1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear Transformation and Isomorphism

  1. Dec 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Given the transformation fh : R 3 → R 3 defined by fh(x, y, z) = (x−hz, x+y −hz, −hx+z), where h ∈ R is a parameter.
    a) Find, for all possible values of h, Ker(fh), Im(fh), their bases and dimensions.
    b) Is fh an isomorphism for some value of h?
    2. Relevant equations
    Ax=o

    3. The attempt at a solution
    [ x−hz ]
    [ x+y −hz ]
    [ −hx+z ]

    Matrix associated to the linear transformation =

    [ 1 0 -h ]
    [ 1 1 -h ]
    [ -h 0 1 ]


    Reduced row echelon form =
    [ 1 0 -h ]
    [ 0 1 0 ]
    [ -h 0 1 ]

    x - hz = 0
    y=0
    -hx+z=0

    Therefore h = 1.

    ker(fn)

    [ 1 0 -1 ]
    [ 0 1 0 ] = 0
    [ -1 0 1 ]

    ker(fn) = span { (1,0,-1) , (-1,0,1) } = range(ker(fn))

    To find the Im(fn) we will firstly set up the original matrix with h=1

    [ 1 0 -1 ]
    [ 1 1 -1 ]
    [ -1 0 1 ]

    Then transpose =

    [ 1 1 -1 ]
    [ 0 1 0 ]
    [ -1 -1 1 ]

    Row reduced echelon form of the transposed matrix =

    [ 1 0 -1 ]
    [ 0 1 0 ]
    [ 0 0 0 ]

    Im(fn) = span { (1,0,-1) }

    Dimension of the ker(fn) and Im(fn) = 1

    Because this is the only vector in the Im(fn) it also = the range(fn). I'm still a bit confused about what the range of a linear transformation is and how to find it.

    I'm not really sure where to start with part b) of the question...
    The only time the matrix will = 0 is when x,y,z = 0 (trivial solution) so I would say it is an isomorph, but I'm not totally sure.

    [ 1 0 -1 ]
    [ 1 1 -1 ]
    [ -1 0 1 ]
     
  2. jcsd
  3. Dec 9, 2015 #2
    Call ##{\cal B} = \{ e_1, e_2, e_3 \}## the canonical base of ##\mathbb{R}^3##. Can you show that if ##h\notin \{-1,1\}##, then ##f_h({\cal B}) ## is linearly independent? Then why is ##f_h({\cal B})## is a base of ##\mathbb{R}^3##? Why does it prove that ##f_h## is an automorphism of ##\mathbb{R}^3## ?

    If ##h\in\{-1,1\}##, then ##f_h({\cal B})## is linearly dependant but generates ##\text{Im}(f_h)##. Can't you easily find a free subfamily that generates ##\text{Im}(f_h)##? What does it tell you about the rank of ##f_h## ?
     
  4. Dec 9, 2015 #3
    e1 e2 e3
    [ 1 0 -h ]
    [ 1 1 -h ]
    [ -h 0 1 ]

    iff h = 2

    [ 1 0 -2 ]
    [ 1 1 -2 ]
    [ -2 0 1 ]

    fh(B) is linearly independent if we row reduce the matrix above.

    The Im(Fh) that I calculated above was linearly dependent.

    Rank of fh is 1. for Im(fh) rank = 2

     
  5. Dec 9, 2015 #4
    Unless I'm wrong, ##f_h(e_1) = (1,1,-h)##, ##f_h(e_2) = (0,1,0)##, and ##f_h(e_3) = (-h,-h,1)##.
    First of all, at what condition ##f_h(e_1),f_h(e_2)## and ##f_h(e_3)## are linearly independant ? Unless you prove me wrong, this condition is ##h\notin \{ -1,1 \}##. Do we agree at that point ?
     
  6. Dec 9, 2015 #5
    They are linearly independent if h = 2
     
  7. Dec 9, 2015 #6
    Ok, but they are linearly independent in many more cases
     
  8. Dec 9, 2015 #7
    Ok let's say we agree. The important result here, and it must be somewhere in your algebra notes in a more general form is :

    ##f_h## is bijective if and only ##\{ f(e_1), f(e_2), f(e_3) \}## is a base of ##\mathbb{R}^3##.

    So that for ##h\notin \{-1,1\}##, your function is bijective, so it's kernel is ##\{0\}##, it's image is ##\mathbb{R}^3##. Bases and dimensions for the image and the kernel are very easy.

    The less easy case is when ##h\in\{-1,1\}##. Here ##f_h## is not bijective. ##( f(e_1), f(e_2), f(e_3) )## is a linearly dependent family that spans the image of ##f_h##. You can find a linearly independent subfamily that also spans the image of ##f_h##, that is to say a base of ##\text{Im}(f_h)##.
     
  9. Dec 29, 2015 #8
    revisiting an old post. Hope this is ok...

    ker (fh) = span { ( 1,0,-1) , (-1,0,1) } dimension = 2
    iff h=1
    [ 1 0 -1 ]
    [ 1 1 -1 ]
    [ -1 0 1 ]

    Transpose the matrix (above) and row reduce. We get:

    [ 1 0 -1]
    [ 0 1 0 ]
    [ 0 0 0 ]

    Im ((fh) = span { ( 1,0,-1) } dimension = 1
     
  10. Dec 29, 2015 #9

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Why bother with matrices and echelon forms, etc.? The kernel is just the set
    [tex] \text{Ker}(fh) = \{(x,y,z): x-hz=0,x+y-hz=0, z-hx=0 \}, [/tex]
    which can easily be solved/analyzed without using any matrix tools (which actually add nothing to the analysis). The image is
    [tex] \text{Im}(fh) = \{(p,q,r): x-hz=p, x+y-hz=q, z-hx=r \:\text{for some} \: x,y,z \}. [/tex]
    Again, solving by high-school algebra is as fast a way as any.

    Once you have solved both these questions the answers to the others drop out almost immediately.

    Your conclusions about the value of h, etc., are ill-considered. For instance, in your first analysis you conclude that h = 1. Well, who says you need h = 1? What happens if you do not have h = 1? Nobody forces you to have h = 1 in the original transformation!
     
  11. Dec 29, 2015 #10
    No one forces me to have x=1 either though do they? That's where I get confused. If one variable can take on any value then why can't another?
     
  12. Dec 29, 2015 #11

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Well, if you have h = 1, then Ker(fh) and Im(fh) have a certain form. If you have h = -1, then Ker(fh) and Im(fh) have some other form. And, if you have h ≠ -1,1, still other forms of Ker(fh) and/or Im(fh) occur. Since h is allowed to be anything, you need to figure these out in order to be sure you have covered all possibilities, and to be able to describe what happens as h varies.

    And, while it is true that when h = 1 nobody forces you to have some particular value for x (for example), the issue is that if your point (x,y,z) is in Ker(f1), then you are, indeed, forced to take certain values; if you fail to take those values your point fails to fall into the set Ker(f1).
     
    Last edited: Dec 29, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Linear Transformation and Isomorphism
Loading...