Let T be a linear transformation of R^4 into R^4 for which: T([0 1 -4 7]) = [29
ID: 3112721 • Letter: L
Question
Let T be a linear transformation of R^4 into R^4 for which: T([0 1 -4 7]) = [29 14 14 -14], T([2 1 3 7]) = [52 39 26 26], T([4 1 0 7]) = [45 34 18 18], T([1 1 1 1]) = [9 13 5 5]. Find the matrix for T. That is, find the matrix A for which T (u) = Au. Find the column space for T. That means, find a basis for the column space for A. Find the null space for T. That means, find a basis for the mull space for A. Find the row space for T. That means, find a basis for the column space for the transpose of A. Find T ([5 1 5 1]). Find all vectors w = [W_1 W_2 W3 W_4] for which T ([W_1 W_2 W_3 W_4]) = [2 3 -1 7] or prove none exists. Find all vectors w = [W_1 W_2 W_3 W_4] for which T ([W_1 W_2 W_3 W_4]) = [2 3 7 7] or prove none exists.Explanation / Answer
1.We know that the standard matrix (A) of T has columns which are images of e1 = (1,0,0,0)T, e2 = (0,1,0,0)T, e3= ( 0,0,1,0)T, and e4 = (0,0,0,1)T. To determine A, we have to express e1,e2,e3,e4 as linear combinations of v1 = (1,0,-4,7)T , v2 =(2,1,3,7)T ,v3 =( 4,1,0,7)T and v4 = (1,1,1,1)T.
Let B be the 4x8 matrix with v1,v2,v3,v4, e1,e2,e3,e4 as columns. In order to express e1,e2,e3,e4 as linear combinations of v1,v2,v3,v4 , we will reduce B to its RREF as under:
Interchange the 1st row and the 2nd row; Add 4 times the 1st row to the 3rd row
Add -7 times the 1st row to the 4th row; Multiply the 2nd row by 1/2
Add -7 times the 2nd row to the 3rd row; Multiply the 3rd row by -1/10
Multiply the 4th row by -1/6; Add 3/20 times the 4th row to the 3rd row
Add -1/2 times the 4th row to the 2nd row; Add -1 times the 4th row to the 1st row
Add -2 times the 3rd row to the 2nd row; Add -1 times the 3rd row to the 1st row
Add -1 times the 2nd row to the 1st row
Then the RREF of B is
1
0
0
0
-3/20
23/120
-1/10
7/120
0
1
0
0
-1/5
-2/15
1/5
2/15
0
0
1
0
7/20
-9/40
-1/10
-1/40
0
0
0
1
0
7/6
0
-1/6
Now, it is apparent that e1 = -(3/20)v1 -(1/5)v2 +(7/20)v3 ; e2 =(23/120)v1 -(2/5)v2 - (9/40)v3 +(7/6)v4; e3 = - (1/10)v1 +(1/5)v2 -(1/10)v3 and e4 = (7/120)v1 +(2/15)v2 -(1/40)v3-(1/6)v4.
Since T is a linear transformation, it preserves vector addition and scalar multiplication. Therefore, we have T(e1 )= -(3/20) T(v1) -(1/5)T(v2) +(7/20) T(v3) = -(3/20)(29,14,14,-14)T -(1/5)(52,39,26,26)T+ (7/20)(45,34,18,18)T=(-87/20,-21/10,-21/10,21/10)T+(-52/5,-39/5,-26/5,-26/5)T+ ( 63/4,119/10,63/10,63/10)T = (1,2,-1,16/5)T.
Similarly, T(e2) =(23/120)T(v1) -(2/5)T(v2) - (9/40)T(v3) +(7/6)T(v4) = (23/120)(29,14,14,-14)T- (2/5)(52,39,26,26)T –(9/40) (45,34,18,18)T+(7/6)(9,13,5,5)T =(667/120,161/60,161/60,-161/60)T –(104/5,78/5,52/5,52/5)T–(81/8,153/20,81/20,81/20)T+(21/2,91/6,35/6,35/6)T = (223/15,-27/5,-89/15,-113/10)T;
T(e3)=-(1/10)T(v1)+(1/5)T(v2)-(1/10)T(v3)=-(1/10)(29,14,14,-14)T+(1/5)(52,39,26,26)T-(1/10)(45,34,18,18)T =(-29/10,-7/5,-7/5,7/5)T+(52/5,39/5,26/5,26/5)T+(-9/2,-17/5,-9/5,-9/5)=(3,3,2,24/5)T;
T(e4)= (7/120)T(v1)+(2/15)T(v2)-(1/40)T(v3)-(1/6)T(v4)=(7/120)(29,14,14,-14)T+(2/15)(52,39,26,26)T- (1/40)(45,34,18,18)T–(1/6)(9,13,5,5)T = (203/120,49/60,49/60,-49/60)T+(104/15,78/15,52/15,52/15)T + (-9/8,-17/20,-9/20,-9/20)T + ( -3/2,-13/6,-5/6,-5/6)=(6,3,3,41/30)T.
Then A =
1
223/15
3
6
2
-27/5
3
3
-1
-89/15
2
3
16/5
-113/10
24/5
41/30
The RREF of A is I4. Hence the columns of A arelinearly independent. Thus, a basis for Col(A) is {(1,2,-1,16/5)T, (223/15,-27/5,-89/15,-113/10)T, (3,3,2,24/5)T, (6,3,3,41/30)T}.
The null space of A consists of all the vectors satisfying the condoition AX = 0. Since the RREF of A is I4, therefore, the equation AX = 0 has a trivial solution only. Hence Null(A) has only the zero vector.
The RREF of AT is I4. Hence the rows of A arelinearly independent.Thus a basis foir Row(A) is { (1,223/15,3,6),(2,-27/5,3,3),(-1,89/15,2,3),(16/5,-113/10,24/5,41/30).
Let C be the matrix with v1,v2,v3,v4 and u = (5,1,5,1)T as columns. The RREF of C is
1
0
0
0
12/29
0
1
0
0
8/29
0
0
1
0
-40/29
0
0
0
1
169/29
Then u = (12/29)v1+(8/29)v2+(-40/29)v3+(169/29)v4 so that T(5,1,5,1)T= T(u) = (12/29)(29,14,14,-14)T+ (8/29) (52,39,26,26)T+(-40/29)(45,34,18,18)T+(169/29)(9,13,5,5)T = (348/29,168/29,168/29,-168/29)T +(416/29,312/29,208/29,2108/29)T+(-1800/29, -1360/29,-720/29,-720/29)29)T + (1521/29,2197/29,845/29,845/29)T = ( 485/29,1317/29,501/29, 2065/29)T
Let X = (w1,w2,w3,w4)T . If T(X) = (2,3,-1,7)T, then AX =(2,3,-1,7)T so that X = A-1(2,3,-1,7)T. Now, A-1=
-2058/29779
18392/29779
-83163/208453
-36810/208453
1545/29779
-2955/29779
-4635/208453
8100/208453
5350/29779
-23821/29779
43508/208453
106110/208453
-1197/29779
16167/29779
513/29779
-9570/29779.
Therefore X = (w1,w2,w3,w4)T can be calculated.
Similarly, X = A-1(2,3,7,7)T can be calculated
1
0
0
0
-3/20
23/120
-1/10
7/120
0
1
0
0
-1/5
-2/15
1/5
2/15
0
0
1
0
7/20
-9/40
-1/10
-1/40
0
0
0
1
0
7/6
0
-1/6
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