1. State the definitition of a vector space. 13 1 5 13 8 4 -4 2 -2 12 14 1 0 0 -
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1. State the definitition of a vector space. 13 1 5 13 8 4 -4 2 -2 12 14 1 0 0 -1 0 11 0 1 0 1 0-1 2. LetAs| 5 4 3 8-1-91, and B=10 0 1 3 0-3 6 5 4 11 9 -2 2 6 5 19 8 -11 Given that A and B are row equivalent, answer the following questions. For parts (d)-(e), give a brief explanation for your answer. (a.) Find the dimension and a basis for the null space of A, N(A) b.) Find the dimension and a basis for the row space of A, R(A). (c.) Find the dimension and a basis for the column space of A, CS(A) 3. Let B' = {(1,0,-1), (0,1,1), (2-1-2) be a basis for R3. (a.) Write down the standard basis, S, for R3 (b) If B'-(2, 1,5), what is z? 4. Without performing row operations (or doing any other work), give the reason that each of the following sets is not a basis for Ps, the set of polynomials of degree s 3. 5. Consider the following set of 2 x 2 matrices in M22 (a.) What is the dimension of M2a? b.) What is the standard basis, S, for M22? c.) Show that B is not a basis for M22, and justify your answer.Explanation / Answer
1.A vector space is a set of vectors, which which is closed under vector addition and scalar multiplication, where the scalars belong to a particular field. The operations of vector addition and scalar multiplication have to satisfy certain axioms as under:
I. For all X, Y V, X+YV( closure under vector addition).
II. For all X, Y , X+Y = Y+X ( commutativity of vector addition).
III. For all X, Y, Z , (X+Y)+Z=X+(Y+Z) (Associativity of vector addition).
IV. For all x, 0+X = X+0 = X ( Existence of Additive identity)
V. For any X, there exists a -X such that X+(-X)= 0 (Existence of additive inverse)
VI. For any scalar k and the vector v V, the vector kv V( closure under scalar multiplication).
VII. For all scalars r and vectors X,Y, r(X+Y)=rX+rY (Distributivity of vector addition).
VIII. For all scalars r,s and vectors X , (r+s)X=rX+sX (Distributivity of scalar addition).
IX. For all scalars r,s and vectors X, r(sX)=(rs)X( Associativity of scalar multiplication).
X. For all vectors X, 1X=X ( Existence of Scalar multiplication identity).
2. (a)The null space of A is the set of solutions to the equation AX = 0. If X = (x,y,z,u,v,w)T,then in view of A and B being row equivalent, this equation is equivalent to x-u+w = 0 or, x = u-w…(1), y+u-w = 0 or, y = -u+w…(2), z+3u-3w=0 or, z=-3u+3w…(3) and v+w=0 or, v= -w…(4).Then X = (u-w,-u+w,-3u+3w, u, -w,w)T = u(1,-1,-3,1,0,0)T+w(-1,1,3,0,-1,1)T. Thus, {(1,-1,-3,1,0,0)T,(-1,1,3,0,-1,1)T } is a basis for the null space of A and the dimension of the null space of A is 2.
(b).The set{(1,0,0,-1,0,1), (0,1,0,1,0,-1),(0,0,1,3,0,-3),(0,0,0,0,1,1)} is a basis for RS(A). The dimension of RS(A) is 4.
(c ). A scrutiny of B reveals that the first 3 and the 5th columns of A are linearly independent and the remaining 2 columns of A are linear combinations of these 4 columns.Thus {(1,4,5,6,2)T,(3,-4,4,5,6)T , (1,2,3,4,5)T,(13,12,-1,9,8)T} and { (1,0,0,0,0)T,(0,1,0,0,0)T,(0,0,1,0,0)T,(0,0,0,1,0)T} are 2 bases for CS(A). Its dimension is 4.
3. (a). The standard basis for R3 is {e1,e2,e3} = {(1,0,0)T,(0,1,0)T,(0,0,1)T}.
(b).If [x]B’=(2,1,5)T, then x = 2(1,0,-1)T+1(0,1,1)T+5(2,-1,-2)T = (7,-4,-11)T.
4.(a). The dimension of P3 is 4 so that the set S containing 5 vectors cannot be a basis for P3.
(b). The dimension of P3 is 4 so that the set S containing 3 vectors cannot be a basis for P3.
5.(a). The dimension of M2,2 is 4.
(b). The standard basis for M2,2 is {E11,E12,E21,E22} where each Eij is a 2x2 matrix with 1 in the ijth position and zeros elsewhere.
(c ). The 4th matrix in the set B is the sum of the first 3 matrices so that B is linearly dependent and hence cannot be a basis for M2,2 .
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