- #1
forty
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Let T: R3 -> M(2,2) be the linear transformation given by
T(x,y,z) = [ z ...-z ]
.....[ 0 ... x-y]Fix bases B = {(1,0,0),(0,1,0),(0,0,1)} and C = { [1 0] , [0 1] , [0 0] , [0 0] }
............[0 0]...[0 0]...[1 0]...[0 1]for R3 and M(2,2) respectivelya) Find the matrix [T]c,b of T with respect to the bases B and C.
b) Use the matrix from part (a) to find the basis for Im(T)
c) Use the matrix from part (a) to find the basis for Ker(T)
For parts b and c I'm pretty sure that's just finding the column space (part b) and solution space (part c) of the matrix.
As for a its more so to do with my own confusion when it comes to bases.
I worked out the matrix to be [ 0 0 0 1 ]
...(3 by 4 matrix)...[ 0 0 0 -1]
.......[ 1 -1 0 0]
The way i got this was by taking each vector from B applying the transformation then writing it in terms of the basis C and then writing the coefficients of each matrix in the rows of the above matrix. I hope that even makes sense >_>.
Any help greatly appreciated :)
P.S. sorry for the garbagety matrices :SEDIT: Maybe its the transpose of that matrix... I am lost >.<
T(x,y,z) = [ z ...-z ]
.....[ 0 ... x-y]Fix bases B = {(1,0,0),(0,1,0),(0,0,1)} and C = { [1 0] , [0 1] , [0 0] , [0 0] }
............[0 0]...[0 0]...[1 0]...[0 1]for R3 and M(2,2) respectivelya) Find the matrix [T]c,b of T with respect to the bases B and C.
b) Use the matrix from part (a) to find the basis for Im(T)
c) Use the matrix from part (a) to find the basis for Ker(T)
For parts b and c I'm pretty sure that's just finding the column space (part b) and solution space (part c) of the matrix.
As for a its more so to do with my own confusion when it comes to bases.
I worked out the matrix to be [ 0 0 0 1 ]
...(3 by 4 matrix)...[ 0 0 0 -1]
.......[ 1 -1 0 0]
The way i got this was by taking each vector from B applying the transformation then writing it in terms of the basis C and then writing the coefficients of each matrix in the rows of the above matrix. I hope that even makes sense >_>.
Any help greatly appreciated :)
P.S. sorry for the garbagety matrices :SEDIT: Maybe its the transpose of that matrix... I am lost >.<
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