zeion
Feb13-10, 12:23 PM
1. The problem statement, all variables and given/known data
Determine whether the following mappings f is onto or one-to-one. Is f an isomorphism?
a) f maps R2 into R2 and is defined by f(x,y) = (x-2y, x+y)
b) f maps R2 into R3 and is defined by f(x,y) = (x, y, x+y)
i) f maps R3 into P2(R), defined by f(a1, a2, a3) = a2 - a3x + (1-a1)x^2
j) f maps Rmxn into Rnxm and is defined by f(A) = A^T (transpose of A) for all A in Rmxn
2. Relevant equations
3. The attempt at a solution
Can I solve this by forming the matrix of the output?
ie I have x-2y = 0 and x+y = 0 as a matrix and row reduce, I get a 2x2 identity matrix and therefore
1) There are always solution for any augmented matrix, so f is onto
2) There are no parameters in the matrix and both variables always have a unique solution for every augmented matrix, so f is one-to-one
Is that right?
For b)
I get form the 3x2 matrix and row reduce, then a have third row of zero and no parameter, therefore f is not onto since there will always be some component of R3 that has no solution (hence the zero row). But f is 1-1 since both variables always have a unique solution for any augmented matrix (hence it has 2 leading ones).
Does that explanation seem right?
For i)
I form the matrix and row reduce to a I3x3, therefore f is onto and 1-1.
For j)
f is one-to-one since it forms the transpose of A and every row of A is mapped to every column of the image once.
It is onto since this applies to every matrix of Rmxn, therefore transposed into every matrix of Rnxm.
Determine whether the following mappings f is onto or one-to-one. Is f an isomorphism?
a) f maps R2 into R2 and is defined by f(x,y) = (x-2y, x+y)
b) f maps R2 into R3 and is defined by f(x,y) = (x, y, x+y)
i) f maps R3 into P2(R), defined by f(a1, a2, a3) = a2 - a3x + (1-a1)x^2
j) f maps Rmxn into Rnxm and is defined by f(A) = A^T (transpose of A) for all A in Rmxn
2. Relevant equations
3. The attempt at a solution
Can I solve this by forming the matrix of the output?
ie I have x-2y = 0 and x+y = 0 as a matrix and row reduce, I get a 2x2 identity matrix and therefore
1) There are always solution for any augmented matrix, so f is onto
2) There are no parameters in the matrix and both variables always have a unique solution for every augmented matrix, so f is one-to-one
Is that right?
For b)
I get form the 3x2 matrix and row reduce, then a have third row of zero and no parameter, therefore f is not onto since there will always be some component of R3 that has no solution (hence the zero row). But f is 1-1 since both variables always have a unique solution for any augmented matrix (hence it has 2 leading ones).
Does that explanation seem right?
For i)
I form the matrix and row reduce to a I3x3, therefore f is onto and 1-1.
For j)
f is one-to-one since it forms the transpose of A and every row of A is mapped to every column of the image once.
It is onto since this applies to every matrix of Rmxn, therefore transposed into every matrix of Rnxm.