37_hwc1SolnsODDA

37_hwc1SolnsODDA - SOLUTION (a) It is easiest to start with...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
column of AB is the linear combination of the columns of A with coefficients coming from B 1 , 1 B 2 , 1 B 3 , 1 , the first column of B . Therefore, v 2 + v 3 = B 1 , 1 v 1 + B 2 , 1 v 2 + B 3 , 1 v 3 , so we can always take B 1 , 1 = 0, B 2 , 1 = 1 and B 3 , 1 = 1. In the same way we find B 1 , 2 = 0, B 2 , 2 = 1 and B 3 , 2 = 0, and also B 1 , 3 = 1, B 2 , 3 = 1 and B 3 , 3 = 0. Thus, B = 0 1 1 1 1 1 1 0 0 . 5.7 (a) Let A be the n × n diagonal matrix A = diag( a 1 , a 2 , . . . , a j ) in which, for some i with 1 i n , a i = 1, but a j = 0 for j 6 = i . In other words, A i,i = 1, and every other entry is zero. Let B be any n × n matrix. Describe the rows of AB and the columns of BA . (b) Is there any non-diagonal n × n matrix B that commutes with A (that is AB = BA ) for all n × n diagonal matrices A ? (We already know that each diagonal n × n matrix B commutes with every diagonal matrix. The question is: are there any others?) Give an example, or explain why not.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: SOLUTION (a) It is easiest to start with the column of BA . By V.I.F. 3 , which is the denition of matrixmatrix multiplication, (column j of BA ) = B (column j of A ) . Since (column j of A ) = e i if j = i if j 6 = i , (column j of BA ) = B e i if j = i if j 6 = i , Since B e j is the j th column of B , we have (column j of BA ) = (column j of B ) and all other columns of BA are zero . This tells us what the columns of BA are. To see what the rows of AB are, note that ( AB ) t = B t A t and A t = A . Thus, ( AB ) t = B t A . 21 /september/ 2005; 22:09 38...
View Full Document

Ask a homework question - tutors are online