# hmw4 - Solutions to Homework 4 Math 110 Fall 2006 Prob...

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Solutions to Homework 4. Math 110, Fall 2006. Prob 2.3.13. Let A = [ a ij ]. Then A t = [ a ji ], and tr( A ) = n i =1 a ii = n j =1 a jj = tr( A t ) . The elements of A = [ a ij ], B = [ b ij ], AB = [ c ij ] and BA = [ d ij ] are connected by the formulas c ij = n k =1 a ik b kj , d ij = n k =1 b ik a kj . So, tr( AB ) = n i =1 c ii = n i =1 n k =1 a ik b ki = n i =1 n k =1 b ik a ki = n i =1 d ii = tr( BA ) . The second-last equality is obtained by interchanging indices i and k . Prob 2.3.15. Suppose A has n columns { A i : i = 1 , . . . , n } , and write A = [ A 1 , A 2 , . . . , A n ]. Then MA = [ MA 1 , . . . , MA n ], i.e., the columns of MA are column-vectors MA 1 , . . . , MA n . So if A j = i = j a i A i , then MA j = M ( i = j a i A i ) = i = j a i MA i , i.e., the j th column of MA is a linear combination of its other columns with the same coefficients. Prob 2.3.17. For any vector x , we have x = Tx + ( x - Tx ). Note that Tx is in the range R ( T ) of T and x - Tx is in the kernel N ( T ), since T ( x - Tx ) = Tx - T 2 x = Tx - Tx = 0 . This shows that V = R ( T ) + N ( T ). By the Rank-Nullity theorem, we can therefore conclude that this sum is direct (see Prob. 2.1.35a). Also note that T acts as the identity on R ( T ).

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