Combined linear transformations

[Niles]

[SOLVED] Combined linear transformations

Homework Statement

I have a linear transformation L : R^3 -> R^3 represented by a matrix A. I also have another linear transformation S : R^3 -> R represented by a matrix B.

The dimensions of the matrix A must be 3x3 and for B it is 1x3. I have to find the rank and the nullity of the linear transformation S o L.

I don't know if it's necessary, but the linear transformations are:

L(x,y,z) = (6x-3y-2z , 14x-7y-4z , -5x+3y+3z) and

S(x,y,z) = x+y+z.

The Attempt at a Solution

The transformation S o L is linear, because S and L is linear and the matrix C that represents S o L is given by B*A. This product gives me a 1x3 matrix and the rank and nullity of this matrix is what is being asked for. But how (or is it even possible) to find the rank and nullity of C?

Thanks for all your help,

sincerely Niles.

 

[HallsofIvy]

If you were to write the linear transformations as matrices, then C= S o L would be the product of those matrices. You could find the that product matrix just as you say. It certainly is possible to find the rank and nullity of any matrix. The nullity is, by definition, the dimension of the kernel. Here C will be a matrix having 1 row and 3 "columns", 3 numbers in that 1 row. Let <x, y, z> be a column matrix (3 rows, 1 column) and take the product Cx which will, of course, be a single number. If <x, y, z> is in the kernel, then that number must be 0. You have one equation in 3 unknown numbers x, y, z. The nullity is the dimension of the solution space.

You could, instead, write the formula for the compostion. It will be less confusing to change the "labels" for the variables: let (u, v, w)= L(x, y, z) so S(u, v, w)= u+ v+ w= (6x-3y-2z)+ (14x-7y- 4z)- (-5x+ 3y+ 3z)= 15x- 7y- 3z. (That is the one equation in 3 unknown numbers I mentioned above.) The kernel consists of all x, y, z, such that 15x- 7y- 3z= 0. It should be easy to see that you can solve that for one of the variables in terms of the other 2. The kernel has dimension 2 and the nullity is 2.

You should know that the sum of the rank and nullity must be the dimension of the domain space, here 3. The rank is the dimension of the "image" which is a subspace of the range space. Since the range space, here R¹, has dimension 1, the rank can only be 0 or 1. If S o L took every vector in R³ to 0, then the rank would be 0 and the nullity 3. Here, that clearly does not happen so the rank is 1 and the nullity 2.