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Homework Help: Finding the matrix for a linear transformation

  1. Aug 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Let V = Span{(1,1,0), (1,2,3)}. Define a linear transformation L: V => R^3 by L(1,1,0) = (1,0,0) and L(1,2,3) = (0,1,0). For any (x,y,z) element of V find L(x,y,z)

    2. Relevant equations

    3. The attempt at a solution
    It seems like there should be some straightforward approach to finding the matrix for the transformation. But I keep getting lost. So the standard thm is :

    Let L be a linear transformation from V to W and let

    S = {v1, ... ,vm} and T = {w1, ... ,wn}

    be bases for V and W respectively. Then the matrix A representing L with respect to the bases S and T has ijth component that is the jth coordinate of the vector L(vi).

    But then L(1,1,0) = (1,0,0) = 1e1 + 0e2 + 0e3 and
    L(1,2,3) = (0,1,0) = 0e1 + 1e2 + 0e3 (where the ei's are the standard bases). But then I just have

    1 0
    0 1
    0 0

    which won't even multiply with a 3 term vector.

    Another approach I tried and I think worked is just to note any element of V will look like a(1,1,0) + b(1,2,3). Then L( a(1,1,0) + b(1,2,3) ) = aL(1,1,0) + bL(1,2,3) = a(1,0,0) + b(0,1,0). So for any x,y,z I just need to find the corresponding a and b. So any x,y,z element of V must have the relationship that a + b = x, a + 2b = y, and 3b = z for some a,b. This last equation implies b = z/3. Next I just need to find a: a + b = x -> a = x - b and the previous result of b = z/3 -> a = x - z/3. So then I want to say that for any x,y,z element of V. L maps it to (x - z/3, z/3, 0). Does this work? Really, I just want to figure out the direct approach.
  2. jcsd
  3. Aug 19, 2012 #2
    The most direct approach would be write down matrix A with indefinite components, then substitute the known transformations, and solve the resultant linear system.
  4. Aug 19, 2012 #3
    What can you tell about a matrix which columns are T(e_1), T(e_2) and T(e_3)?
  5. Aug 19, 2012 #4
    You've gotten off to a great start!

    One question about first-- how can V have an (x,y,z) element, given that it only has dimension 2?

    Let's use your idea about how any element of V will look like a(1,1,0)+b(1,2,3). Now, let's given (1,1,0) the name f, and (1,2,3) the name g. This simplifies writing them out, and lets us say that V is the set of all possible linear combinations of f and g-- that is, V is the set of all possible af+bg.

    But what does af+bg mean in terms of the standard basis for R^3? Well, like you said: x=a+b, y=a+2b, and z=3b (where (a,b) is the coordinate of a point with respect to the basis of V).

    So L(v)=[Matrix that creates that system of equations] [a b]^T = [x y z]^T (where [...]^T means the transpose of that matrix).

    What would you do know to find the matrix to create that system?

    EDIT: One way to think about what that matrix should even look like-- if [a b]^T is 2x1, and [x y z]^T is 3x1, what must A (the matrix you're looking for) be?
    Last edited: Aug 19, 2012
  6. Aug 19, 2012 #5
    Okay, in order of your responses:

    Do you mean setting it up for each of the basis vectors? So letting A be a general 3x3:

    a11 a12 = 1
    a21 a22 = 0
    a31 a32 = 0

    a11 2a12 3a13 = 0
    a21 2a22 3a23 = 1
    a31 2a32 3a33 = 0

    But then I would have 9 unknowns but only 6 equations. Would I just calculate it for some 3rd vector which is a combination of the two basis vectors to get an additional 3 equations? It seems like it would be tedious to solve for all these

    That would be the identity matrix? I'm not sure how that would play in.


    Okay, so I am with you for the most part. In the last paragraph of my post I did the same thing but I don't think I wrote it out very clearly (about concluding that the output of L just maps the scalar multiples of the basis of V).

    Then you say "If we always want z=0, what do you think the third column of your matrix should be? Note also what z must always be in V... you have a basis of only two vectors, so what would the third component always be?"

    It still isn't apparent to me what I can conclude about the third column of the transformation matrix. I understand that the dot product of any vector and the 3rd row of the matrix must equal 0 since the output will always be 0 in the 3rd component. But it seems like z can play a role. ie the fact that the basis for V only has one vector with a z component and it is a 3 means that z must always be a factor of 3.

    Thanks for your hints everybody! If you can give some more that would be great, I still don't quite see any of your hinted approaches. In addition I have a couple of specific questions.

    1) is my conclusion in the last paragraph of my first post correct. Will L(x,y,z) -> (x - z/3, z/3, 0) in general?

    2) Can this problem be approached directly from the theorem I stated in my original post. Can we generate the matrix just by applying the transformation to the basis vectors of V and writing the result in terms of the basis vectors of R^3?

    Thanks again!
  7. Aug 19, 2012 #6
    Oops, you edited while I was writing that out, I will read your new one again and reply. Ignore what I said for now Schaefera
  8. Aug 19, 2012 #7
    So the matrix needs to be a 3x2. From what you said I would want to just take the coefficients and get

    1 1
    1 2
    0 3

    But this seems like this is giving the relationship between x,y,z for that vector to be in the plane formed by (1,1,0) and (1,2,3). Not the result of L(x,y,z). I assume I am misinterpreting?
  9. Aug 19, 2012 #8
    Perhaps I could have been a bit more clear-- you're right about that 3x2 matrix. Now, that is how you can justify getting to your system. You can solve that system for a and b, and what you did it correct. BUT: vectors in V need only two components, so my point was that the three components of R^3 are all kind of wrapped up together into 2 components for V. So your answer is right, but leave out the third component you called 0.

    I would think of L(x,y,z) as kind of doing the transformation backwards. That's why you set up the system, then solve in terms of the variables a and b. just remember that you only need 2 components after L is applied.
  10. Aug 19, 2012 #9
    This is a point I keep being confused about. I still find it weird when a Linear Transformation is declared to map from some space to another (lets say V --> W) when in reality it maps to a subspace of W. If I left off the z component, then I would have though it isn't mapping to R^3, it is mapping to R^2. Don't I need to keep the 0 just so it is in R^3. I guess I have been assuming that a vector of n components is not an element of a space of R^m where m>n. Is that not true?
  11. Aug 19, 2012 #10
    Correct, but that is because the problem is formulated in terms of a 2D-subspace of 3D-vector space, and the map is defined only for the subspace. If you take a vector not in the subspace and act with the map on it, the result is undefined. So you are free to define that as you wish. You could, for example, require that any vector not in the subspace be mapped to zero. Or you could require that the matrix be symmetric. Or you could just leave three independent parameters in the solution.

    This will give a linearly dependent equation - useless. You need an independent condition as remarked above.

    Well, it was you who asked for a direct approach. The equations are actually quite simple, it won't take long.
  12. Aug 19, 2012 #11
    You would be going between 2D and 3D vector spaces. So you have a point (x,y,z) in the R^3 case, but only (a,b) in the V case. You're not leaving off a z component, sorry if I said that confusingly. You just don't need a 3rd component within V because it's only a 2D surface.
  13. Aug 19, 2012 #12
    Okay, I see what you mean. I think I was just thrown off because the question was for any (x,y,z) element of V. I think I just need more time working with non standard bases.
  14. Aug 19, 2012 #13
    It is a 2D plane, but it is still defined in terms of 3D elements. The transformation matrix must be 3x3. What could be done, is map V into a 2D space, so that (1, 1, 0) -> (1, 0) and (1, 2, 3) -> (0, 1), then map that back into 3D space, (1, 0) -> (1, 0, 0) and (0, 1) -> (0, 1, 0). That would be 2x3 and 3x2 matrices, respectively, and the resultant 3x3 matrix would be the product of the two.
  15. Aug 19, 2012 #14
    I think I'm starting to see what you mean but I am still daunted by solving those 6 equations. Is the approach to set up a giant 6 by 10 (one more than 9 with the adjoint) matrix and row reduce it?!
  16. Aug 19, 2012 #15
    But we're talking about different bases. In V, a basis of 2 vectors, you only need 2 components to describe any point. So it should be a 2x3 matrix!
  17. Aug 19, 2012 #16
    Yes. Most of the elements in the matrix will be zeros, so it is not really as hard as it seems.
  18. Aug 19, 2012 #17
    I think Voko is speaking from the standpoint that the question was concerned with the standard basis. For any (x,y,z) element of V.... so even though any such point is a two 2D combination of the basis vectors the transformation is concerned with that in the standard basis. But I obviously don't know much and am still pretty confused about the whole thing.
  19. Aug 19, 2012 #18
    If you really want to convince yourself of it, ask what would the 3rd component in the basis of V even mean? You have no more linearly independent vectors to multiply... what would it do? You see, it only has two components in V!
  20. Aug 19, 2012 #19
    The elements of V are 3D vectors. You cannot just ignore their components because they all happen to be in a 2D subspace. If you want to make use of the fact that they are in 2D subspace, you need to "identify" them with 2D vectors - by mapping them into a space whose elements are 2D, as I suggested.
  21. Aug 19, 2012 #20
    Okay Voko, I am still having trouble on the direct approach. I tried row reducing but I didn't to where I can solve explicitly for things. But I feel like you were implying that it would be a general answer. But I'm not sure how this ties back to actually use it to answer L(x,y,z). What approach would you take? Would you do the direct?
  22. Aug 19, 2012 #21
    But not with respect to the base of V, is my point. They are 3D in that they are in space, but if we take the basis of V, then they really aren't simply because they all lie within the plane spanned by the basis vectors of V. We need 3 components to talk about their position relative to the standard basis of R^3, but only 2 relative to the basis of V.

    After all, what would the 3rd component relative to V even mean-- what basis vector would that multiply? (The answer is none! :rolleyes:)
  23. Aug 19, 2012 #22
    I have already tried the direct approach. It really is a breeze. Denote the matrix elements a, b, c, d, e, f, g, h, i (row after row), you will get six easy equations. Three of the variables will be independent, express the rest in terms of them. Done.
  24. Aug 19, 2012 #23
    It is true that when elements of V are represented in the basis of V, only two coefficients are required. But they are still R^3 entities, so a matrix acting on them will have to be 3x3. Or you have to map them into R^2 first, as I have already mentioned.
  25. Aug 19, 2012 #24
    Sorry to be slow at this. So I then got c,f and i to be the independent variables.

    a = 2 + 3c
    b = -1 -3c
    c = c
    d = 3f
    e = -3f
    f = f
    g = 3i
    h = -3i
    i = i

    But then I reform the matrix and left multiply by a general (x,y,z)? I have lost sight of where this is going
  26. Aug 19, 2012 #25
    I have the same result, except d = 3f - 1, e = 1 - 3f. Check your computations.

    If you don't need a general result, the easiest would then be to let c = f = i = 0.
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