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Math Help - Finding a linear map with a kernal equal to a subspace

  1. #1
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    Finding a linear map with a kernal equal to a subspace

    if V= (a,a+b,a+b+c,a+b+c+d,a+b+c+d), a,b,c,d are all real numbers

    how can we find a linear map from R^5 to R^4 with a kernal equal to V?

    I can calculate the basis for V, and it can be shown V is a subspace of R^5.

    Any hints on how to start the question, or any intuition behind it at all?

    Thanks
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    Re: Finding a linear map with a kernal equal to a subspace

    rank-nullity therorem. think about what the matrix for such a linear map must be. for example, if your basis is {v1,v2,v3,v4}, you must have:

    L(v1) = L(v2) = L(v3) = L(v4) = 0.

    the problem is easier than it looks.
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  3. #3
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    Re: Finding a linear map with a kernal equal to a subspace

    Quote Originally Posted by Deveno View Post
    rank-nullity therorem. think about what the matrix for such a linear map must be. for example, if your basis is {v1,v2,v3,v4}, you must have:

    L(v1) = L(v2) = L(v3) = L(v4) = 0.

    the problem is easier than it looks.
    Thanks for your reply.

    I'm finding linear algebra extremely hard, could you please elaborate on your answer?
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  4. #4
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    Re: Finding a linear map with a kernal equal to a subspace

    the rank-nullity therorem states:

    for a linear transformation L, defined on a vector space V of dimension n:

    n = dim(ker(L)) + rank(L)

    (dim(ker(L)) is the dimension of the kernel, or null space of L, and is also called: "the nullity of L". rank(L) is the dimension of the image, or range, of L).

    in this case, we have: 5 = 4 + 1 (the null space of L is the subspace V of R5).

    so the rank of L is 1. this means there is just ONE element in any basis for the image of L. this also means we can construct a matrix for L (in some basis) with just one linearly independent column.

    i presume the basis you found for V was:

    {(1,1,1,1,1),(0,1,1,1,1),(0,0,1,1,1),(0,0,0,1,1)}.

    so to construct a matrix for L we have:

    \begin{bmatrix}0&0&0&?&?\\0&0&0&?&?\\0&0&0&?&? \\0&0&0&?&? \end{bmatrix}

    what can the ?'s possibly be?
    Last edited by Deveno; December 18th 2012 at 04:32 AM.
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  5. #5
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    Re: Finding a linear map with a kernal equal to a subspace

    Quote Originally Posted by Deveno View Post
    the rank-nullity therorem states:

    for a linear transformation L, defined on a vector space V of dimension n:

    n = dim(ker(L)) + rank(L)

    (dim(ker(L)) is the dimension of the kernel, or null space of L, and is also called: "the nullity of L". rank(L) is the dimension of the image, or range, of L).

    in this case, we have: 5 = 4 + 1 (the null space of L is the subspace V of R5).

    so the rank of L is 1. this means there is just ONE element in any basis for the image of L. this also means we can construct a matrix for L (in some basis) with just one linearly independent column.

    i presume the basis you found for V was:

    {(1,1,1,1,1),(0,1,1,1,1),(0,0,1,1,1),(0,0,0,1,1)}.

    so to construct a matrix for L we have:

    \begin{bmatrix}0&0&0&?&?\\0&0&0&?&?\\0&0&0&?&? \\0&0&0&?&? \end{bmatrix}

    what can the ?'s possibly be?
    is the dimension of the null space L 4 because the basis for V is 4 vectors?

    Why is the null space of L the subspace of V?

    How did you construct that matrix?

    When finding a matrix, should the kernal equal V or the basis of V? is this the same thing?

    Apologies if these are elementary questions but I'm finding this hard and really appreciate your help. I have never been taught the rank nullity theorem and have been given this question.
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    Re: Finding a linear map with a kernal equal to a subspace

    Quote Originally Posted by darren86 View Post
    is the dimension of the null space L 4 because the basis for V is 4 vectors?
    yes, dimension is the number of vectors is ANY basis.

    Why is the null space of L the subspace of V?
    the null space of L IS V. V itself is a subspace of R5. we are given by the problem itself that ker(L) = V. we're supposed to find such an L.

    How did you construct that matrix?
    ask yourself, is (1,0,0,0,0) in V? can we find a,b,c,d so that (1,0,0,0,0) = (a,a+b,a+b+c,a+b+c+d,a+b+c+d)? well if we can, then a = 1. but since a+b = 0, b = -1. since a+b+c = 0, and a+b = 0, c = 0. finally, d = 0 (similar reasoning).

    that is:

    (1,0,0,0,0) = (1,1,1,1,1) - (0,1,1,1,1).

    if we make a matrix for L (in the standard basis), then since (1,0,0,0,0) is in V, the first column for L (which is L(1,0,0,0,0)) must be all 0's.

    convince yourself that (0,1,0,0,0), (0,0,1,0,0) are both in V, and that (0,0,0,1,0) and (0,0,0,1,1) are NOT (hint: look at the last two coordinates).

    When finding a matrix, should the kernal equal V or the basis of V? is this the same thing?

    Apologies if these are elementary questions but I'm finding this hard and really appreciate your help. I have never been taught the rank nullity theorem and have been given this question.
    you are mis-reading the question. the question says:

    FIND a linear transformation L WITH kernel V. V is a 4-dimensional subspace of R5. so L is going to take MOST of R5 to the 0-vector of R4.

    as you can see by the head-start i gave you, the matrix for L will take any vector (x,y,z,0,0) to 0. what do we need to happen, for L(0,0,0,1,1) = (0,0,0,0)? in particular if:

    L = \begin{bmatrix}0&0&0&l_1&?\\0&0&0&l_2&?\\0&0&0&l_3  &?\\0&0&0&l_4&? \end{bmatrix}

    what does that last column need to be?
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  7. #7
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    Re: Finding a linear map with a kernal equal to a subspace

    Quote Originally Posted by Deveno View Post
    yes, dimension is the number of vectors is ANY basis.



    the null space of L IS V. V itself is a subspace of R5. we are given by the problem itself that ker(L) = V. we're supposed to find such an L.



    ask yourself, is (1,0,0,0,0) in V? can we find a,b,c,d so that (1,0,0,0,0) = (a,a+b,a+b+c,a+b+c+d,a+b+c+d)? well if we can, then a = 1. but since a+b = 0, b = -1. since a+b+c = 0, and a+b = 0, c = 0. finally, d = 0 (similar reasoning).

    that is:

    (1,0,0,0,0) = (1,1,1,1,1) - (0,1,1,1,1).

    if we make a matrix for L (in the standard basis), then since (1,0,0,0,0) is in V, the first column for L (which is L(1,0,0,0,0)) must be all 0's.

    convince yourself that (0,1,0,0,0), (0,0,1,0,0) are both in V, and that (0,0,0,1,0) and (0,0,0,1,1) are NOT (hint: look at the last two coordinates).



    you are mis-reading the question. the question says:

    FIND a linear transformation L WITH kernel V. V is a 4-dimensional subspace of R5. so L is going to take MOST of R5 to the 0-vector of R4.

    as you can see by the head-start i gave you, the matrix for L will take any vector (x,y,z,0,0) to 0. what do we need to happen, for L(0,0,0,1,1) = (0,0,0,0)? in particular if:

    L = \begin{bmatrix}0&0&0&l_1&?\\0&0&0&l_2&?\\0&0&0&l_3  &?\\0&0&0&l_4&? \end{bmatrix}

    what does that last column need to be?
    could the 4th column be (0,0,0,1) and the last (0,0,0,-1), so the matrix sends all of R^4 to zero and also some vectors in R^5 (the ones where the last 2 elements of those vectors are equal)?
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  8. #8
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    Re: Finding a linear map with a kernal equal to a subspace

    indeed, such an L would send (a,a+b,a+b+c,a+b+c+d,a+b+c+d) to:

    first coordinate: 0a + 0(a+b) + 0(a+b+c) + 0(a+b+c+d) + 0(a+b+c+d) = 0
    second coordinate: 0a + 0(a+b) + 0(a+b+c) + 0(a+b+c+d) + 0(a+b+c+d) = 0
    third coordinate: 0a + 0(a+b) + 0(a+b+c) + 0(a+b+c+d) + 0(a+b+c+d) = 0
    fourth coordinate: 0a + 0(a+b) + 0(a+b+c) + 1(a+b+c+d) + (-1)(a+b+c+d) = 0

    however, it does NOT send all vectors of the form (0,0,0,0,k) to 0, but to (0,0,0,-k) (and it sends vectors of the form (0,0,0,k,0) to (0,0,0,k)).

    but that's ok, the range of L is a 1-dimensional subspace of R4, which is what we need.
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  9. #9
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    Re: Finding a linear map with a kernal equal to a subspace

    Quote Originally Posted by Deveno View Post
    indeed, such an L would send (a,a+b,a+b+c,a+b+c+d,a+b+c+d) to:

    first coordinate: 0a + 0(a+b) + 0(a+b+c) + 0(a+b+c+d) + 0(a+b+c+d) = 0
    second coordinate: 0a + 0(a+b) + 0(a+b+c) + 0(a+b+c+d) + 0(a+b+c+d) = 0
    third coordinate: 0a + 0(a+b) + 0(a+b+c) + 0(a+b+c+d) + 0(a+b+c+d) = 0
    fourth coordinate: 0a + 0(a+b) + 0(a+b+c) + 1(a+b+c+d) + (-1)(a+b+c+d) = 0

    however, it does NOT send all vectors of the form (0,0,0,0,k) to 0, but to (0,0,0,-k) (and it sends vectors of the form (0,0,0,k,0) to (0,0,0,k)).

    but that's ok, the range of L is a 1-dimensional subspace of R4, which is what we need.
    Thanks I think I kind of get the question now!
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    Re: Finding a linear map with a kernal equal to a subspace

    Hi, I've got another similar problem:

    If U=<(1,-1,1,-1,1), (1,2,2,2,1), (0,-2,2,-2,0), (1,0,0,0,1)>

    show that there is no linear map from R^5 to R with kernal equal to U.

    Ive attempted this by finding the set of equations that has solution in U, that is x1-x5=0, and x2-x4=0. So when looking for a matrix for the map am I just looking to send all vectors in R^5 which satisfy those equations to 0?

    because i think if we have the matrix (2,-1,0,1,-2) it will send all vectors that satisfy the equations to zero??
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  11. #11
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    Re: Finding a linear map with a kernal equal to a subspace

    Again, Rank nullity Thrm.

    If Dim(R) = 1, then Dim(Ker) = 4

    Dim(R) + Dim(Ker) = Dim( \mathbb{R}^5) = 5.
    So Dim(Ker) = 4.
    Which means, it should have four linearly independent vectors.

    Look at  \begin{bmatrix} 1 \\ -1 \\ 1 \\ -1 \\ 1 \end{bmatrix} + -\frac{1}{2} \begin{bmatrix} 0 \\ -2 \\ 2 \\ -2 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} So, we already see that our of the 4 vectors for the kernel, one is linearly dependent, meaning that the  Dim(Ker) \leq 3 . Since Dim(Ker) = 4 for there to be a linear map with such a kernel, it must be that there exists no such linear map.
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  12. #12
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    Re: Finding a linear map with a kernal equal to a subspace

    the rank-nullity theorem is your friend, if you don't know it, by all means stop whatever it is you're doing and go out and buy a copy! now!

    in this case, what this means is:

    5 = 4 + 1.

    so if such an L exists, it has to be ONTO R. the matrix for such an L would have to be a 1x5 matrix (a single row). that is:

    L = [a b c d e]

    note that since (1,0,0,0,1) is in the kernel, L(1,0,0,0,1) = 0, so a+e = 0. so L must be of the form [a b c d -a].

    note that since (0,-2,2,-2,0) is in the kernel, (0,-1,1,-1,0) = (1/2)(0,-2,2,-2,0) is as well (since L is linear).

    therefore -b+c-d = 0, that is: d = c-b. so L is of the form: [a b c c-b -a].

    since L(1,2,2,2,1) = 0, a+2b+2c+2(c-b)-a = 0, that is: 4c = 0, so c = 0, so we know L is of the form [a b 0 -b -a].

    hmm....what is going on here?

    well, the set you started with is linearly dependent. for example: (0,-2,2,-2,0) = 2(1,-1,1,-1,1) - 2(1,0,0,0,1). this tells us that dim(ker(L)) ≤ 3, so we have:

    5 = 3 + dim(im(L)). but since dim(im(L)) ≤ 1, this just isn't possible.

    now, you may be thinking: what's wrong with L = [2 -1 0 1 -2] (this is a choice of a = 2, b = -1)?

    well, it's true that every element of <(1,-1,1,-1,1), (1,2,2,2,1), (0,-2,2,-2,0), (1,0,0,0,1)> is in ker(L). but: have you considered that ker(L) might be BIGGER?

    for example (1,0,0,-2,0) is clearly in ker(L), 2*1 + -1*0 + 0*0 + 1*-2 + -2*0 = 2 + 0 + 0 - 2 + 0 = 0

    is (1,0,0,-2,0) in <(1,-1,1,-1,1), (1,2,2,2,1), (0,-2,2,-2,0), (1,0,0,0,1)>? suppose it were:

    then (1,0,0,-2,0) = a(1,-1,1,-1,1) + b(1,2,2,2,1) + c(0,-2,2,-2,0) + d(1,0,0,0,1) = (a+b+d,-a+2b-2c,a+2b+2c,-a+2b-2c,a+b+d), that is:

    a+b+d = 1
    -a+2b-2c = 0
    a+2b+2c = 0
    -a+2b-2c = -2
    a+b+d = 0

    now ask yourself, how can a+b+d be 1 and 0 at the same time? it can't. so your proposed L doesn't "just" zero out <(1,-1,1,-1,1), (1,2,2,2,1), (0,-2,2,-2,0), (1,0,0,0,1)>, it takes out a LOT more vectors, as well

    (as another example, (0,2,3,0,-1) is also in the kernel of L = [2 -1 0 1 -2] but cannot be written as a linear combination of your spanning vectors).

    it turns out (you may want to verify this), that if U = <(1,-1,1,-1,1), (1,2,2,2,1), (0,-2,2,-2,0), (1,0,0,0,1)>, that dim(U) = 3, while your L has a kernel of dimension 4, meaning it's "bigger" than U.

    you found a map with kernel CONTAINING U, not EQUAL to it.
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    Re: Finding a linear map with a kernal equal to a subspace

    Quote Originally Posted by Deveno View Post
    rank-nullity therorem. think about what the matrix for such a linear map must be. for example, if your basis is {v1,v2,v3,v4}, you must have:

    L(v1) = L(v2) = L(v3) = L(v4) = 0.

    the problem is easier than it looks.
    I don't understand this. L is FROM R^5 to R^4 so a basis that L can be applied to must have 5 vectors, not 4.
    Last edited by HallsofIvy; December 22nd 2012 at 02:09 PM.
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    Re: Finding a linear map with a kernal equal to a subspace

    the vectors vj all live in R5, they are just a basis for the kernel V (an unfortunate choice of letter to symbolize a kernel, i agree).

    yes, the kernel of a map doesn't totally determine it (except for the 0-map) which is why the matrix for L had question marks in it.

    V is in this case a subspace of R5. as a subspace, it has a basis with 4 elements in it (because the spanning (generating) vectors in this case happen to be LI).

    3 of the standard basis vectors happen to be in V. and since the last 2 coordinates of any element of V are always the same (in the standard basis) we must have L(0,0,0,1,0) + L(0,0,0,0,1) = (0,0,0,0).

    (this also follows directly from (0,0,0,1,1) is in ker(L), and the linearity of L).

    since:

    (1,0,0,0,0) = 1(1,1,1,1,1) - 1(0,1,1,1,1)
    (0,1,0,0,0) = 1(0,1,1,1,1) - 1(0,0,1,1,1)
    (0,0,1,0,0) = 1(0,0,1,1,1) - 1(0,0,0,1,1)

    it's clear that L(a,b,c,0,0) must be 0 for any choice of a,b, and c. that means L(e4) completely determines L (ordinarily this wouldn't be so, but we know L is of rank 1).
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    Re: Finding a linear map with a kernal equal to a subspace

    Quote Originally Posted by darren86 View Post
    if V= (a,a+b,a+b+c,a+b+c+d,a+b+c+d), a,b,c,d are all real numbers
    Hmm, if a,b,c,d are all real numbers, isn't V then 1-dimensional?
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