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Homework Help: Linear Algebra - isomorphism questions

  1. Feb 13, 2010 #1
    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.
     
    Last edited: Feb 13, 2010
  2. jcsd
  3. Feb 14, 2010 #2

    HallsofIvy

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    Science Advisor

    Yes, although you don't need to do it with matrices. If a and b are any two numbers then we can solve f(x,y)= (a,b) as x- 2y= a, x+ y= b for x= (a+2b)/2, y= (a+b)/3, so this is onto. Also if [itex]f(x_1,y_1)= f(x_2,y_2)[/itex], [itex]x_1- 2y_1= x_2- 2y_2[/itex], [itex]x_1+ y_1= x_2+ y_2[/itex], then [itex]x_1- x_2= y_2- y_1[/itex] from the first equation and [itex]x_2- x_1= -2(y_2- y_1)[/itex]. Then [itex]y_2- y_1= -2(y_2- y_1)[/itex] which leads to [itex]y_1= y_2[/itex] and then [itex]x_1= x_2[/itex] showing that this function is one-to-one.

    You still need to show that f((a,b)+ (c,d))= f((a,b))+ f((c,d)) and f(r(a,b))= rf(a,b) for all real numbers a, b, c, d, and r to show that this is an isomorphism.

    You need to say a bit more. You can use the fact that the transpose is "dual". That is, that
    (AT)T= A.

    This function is onto because, given any matrix B, it is the transpose of BT. It is one-to-one because, if
    AT= BT
    then, taking the transpose of each side, A= B.

    And, again, just showing that a function is one-to-one and onto is NOT enough to show that it is an isomorphism. You must also show that the function is a homormorphism- that it "preserves" the operations.
     
    Last edited by a moderator: Feb 14, 2010
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