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Can't understand technique for computing transformation matrix

  1. Mar 9, 2012 #1
    My linear algebra textbook presents a technique for computing the transition matrix from an old basis B' to a new basis B. Apparently if you set up an augmented matrix with B on the left and B' on the right, then put B into RREF, the resulting matrix on the right after those row operations will be the transition matrix from B' to B. The book doesn't explain why this works and I can't figure out why myself. Can anyone help me out? Thanks.
     
  2. jcsd
  3. Mar 9, 2012 #2

    chiro

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    Hey coreluccio.

    The change of basis for normal linear algebra (not tensors) just changes a basis from one linear basis to another. Let's look at this in a bit more detail:

    Consider you have c in basis A and want to convert it to basis B.

    So given c in basis A we want to find c' which is in basis B.

    This means we have to find c' where:

    c' = Tc.

    Now c is in basis A. Lets now consider a universal basis U where we have the following and z is vector in the universal space. This means given z:

    c = Az
    z = A^-1c
    c' = Bz

    So since we want to find c' in terms of c, we just use the following substitution:

    c' = Bz = B(A^-1)c = B x A^-1 x c

    So why does this work?

    The reason this works is that for A to be a basis (as with B) then A must be invertible as with B. That means that A and B have to be square (which they are because they are basis) and they are invertible which gaurantees A^-1 and B^-1.

    By assuming that z is a universal basis (consider identity matrix), we just use these properties to derive the change of basis and from this we get the following change of basis formula.
     
  4. Mar 10, 2012 #3
    I still don't really understand.

    For example, say I have a vector w where its coordinates relative to the standard basis in 2-space are (1,1)

    Now say I have a new basis that I would like to map my coordinates of w to. This basis consists of the vectors u1 = (2,-4), u2 = (3,8).

    If I set up an augmented matrix of the form (left side is new basis, right is old)

    [2 3 |1 0]
    [-4 8 |0 1]

    Then row reduce this matrix so the left side is the identity matrix, the ride side will be the transformation matrix. I'm lost as to why this works. I think I am missing something really simple but I don't know what it is.
     
  5. Mar 10, 2012 #4

    chiro

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    Hey coreluccio.

    Again its a lot easier to understand my example above then try to just apply a formula. I'll rephrase it in a manner that is a lot clearer.

    Like before z is a vector in a universal basis (identity matrix i.e. standard x,y,z axis), c is the transformed vector in the basis A and c' is the vector in basis B.

    Now let me explain what is going on here: this is the key to understanding why I gave you the answer I did above.

    First lets look at the equation c = Az.

    What this is doing is its converting the vector z in the normal i,j,k (or x,y,z whatever you are familiar with) basis to the vector c that is in the basis A.

    What is A? It's just a matrix that represents the basis information for A. The columns are the basis vectors just like the identity matrix has the basis vectors (1,0,0) (0,1,0) and (0,0,1). In A, your basis vectors are just the columns of A and you can think about them in the exact same way as you think of the normal basis in terms of i,j,k vectors.

    So basically c=Az transfers z from the identity basis to the A basis. c'=Bz does the exact same thing except it transfers z from identity basis to the B basis.

    Now all you want to do is go from basis A to B and mathematically all you are doing is using the substitution z = A^-1c which gives you c' = B x A^-1 x z.

    Once you have this then you can use all your matrix theory to get the values of the transformed vectors in whatever space you want.

    Now lets answer your question about row-reduction.

    Row-reduction in the form you are talking about is used to find the inverse. So if we have say Ax = b and we want to find A^-1 then what we do is setup your equation like the following:

    [-3 4 | 1 0]
    [ 1 0 | 0 1]

    and then get the LHS matrix to be the identity. This is basically the same as finding the inverse of the LHS. If you want to understand why think about the following:

    Lets say we find the inverse of A to be A^-1 and it exists. Now lets say we have our augmented system [A | I] like above. What will happen when we multiply both sides by A^-1 if it exists? We will get [A | I] x A^-1 = [I | A^-1]. This is why we do the row reduction: all we are doing is looking for the inverse!

    This is just the process of finding the inverse which means that again if we have Ax = b and we have A and b then we want to find x and to do that we need to find the inverse.
     
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