Question

how did we get the left null space please use simple way

6% 0-0, 1:44 AM Fri May 17 , Calc 4 4 Exaimi 3 solutions Math 250B Spring 2019 1. Let A 2 6 5 (a) Find bases for and the dimensions of the four fundamental subspaces. Solution Subtract row onc from row 2, then 8 times row 2 from row 3, then 5 timcs rovw 2 fro row. Finally, divide row1 by 2 to get the row reduced echelon for of A: 1 1/2 0-1/2 A basis for C(Ат) is the first two rows of R: a basis for C(A) is columns 1 and 3 of A. The special solutions are (-1/2,1,0,0), 12,0,11) And, since -row 1 row 2 - (1/8) row 3 gives a row of zeros, 1, 1/8) is a basis for N(A. The following table gives bascs ad disios of the four suspaes basis dinension C(A)26 C(A) 1/2 0 -1/2],0 01)2 N(4) (b) Find a condition on b so that Ax b has a solution Solution. b must be in the column space, that is the plane spanned by (2,2, 0), 5, 6, 8) Alternatively, when one docs EROs to rod b to cchclon for, the las row is o 0008(b2-i]. Thus, a condition for b to be in the column space is ( Find the coplete solution when b- 6 Solution. In this case b is the third column of A, so a particular solution is Xp-(0,0,1,0) Thus, the complete solution is 4341 2,4 free X- 1:44 AM Fri May 17 Calc 4 4 1. (a) Let S be the subspace of R3 that consists of all solutions to the equation 2x -3y0. Detcrminc a basis for S, and hence, find dim S] Solution. One way to find a basis is to take the special solutions obtained by setting y- 1,0 and y 0,z 1. We get s (3/2,1,0), s2 (-1/2,0, 1). If we scale these by 2 we will get the basis A basis contains two vectors, so dimIS]-2. (b) Find a 3 × 3 matrix A whose oolumn space is given by the plane 2 -3y + z = 0. Describe the left ulls N(A tsatrix Solutio. We can ptt vetors above as tw ofthe coluns and make the remainin,g colun any linear cobion of these. One exaple is A=12 0 0 l'he left nullspace is the orthogonal complement of the column space, which in this case is the pine. The orthogonal complement of a plane is the normal line 80 N(A1 İs the normal line to the plane passing through the origin. In particular, 2 How NiA')-span BONUS Suppose A and B are 3 x 3 matrices and AB ranks of A and B cannot both be equal to two. 0 is the zero matrix. Explain why the Solution. Suppose A has rank 2, Then the dimension of the nullspace must be one. If AB = 0, then all three columns of B are in the aof A, and hence muste scalar tiples of Some vector. Thus, the comn spae of B is dine or zer. So the rank of canmot be two. Likewise, if B has rank 2, B has 2 LI colums.If AB 0, then both of these columns are in NiA). Thus, N(A) must have dimension at least 2, so the rank of A can be at most one

Answer 1

5 6 1 5 3 2 0 26 2 2 o 0

2 0 2 6 3 2 o 0 2 6 2.