Linear Transformation with Respect to Given Bases

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SUMMARY

The discussion focuses on the linear transformation T: R3 -> M(2,2) defined by T(x,y,z) = [ z ... -z; 0 ... x-y]. The participants analyze the transformation with respect to the bases B = {(1,0,0),(0,1,0),(0,0,1)} and C = { [1 0], [0 1], [0 0], [0 0] }. The matrix representation [T]c,b was determined to be a 3x4 matrix: [ 0 0 0 1; 0 0 0 -1; 1 -1 0 0]. The discussion also addresses finding the image and kernel of the transformation, confirming that these correspond to the column space and solution space of the matrix, respectively.

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  • Understanding of linear transformations in vector spaces
  • Familiarity with matrix representation of transformations
  • Knowledge of basis vectors and their role in linear algebra
  • Concepts of image and kernel in linear mappings
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Students and professionals in mathematics, particularly those studying linear algebra, as well as educators looking to clarify concepts related to linear transformations and matrix representations.

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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 >.<
 
Last edited:
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Yes, the way you found the matrix certainly makes sense and, in fact, is the best way to do it. It is also true that "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."

Now, since you have done part (a) and know how to do parts (b) and (c), what is your question?
 
So the first matrix i got is correct?

[ 0 0 0 1 ]
[ 0 0 0 -1]
[ 1 -1 0 0]or is is the transpose of it?
 

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