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Homework Help: Find Linear operator [L] and compute

  1. Jul 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Let L be a linear operator such that:
    L[1, 1, 1, 1] = [2, 1, 0, 0]
    L[1, 1, 1, 0] = [0, 2, 1, 0]
    L[1, 1, 0, 1] = [1, 2, 0, 0]
    L[1, 0, 1, 1] = [2, 1, 0, 1]

    a) Find [L]

    b) Compute L[1, 2, 3, 4]

    2. Relevant equations

    3. The attempt at a solution
    I used another post on this forum to work on part b (I think) without actually using [L] itself. I expressed [1, 2, 3, 4] as a linear combination of the other vectors (a*[1, 1, 1, 1], b*[1, 1, 1, 0], c*[1, 1, 0, 1], and d*[1, 0, 1, 1]) but I'm not sure where to go from there. Anyways, wouldn't it be easier to just find [L] like it asks and use that? Thanks in advance.
  2. jcsd
  3. Jul 28, 2011 #2
    Any ideas to get me started at least?
  4. Jul 28, 2011 #3
    OK, if you want to calculate only the value of L at (1,2,3,4) then your idea would be worth a try, however, as you have to find L explicitly, you should use that L can be reagarded via some representation as a 4 x 4 matrix, i.e., 16 entries. By calculating with 16 variables, you would ultimately obtain a solution, but I think, it is to boring. So you could try to use the expressions above as well as the fact that L is a linear operator to calculate L acting on each of the basis vectors (1,0,0,0), (0,1,0,0), (0,0,1,0), (0, 0, 0, 1). But I am not quite sure about that. The system of linear equations may be boring to perform but it will ultimately give you the solution. Sp prepare a huge mug of coffee ;)
  5. Jul 28, 2011 #4


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    Homework Helper

    You suggestion to find L(1,2,3,4) is absolutely correct. Dongo's suggestion of finding L is correct, it will be a 4x4 matrix.
  6. Jul 28, 2011 #5
    Have you checked that the vectors (1,1,1,1), (1,1,1,0), (1,1,0,1), and (1,0,1,1) actually form a basis? We need to know that first.

    Once that is done, then it's quite simple to find the matrix representation of L. In fact, the columns of [L] have a very straitforward formula...

    Here's a hint. Suppose [itex]V = \left(v_1,v_2,v_3,v_4\right)[/itex] is a basis for the vector space. Look at how L operates on a linear combination [itex]c_1 v_1 + c_2 v_2 + c_3 v_3 + c_4 v_4 [/itex] remembering the properites of a linear transformation.
  7. Jul 28, 2011 #6
    Hey thanks for the replies.

    Somehow I found L. Although I'm not 100% clear on how I did it...

    I started with a matrix of the original vectors that L was acting on, times some 4x4 matrix, and set it equal to a matrix with the solutions of L(v1), L(v2), etc... but transposed.

    |1 1 1 1||a b c d| = |2 1 0 0|
    |1 1 1 0||e f g h | = |0 2 1 0|
    |1 1 0 1||i j k i | = |1 2 0 0|
    |1 0 1 1||m n o p| = |2 1 0 1|

    Solving gave me the matrix:
    |-1 3 1 1|
    | 0 0 0 -1|
    | 1 -1 0 0|
    | 2 -1 -1 0|

    Which, after transposing, is [L]. (I checked)

    In my mind I thought I was doing what I read from y'all. But now I can't figure out what I really did. Seems kinda crazy. Thoughts?
  8. Jul 28, 2011 #7
    OK, suppose

    [tex] B = \left( (1,1,1,1),(1,1,1,0),(1,1,0,1),(1,0,1,1) \right)[/tex]

    is, in fact, a basis (we still need to check that!). The matrix [L] that you just computed is allegedly the matrix representation of L with respect to B. Let [itex]v_1[/itex] represent that first basis vector. Its coordinate vector [itex] \left[v_1 \right][/itex] with respect to B is really easy to compute, right? It's just [itex](1,0,0,0)^T[/itex].

    Well we have an easy check then. Namely, [itex][L][v_1][/itex] should equal the coordinate vector of (2,1,0,0) with respect B.

    Does it?
  9. Jul 28, 2011 #8
    Yes it is a basis.

    But I'm not sure I follow. In that case wouldn't [L] just be the columns of L(v1), L(v2), ... L(vn)??

    I think my problem is that I keep thinking of this as matrix multiplication when it's not
  10. Jul 28, 2011 #9

    Jackpot. However, we have to be a bit careful: The columns of [L] are the COORDINATE VECTORS of [itex]L(v_1),\ldots,L(v_n)[/itex] with respect to B. The vectors [itex]L(v_1),\ldots,L(v_n)[/itex] alone are just 4-tuples and it doesn't make much sense to say that they are columns of a matrix...

    There are indeed quite subtle ideas going on here. We need to differentiate between the vector space [itex]V[/itex] of 4-tuples (on which L is a linear operator) and the vector space [itex]F^4[/itex] of column vectors with 4 entries.

    (PhysicsForums doesn't have the LaTeX commands I'm used to for typesetting matrices, so I attached a quick explanation as a .pdf file. See it.)

    What is really happening is that when a basis for V is fixed, it allows you to translate the problem from V to [itex]F^4[/itex] (via the basis and coordinate vectors), perform the computations (matrix multiplications) in [itex]F^4[/itex], then translate the answer back to V (via the basis and coordinate vectors).

    So, the bulk of the computations in solving this problem (at least in part a) are calculating coordinate vectors...

    Attached Files:

  11. Jul 28, 2011 #10


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    Another thing you could do is look for a combination where:
    Like wise for the other basis vectors and compute
    This will give you an easier time of constructing the matrix representation of L.
  12. Jul 29, 2011 #11
    This will indeed be the easiest way to solve this problem.
  13. Jul 29, 2011 #12
    I really appreciate all the help with this one y'all.

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