions to homework is obvious that the rank of the matrix is 2 (as the two rows and thefirst two columns are linearly independent). The two linearly independentrows span the row space ( , they form a basis for the row space), whichis obviously 2 dimensional. The first two, linearly independent columns,span the column space ( , they form a basis for the column space), whichis obviously 2 dimensional. The nullspace is obviously 1 dimensional andspanned by the canonical coordinate vectoreT1(0 0 1)Tso that a basis forthe nullspace is given by{e1}.Meyer (a) To show x 1 x 2, we have, x 21=(n i=1|xi|)2=n i=1|xi|2+ i6=j|xi||xj|= x 22+CwhereC 0. So x 21 x 22, x 1 x 2. To show x 2 x , we have, x 22=n i=1|xi|2 maxj|xj|2= x 2 so x 2 x.

2 (b) To show x 1 n x 2, we have, x 1=eT|x| e 2 x 2= n x 2whereeis the vector of all 1 s, and|x|is the vector whoseithcomponent is|xi|. To show x 2 n x , we have, x 22=n i=1|xi|2 n i=1maxj|xj|2=nmaxj|x|2j=n x 2 so x 2 n x .1 To show x 1 n x , we have, x 1=n i=1|xi| n i=1maxj|xj|=nmaxj|xj|=n x . that we (and Meyer) define the inner product to be linear in thesecond recall that x1,x2 = x2,x1 .(a) x1, x2 = x1,x2 = x1,x2 = x2,x1 = x2, x1 = x1,x2 .(b) 1x1+ 2x2,x = x, 1x1+ 2x2 = 1 x,x1 + 2 x,x2 = 1 x1,x + 2 x2,x Note that in part (b) we obtain the property ofadditivity in the firstargumentas a special case (just take 1= 2= 1). that the operatorsA:X Y,B:X Y, andC:Y Zarelinear and that the adjoint operator,A , is defined by A x1,x2 = x1,Ax2.

3 You also need to recall the properties of the inner product (such as linearityin the second argument) and the additional properties proved in Problem1 above.(a) y, Ax = y,A x = A y, x = A y,x .(b) y,(A+B)x = y,Ax+Bx = y,Ax + y,Bx = A y,x + B y,x A y+B y,x = (A +B )y,x .(c) The fact that ( A+ B) = A + B follows immediate fromproperties (b) and (a), in that order.(d) x,A y = A y,x = y,Ax = Ax,y .(e) z,CAx = C z,Ax = A C z,x .(f) A y,x = y,Ax = y,Ax = A y,x = A y,x .(g) A (y1+y2),x = y1+y2,Ax = y1,Ax + y2,Ax = A y1,x + A y2,x = A y1+A y2,x .(h) Linearity ofA follows from properties (g) and (f), in that that the inner products are defined as, x1,x2 =xH1 x2and y1,y2 =yH1Wy2,1 Whereas many other authors define the inner product to be linear in the first course, in the real vector space case the inner product is linear in both arguments so thatthe distinction andWare hermitian ( , H= andWH=W) and positive definite (and hence both are invertible).

4 Note that in this case their in-verses are also hermitian, positive definite.(a) y,Ax =yHWAx=yHWA 1 x=(( 1 AHW)y)H x= ( 1 AHW)y,x .(b) For r(A) =n,Ahasfull column rankand it must be the case thatm n. SinceAis possibly over determined, we solve the leastsquares problem by enforcing the geometric condition thaty Ax R(A) =N(A ). This yields the normal equations,A Ax=A y .Because r(A) =n, then nmatrixA Aalso has ranknand istherefore invertible. (This fact is consistent with the nullspace ofAbeing trivial, so that the least squares problem must have a uniquesolution.) Thus, we have thatr(A) =n x= (A A) 1A y ,for any value ofy. It must therefore be the case thatr(A) =n A+= (A A) 1A ,whereA = 1 AHWas determined in Part (a).

5 WithWa fullrank hermitian matrix and r(A) =n, it is the case thatAHWA isinvertible and as a consequence the pseudoinverse,A+, isindependentof the weighting matrix ,r(A) =n A+= (AHWA) 1 AHW .(c) For r(A) =m,Ahas full row rank and therefore it must be the casethatn m. Note thatAisonto, and thereforey=Axis solvablefor ally. However, the system is possibly underdetermined, so wewant to look for a minimum norm solution. This requires that weenforce the constraint that any solution toy=Axmust also satisfythe geometric condition thatx N(A) =R(A ). This conditionis equivalent to,x=A ,for some vector .This condition, together with the requirement thatxbe a solution toy=Ax, yields,AA =y .Because r(A) =m, them mmatrixAA also has ranknand is , = (AA ) 1y3which yields the result thatr(A) =m x=A (AA ) 1y ,for ally.

6 Thus,r(A) =m A+=A (AA ) r(A) =mand 1a hermitian full rank matrix, it is the casethat them mmatrixA 1 AHis invertible. With the fact thatA = 1 AHW, this yields the fact that for r(A) =m, the pseudoinverse isindependentof the weighting matrixW,r(A) =m A+= 1AH(A 1AH)