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Basis, kernal, image, transformation, etc (not a hard one!)

  1. Nov 3, 2005 #1
    I hope everyone could understand my matrix notation!

    T: M22 -----> M22; T[a b; c d] = [a+b b+c; c+d d+a]

    I found the the basis for ker(T) to be [1 -1; 1 -1]

    im having problems finding a basis for im(T)? I know that it has to be dim3!

    so far i have:

    im(T) = {[a+b b+c; c+d d+a]/ [a b; c d] in M22} = M22

    How on earth do i get a basis out of that?

    Anybody have any ideas?

    Thanks
     
  2. jcsd
  3. Nov 3, 2005 #2

    Hurkyl

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    If you just tried picking vectors in im T at random, do you think you'll have a pretty good chance of finding three linearly independent ones? Can you think of any way to improve your odds?

    Random guessing is a very underrated technique. :biggrin:

    P.S. in

    im(T) = {[a+b b+c; c+d d+a]/ [a b; c d] in M22} = M22

    I assume that last '=' was supposed to be a subset symbol?
     
  4. Nov 3, 2005 #3
    in honesty, i have no idea what that last = sign means, the book just puts it their on other examples. Im doing at even question, so it has no solution in the solution manual

    Im not gussing about the dim of the image. I KNOW that its 3 since dim(V) = dim(Ker) + dim(Im)

    and i found that the dimension of the kernal is 1, and therefore, dimension of the image has to be 3.

    I just need to find the basis for im(T) = {[a+b b+c; c+d d+a]/ [a b; c d] in M22} and its got be really STUCK :(
     
  5. Nov 3, 2005 #4

    Hurkyl

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    I meant guess a basis!
     
  6. Nov 3, 2005 #5
    does [1 0; 0 0] [0 1; 0 0] [0 0; 1 0] work?
     
  7. Nov 3, 2005 #6

    Hurkyl

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    They're certainly linearly independent. Are they in the image of T? If so, then is there any reason why they wouldn't suffice for a basis? Can you prove that they are a basis?
     
  8. Nov 3, 2005 #7
    Those questions are making this harder :(

    They are independint, but NO, they are not an image. becuase it dose not satisfy m(T) = {[a+b b+c; c+d d+a]/ [a b; c d] in M22}

    hmmmm...

    another GUESS is [1 0; 1 0] [0 1; 1 0] [1 0; 0 1] ?????
     
  9. Nov 3, 2005 #8

    Hurkyl

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    They're questions you'd have to answer anyway, though, aren't they? After all, what is the definition of the image, and what is the definition of a basis for that space? :tongue2:
     
  10. Nov 3, 2005 #9
    oh goodness.. this is getting so confusing now :S

    Listen..

    the question says " in each case, (i) find a basis of kerT and (ii) find a basis of imT. " e) T: M22 -----> M22; T[a b; c d] = [a+b b+c; c+d d+a]

    i found ker(T) and my answer was [1 -1; 1 -1]

    image T is im(T) = {[a+b b+c; c+d d+a]/ [a b; c d] in M22} (By definition)

    so how on earth to i find the basis for [a+b b+c; c+d d+a] such that [a b; c d] are in M22?

    now.... a basis is i) linearly independent, and ii) a spanning set of that space.

    Now i just have no idea how to extract a basis out of that image :S
     
  11. Nov 3, 2005 #10

    Hurkyl

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    Can you find any vectors in im T?
     
  12. Nov 3, 2005 #11
    I don't understand what you mean by that? Perhpas [1 0; 0 0] ?? is that what your trying to ask?
     
  13. Nov 3, 2005 #12

    Hurkyl

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    I want you to give me a vector that is in the image of T, and a proof that it is in the image of T.
     
  14. Nov 3, 2005 #13
    well... [1 1; 1 1] is in imageT. and explanation is since [1 1; 1 1] satisfies [a b; c d] being in M22. how does that sound?

    see.. imT = {T(v)/v in V}

    So wouldn't any vector in M22 satisfy that?
     
  15. Nov 3, 2005 #14

    Hurkyl

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    Ah, I see your problem now: you're confused by the set-builder notation.

    The notation

    [tex]
    \mbox{im} T = \{ T(v) | v \in V \}
    [/tex]

    means the following:

    For any w, w is in im T if and only if the equation w = T(v) has a solution with v in V.
     
  16. Nov 3, 2005 #15

    Hurkyl

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    So, the fact that

    [tex]
    \left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right) \in M_{2,2}
    [/tex]

    tells us that

    [tex]
    T\left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right) \in \mbox{im}\ T
    [/tex]

    or equivalently,

    [tex]
    \left( \begin{array}{cc} 2 & 2 \\ 2 & 2 \end{array} \right) \in \mbox{im}\ T
    [/tex]

    since

    [tex]
    T\left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right)
    = \left( \begin{array}{cc} 2 & 2 \\ 2 & 2 \end{array} \right)
    [/tex]
     
  17. Nov 3, 2005 #16
    Well.. sadly, that notation, I would use in theory... but when doing calculations.. i hardly pay attention to it (and sadly, thats with a lot of other equations too!)

    YES!!! what you said above is what im saying!!! and so would be [3 1; 2 4] etc.....

    I just have no idea what the bais of [a+b b+c; c+d d+a] is.

    i made it so [a b; c d] + [b c; d a] and finding the basis of those two is just easy.

    its [1 0; 0 0] [0 1; 0 0] [0 0; 1 0] [0 0; 0 1] etc... but that dosnt help in finding the basis of the ortininal one equation.

    So what is the basis of [a+b b+c; c+d d+a] ??????? and if i can find that, i found the basis of imT
     
  18. Nov 3, 2005 #17

    Hurkyl

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    I don't know if you're being sloppy or confused, so I'll correct this:

    It doesn't make sense to ask if [a+b b+c; c+d d+a] has a basis: it is not a vector space, it is simply a matrix.

    Now, the set of all matrices that can be written in that form, with a, b, c, d all being real numbers, is a vector space, and it does make sense to ask if it is a vector space.


    Now, back to the question at hand...

    I agree, the matrix [3 1;2 4] is indeed an element of the image of T. (e.g. take a = 2, b = 1, c = 0, d = 2)

    Therefore, the set {[3 1;2 4]} is a linearly independent subset of the image of T with one vector in it.

    Since you know a basis for im T must have three vectors in it, we know that the set {[3 1;2 4]} is one-third of a basis for T, right?

    (This is what I meant by random guessing: just keep picking vectors in im T until you get a basis)
     
  19. Nov 3, 2005 #18
    Yes, im following.

    any M22 matrix is in the image if it sasisfies {[a+b b+c; c+d d+a]/ [a b; c d] is in M22} That is a FACT!

    OKAY... I see how its working out now. lets say that we have [5 3 2 4], then (a=2 b=3 c=0 d=2)

    so far, we have a bais { [1 1; 1 0] [ ] [ ] }

    I suppose the other to parts of the basis is [0 1; 1 1] and [1 0; 1 1]?

    Basis is { [1 1; 1 0] [0 1; 1 1] 1 0; 1 1] }

    Am i getting warmer? Im thinking right now... what about [1 1; 0 1]? why won't that fit it? (other than the fact that i know dim imT = 3 .... ?? It won't fit in because we can only have a max or 3 vectors since exactly 1 of a,b,c,d has to = 0?
     
    Last edited: Nov 3, 2005
  20. Nov 3, 2005 #19

    Hurkyl

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    I can't figure out where you're going wrong...

    [1 1; 1 0] is not an element of im T. Why do you think that it is?
     
  21. Nov 3, 2005 #20
    okay listen... say we have a space with M22 [a a; b b]/a b are real numbers.

    then basis is { [1 1; 0 0] [0 0 1 1] }

    Im trying to apply that to this case.

    And since you showed that {matrix [3 1;2 4] is i an element of the image of T. (e.g. take a = 2, b = 1, c = 0, d = 2) } I see that {1 1; 0 1} fits as part of the basis?

    hmm.. i thought the same thing for [1 1; 1 0]
     
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