Question

find a basis for the range and the rank of the given linear transformation and determine if it is onto.

1) T: R₃[x]→R₂[x] given by T(a+bx+cx²+dx³) = (a+2b+c) + (2a+5b+c+d)x + (2a+6b+d)x².

2)

G).r« 2 ,T(e3) T (e2) 3. Т:R4 M2x2(R) given by T(ei) 2 3 -G ) (G) 2 ,T(е) 4 3 1

Answer 1

1) T: R3x] given by T(a + bx + сx? + dx3) — (а + 2b + с) + (2а + 5b + с + d)x + (2а + 6b + d)x? The associated matrix is given by 1 2 0 1 2 1 1 L2 6 0 1 Note that we have considered the basis {1, x,,x2,x3} for R3[x] and basis {1,x, x2} for R2[x] cx2 dx3) has coordinate vector [a, b, c, d]T. That is, thus (a bx ai a 2b c 2а + 5b + с + d 2 1 1 2 5 1 1 C _ 2 6 0 2а + 6b + d Now we find the range and rank of T as follows: We reduce the matrix into row echelon form 2 1 0 2 1 1 L2 6 0 1 2 01 1 R2 - 2R, 0 1 1 1 2 2 1 -

2 1 C 0 1 1 1 R3 2R2 Lo o 0 As the last matrix is in row echelon form and has rank 3, and we know that dim(R2[x]) has dimension 3, thus the linear transformation T is an onto transformation V the 1st 2nd For a basis of the range see that in the row echelon form the leading elements ar in the 1st 2nd and 4t* column, and 4th column of the original matrix forms a basis for the range space, that is So 2 {1+ 2х + 2x2, 2 + 5х + бх?, х+x?}. we can write a basis for the range space as and in terms of polynomials, ( ) 1 2 ,T(eз) 3. Т:R4 M2x2(R) given by T(ei) ,T(e2) 3 2 2 T(e4) 3 3. The matrix of transformation is given by 1 -1 3 2 2 2 2 3 1 1 L1 3 4 1 We reduce this matrix to reduced ro echelon form n

-1 3 1 3 Row Operation 1: 1 2 2 2 add -l times the 1st row to the 2nd row 3 1 3 -1 2 1 1 2 1 3 1 3 1 3 4 1 4 1 1 -1 3 1 -1 3 Row 0 1 0 1 -1 3 -1 Operation 2: add -l times the 1st row to the 3rd row 1 2 0 1 1 3 2 0 1 3 1 3 4 4 1 1 1 1 -1 3 1 -1 3 Row 0 3 -1 0 add -l times the 1st row to the 4th row 3 Operation 3: 0 1 0 1 2 0 2 0 1 3 0 2 5 4 1 -2 -1 3 3 Row 0 1 0 1 3 -1 3 -1 Operation 4: add -l times the 2nd row to the 3rd row 0 1 0 0 2 C -1 1 0 2 5 0 2 5 2 -2.

-1 3 1 -1 3 Row 0 1 0 1 3 -1 3 Operation 5: add -2 times the 2nd row to the 4th row 0 0 0 0 1 -1 1 1 0 25 2 0 0-1 0 1 -1 3 1 1 -1 3 Row 0 1 3 -1 0 3 -1 Operation 6: multiply the 3rd row by -l 1 0 0 0 0 - 1 1 -1 0 0 -1 0 0 0-1 0 1 1 -1 3 -1 3 Row Operation 7: -1 add 1 times the 3rd row to the 4th row 0 0 0 0 -1 1 1 0 0-1 0 0 0 0 -1 -1 3 -1 3 Row 0 1 3 -1 0 3 -1 Operation 8: multiply the 4th row by -1 V 0 0 0 0 1 -1 1 -1 0 0 0 0 0 1 0 -1 1 1 -1 1 1 3 -1 3 Row 0 1 0 1 3 -1 3 1 add 1 times the 4th row to the 3rd row -1 Operation 9: 0 0 C C 0 0 0 1 0 0 0 1

1 1 1 1 -1 3 -1 3 Row 0 1 0 1 3 3 0 Operation 10: add 1 times the 4th row to the 2nd row 0 0 0 0 1 C 0 0 0 0 0 0 0 1 1 1 1 -1 3 -1 0 Row 3 0 add -3 times the 4th row to the 1st row 0 0 3 0 Operation 0 0 0 0 1 1 0 l1: 0 0 0 1 0 0 0 1 1 -1 0 -1 0 Row 1 0 0 add -3 times the 3rd row to the 2nd row 1 0 Operation 12: 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 110 0 1 1 -1 0 Row 0 1 0 0 0 1 0 0 add 1 times the 3rd row to the 1st row 0 Operation 13: TC 0 01 0 0 0 1 0 0 0 0 0 0 1 1 0 1 0 Row 1 0 0 0 1 0 0 C add -l times the 2nd row to the 1st row Operation 14: 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 As the last matrix is an identity matrix of order 4 so rank is 4, and the given set of matrices forms a basis for the range space.

-1 2 G). () 3 2 3

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