LINEAR ALGEBRA question Suppose T : Rn Rm is a linear transformation and let A b
ID: 3116276 • Letter: L
Question
LINEAR ALGEBRA question
Suppose T : Rn Rm is a linear transformation and let A be its asso- ciated matrix, so that T(v) = Av for all vectors v Rn. Recall that the column space of A, denoted by Col(A) is the subspace of R" spanned by the columns of A. It is the same as the range of T, denoted by Range(T), which is the subspace of all of the vectors of Rm that "get hit" by elements of the domain. The nullspace of A, denoted by Null(A) is the same as the nullspace of T, denoted by Null(T). It is the subspace of R" consisting of all vectors that get mapped to 0. Answer the following questions. You must explain what you are doing and why you are doing itExplanation / Answer
2. (a). I The RREF of the 5x3 matrix M with v1,v2,v3 as columns is
1
0
2
0
1
-3
0
0
0
0
0
0
0
0
0
It means the v3 = 2v1-3v2. Further, if every vector in Null(T) is a linear combination of v1,v2,v3, then v3 itself being a linear combination of v1,v2, every vector in Null(T) is a linear combination of v1 and v2. This implies that { v1,v2 } is a basis for Null(T).Hence the dimension of Null(T) is 2.
(b). As mentioned in part (a) above, { v1,v2 } is a basis for Null(T).
(c ). Range (T) is same as Col(A). However, since A is not known to us, we cannot determine Col(A). However, we can determine Col(AT) as follows. We know that Col (AT) = Null(A). Now, if (x,y,z,u,v)T is an arbitrary vector in Null(A), then(x,y,z,u,v)T.( 1,2,1,2,1)T = 0 or, x+2y+z+2u+v = 0….(1) Similarly, (x,y,z,u,v)T.(-7,-2,-10,10,-13)T = 0 or, -7x-2y-10z+10u-13v = 0 or, 7x+2y+10z-10u+13v = 0…(2). Now, the RREFR of the matrix P =
1
2
1
2
1
7
2
10
-10
13
is
1
0
3/2
-2
2
0
1
-1/4
2
-1/2
Hence the above 2 equations are equivalent to x+3z/2-2u+2v = 0 or, x = -3z/2+2u-2v and y –z/4+2u-v/2 = 0 or, y = z/4-2u +v/2.Then(x,y,z,u,v)T=(-3z/2+2u-2v,z/4-2u+v/2,z,u,v)T= z/4( -6,1,4,0,0)T+u(2,-2,0,1,0)T+ v/2(-4,1,0,0,2)T. Thus {( -6,1,4,0,0)T,(2,-2,0,1,0),(-4,1,0,0,2)T} is a basis for Col(AT).
1
0
2
0
1
-3
0
0
0
0
0
0
0
0
0
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