please help with these 6 practice problems

1 1 .LetA=[2 a space C(A), the null space N (A), the row space, and the left null space of A. . Find a such that rank A =1; and for this value of oz, nd bases for the column 0 2 0 1 1 0 1 3/2 . _ 1 1 1 1 _ 0 1 0 1/2 . Cons1der A 1 _1 _1 _2 and rref (A) 0 0 0 0 0 2 0 1 0 0 0 0 (a) Find rank (A). (b) Find a basis of C(A) and the dimension of C(A). (c) Find a basis of N(A) and the dimension of N(A). 1 1 1 1 1 1 2 1 1 2 .LetA 1 1 _2 0 andb= _1 1 2 2 2 2 (a) Find the complete solution of A9: = b. (b) Find a basis of N(A). 2 . Assume that the system of equations Aw = [ 0 ] has the the complete solution 3 = [ g ] +0: [ _i ] , 2 where a is any scalar. Find A. . Let (11,22, a3,a4 be 4 given vectors in R3. (a) Is {a1,a2,a3, a4} a basis of R3? Justify your answer. (b) Let A = [21 a2 a3 a4] and assume that rank (A) = 3. Is {a1,22,a3,a4} a Spanning set of R3? Justify your answer. .LetSbethespanof{A1 = [(1) g],A2 =[ (1)],A3=[g 1]}. FindabasisofS. In this Lecture: . The null space. . The row space. . The left null space. . The rank. . The complete solution of Ax = b. 9.1 The Null space Definition 9.1 Let A be an m x n matrix. The nullspace of A, denoted by N(A), is the set of all x in In such that Ax = 0. In other words, N(A) is the set of all solutions of the homogeneous system of equations Ax = 0. N(A) is a subspace of R". The following is a proof of this fact. 1. Let x and x2 be any two vectors in N(A), then Axl = 0 and Ax2 = 0. Now, A(x] +x2) = Axl + Ax? = 0 + 0 =0. Hence, x + x2 is also in N(A) 2. Let a be any scalar. Then A(axl ) = aAxl = 00 = 0. Hence, ox is also in N(A). Therefore, N(A) is indeed a subspace of ". Note that A0 = 0, thus the zero vector in R" is in N(A) as should be the case. Note 1 If x E N(A), i.e., if Ax = 0, then obviously x is perpendicular to every row of A. Example 1 Determine N(A), where A = 1 3 : ]. We use Gauss-Jordan to find all solutions of Ax = 0. The augmented matrix is [A|0]= 1 3 6 8]-[08 48] - 68 48 =ref[ 410 ]. LaTex template courtesy of UC Berkeley EECS dept. Disclaimer: These notes have not been subjected to the usual scrutiny reserved for formal publications. They may be distributed outside this class only with the permission of the Instructor. 9-19-2 Lecture 9: The Complete solution of Ax = b Hence, x1 and x3 are basic variables, while 2 is nonbasic. Thus the general solution of Ax = 0 is given by = -2x2, x3 = 0 and x2 is free. 2 2 We go a step further. x = 20 2 is in N(A) if it is of the form x = X 2 : X 2 , where x2 is C3 0 free. The vector ONN s called a special solution of Ax = 0. Now let x2 = a where a is a free variable. Then x is in N(A) if x = a ON for some a. In other words, N(A) is the span of Example 2 Determine the nullspace of A where A = The augmented matrix [ A | 0 ] = -1 1 1 0 1 - 1 Note that the last column in [ A | 0 ] does not affect the calcu- lations, so it can be ignored. rref( A) = 1 0 -1 1 0 1 0 0 0 0 0 0 Thus x1 and x2 are basic variables and x3 and x4 are nonbasic. The general solution of Ax = 0 is given by X1 = X3 - 24, X2 = 0, x3 and x4 are free. We also go a step X3 - X4 -1 further. x = X 2 is in N(A) if it is of the form x = 0 = X3 + 24 CA O , where 3 and x4 are free. The vectors then any x of the form OHOF and HOOK are special solutions of Ax = 0. Also, if x3 = a and x4 = B OHOH + B is in N(A). In other words, N(A) = span { 8 . : , As we have seen so far, the null space (A) and the column space C(A) are two subspaces associated with matrix A. In addition to these two subspace, there are another two subspaces associated with A which we define next.Lecture 9: The Complete solution of Ax = b 9-3 Definition 9.2 The row space of A is the span of the rows of A. In other words, the row space of A is C(AT). Definition 9.3 The left null space of A is the set of all y such that y" A = 0. In other words, the left null space of A is N(A" ). Note: Each matrix A has 4 subspaces associated with it. However, since the row space and the left null space of A can be expressed in terms of C(A" ) and N(A"), the column space and the null space are the two more important subspaces of A. The relations between subspaces of A and the subspaces of rref(A) are summarized in the following: N(A) = N(rref (A)) (null space). C(A) # C(rref (A)) (column space). left null space of A * left null space of rref (A). row space of A = row space of rref (A). For example, let A = 1 1 and thus rref (A) = . . Hence, N(A) = N(AT) = span ( _] ) and C(A) = C(AT) = span ( 1 ). On the other hand, left null space of rref(A) = span ( " ], C(rref( A)) = span row space of rref( A) = span ( ] ], 9.2 The Rank and the Complete Solution of Ax = b Recall that rank A is equal to the number of pivots in rref(A). 1/3 For example, if A = -1 , then rref( A) = 1 -2/3 1 1 0 - 1 0 0 and rref ( AT ) = 0 0 Thus rank A = 2 and rank AT = 2. Theorem 9.4 rank A = rank AT. Note 2 If rank A is equal to the number of columns in A, i.e., if every column of A has a pivot, we say that A has full column rank. If rank A is equal to the number of rows, i.e., if every row of A has a pivot, we say that A has full row rank.9-4 Lecture 9: The Complete solution of Ax = b Note 3 rank A = 0 if and only if A is the zero matrix. Note 4 If A is m x n and if rank A = r, then r 02 11 12 01 02 10 01 1 01 1 4! 0| it a 0| t i 0 2 1 0 0 0 01 Thus A9: = b has no solution since 0:131 + 0.152 + 0:133 = U % 1 for all :81, 2:2 and 3:3. 1 0 Note in this case, C(A)span{ |: 0 :l , |: 2 :| }. l 2 In this case, we know that b is not in C(A). But let's prove that directly. Assume for the moment that lilml'i'lrlitltil- Hence, a = 1, ,3 = % and hence a + 2,8 = 2 75 1. Therefore, this system has no solution for a: and #3. Thus b cannot be written as a linear combination of the columns of A. i.e., b g! C(A). 2 Finally, N (A) = span{ [ 1 :| 2 1 Example 5 In the previous example, let b = |: l :| . Determine a: such that b is in C(A). i.e., Ax = b has a a solution. 9-6 Lecture 9: The Complete solution of Ax = b [A |b ] = 0 2 -1 1 -+ 1 2 o a 0 0 1 -1:474 NN -1 1 a-1 1 0 0 1 0 a - 2 Thus b is in C(A) or Ax = b has a solution iff a = 2. Note that if a = 2, then Ax = b has an infinite number of solutions since N(A) # {0}. In fact in this case, the general solution of Ax = b is given by x1 = 1 + x3, X2 = 2 - 2x3 and x3 is free. Thus, x = ON + 13 ANIN s a particular solution of Ax = b since Also, Ts = I is a special solution as it's in N(A). Summary Let A be m x n and of rank r: 1. If r = n = m, then Ax = 6 has a unique solution, namely z = A-1b. 2. If r = n