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Linear algebra change of bases

  1. Aug 7, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm working on the problems attached.

    2. Relevant equations



    3. The attempt at a solution

    I think I understand how to do problem 4.3 but I'm lost on 4.1 and 4.2, and I'm pretty sure i need to figure those out before I can do 4.3.

    Can someone please try to guide me through 4.1 and 4.2? Thank you
     

    Attached Files:

  2. jcsd
  3. Aug 7, 2010 #2
    For 4.1:

    So your rotation matrix doesn't preserve the length of vectors. What does it do instead? If you don't know, run some vectors through your matrix (that is, multiply the matrix by the vectors). Can you think of a way to rectify this so you have a matrix that does preserve the length of vectors? (Hint: rotate some vectors yourself and see what their coordinates change to)

    For 4.2:

    So now consider the matrix from 4.1. If you had a vector in R2 and you wanted to know what that vector would be in the new basis, what would you do? How could you use the matrix in 4.1 to find it? That would be your change of coordinate basis. Then prove those things about it.
     
  4. Aug 7, 2010 #3
    I'm going to focus on 4.1 first. The rotation matrix length goes from sqrt 2 to 1. I'm not sure how to make it so that the length is preserved.
     
  5. Aug 7, 2010 #4
    Try multiplying the rotation matrix by some positive constant and see what happens.
     
  6. Aug 7, 2010 #5
    wait, which matrix is the rotation matrix? I probably sound really stupid but I'm completely lost
     
  7. Aug 7, 2010 #6
    OK, I kind of breezed past some things, let me be a little more step-by-step.

    We have this "standard basis" in [itex]\mathbb{R}^2[/itex], where the elements are {e1,e2}, and ei is the vector with 1 in the ith position and zero everywhere else.

    But this isn't the only basis for [itex]\mathbb{R}^2[/itex]. In class apparently you discussed a different basis (call it B), made up of the vectors { [1 1]T, [-1 1]T }.

    Now a vector is a vector, but since these are both bases, we know we can write any vector in [itex]\mathbb{R}^2[/itex] as a combination of the basis vectors.

    So suppose we take the vector v = [2 0]T (expressed in the standard basis). This is the same as saying, for example:

    [tex] \vec{v} = 2e_1 + 0e_2 [/tex]

    But we can also express v in the B basis (as [1 -1]T) :

    [tex] \vec{v} = 1b_1 - 1b_2 [/tex]

    These are equivalent, because:

    [tex] b_1 = e_1 + e_2, b_2 = -e_1 + e_2 [/tex]

    So now we see the following: If we have a vector of length 2 on the x-axis of our normal coordinate plane, expressing it in the B basis rotates it 45 degrees (or [itex] \frac{\pi}{4} [/itex] radians) clockwise from (2,0) to (1,-1). This will be the case for any vector we move from the standard basis to the B basis.

    4.1 is concerned with a little problem with these two bases. Often we would like to have it so that if a vector has a certain length in one basis, it has the same length in another basis. So above, our vector has length 2 in the standard basis, but length [itex] \sqrt{2} [/itex] in the basis B.

    So you are supposed to find a different basis (also confusingly called B) made up of two vectors such that we preserve the rotational aspect, but now if a vector has length M in the standard basis, it should also have length M in the new B basis. We can get to matrices in a minute.

    If you're not sure what to do, convert a few vectors from the standard basis to the old B basis and look at their lengths in each basis. A pattern should become clear.
     
  8. Aug 7, 2010 #7
    I tried it with two different vectors for example:
    v = [4 2]T
    In the standard basis I got
    [tex] \vec{v} = 4e_1 + 2e_2 [/tex]
    in the B basis:
    [tex] \vec{v} = 3b_1 - 1b_2 [/tex]

    So the lenfth for the standard basis is sqrt(20), for B it is sqrt(10)
    The ones I tried the number inside the sqrt is half the number of what it is in the standard basis. Is this the type of relationship I'm looking for? or did I mess up somewhere?
     
  9. Aug 7, 2010 #8
    That's right. The length in the standard basis is [itex] \sqrt{2} [/itex] times the length in B.

    So how could you modify the vectors in B so that the ratio is 1:1? (Hint: try scaling them by constants to see what happens.)
     
  10. Aug 7, 2010 #9
    If i take the basis B and multiply each component by 1/sqrt(2) I think that makes the lengths come out the same
     
  11. Aug 7, 2010 #10
    That's right.
     
  12. Aug 7, 2010 #11
    So, for 4.1 the B would be changed to what i said above. That was more simple than I expected it to be. But it makes sense. Thank you

    For 4.2 the matrix P_B would be {[1/sqrt(2) 1/sqrt(2)]^T, [-1/sqrt(2) 1/sqrt(2)]^T} correct? So would I just make that into one matrix and perform the operations on it?
     
  13. Aug 7, 2010 #12
    Yes, that's right. I'll just ramble a second. Now that we know how to change bases by hand, we can look at a change-of-coordinates matrix. Suppose we have a vector expressed in B, [x y]T.

    Now we want to write that vector in E, the standard basis. So we know that that is going to be:

    [tex]
    \left[\begin{array}{cc}x\\y\end{array}\right]_B = \left[\begin{array}{cc}\frac{x}{\sqrt{2}} - \frac{y}{\sqrt{2}}\\\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} \end{array}\right]_E
    [/tex]

    We can write down a matrix for this.

    [tex]
    \left[\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}}\\\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right] \left[\begin{array}{cc}x\\y\end{array}\right]_B = \left[\begin{array}{cc}\frac{x}{\sqrt{2}} - \frac{y}{\sqrt{2}}\\\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} \end{array}\right]_E
    [/tex]

    The matrix on the left is the change-of-coordinates matrix from B to E. You can see that its column vectors are just the vectors in the basis of B. This should make some sense, since multiplying a matrix and a vector is simply taking the ith element of the vector times the ith column of the matrix and adding all of them together for all the i.

    Now the notation PB is a little ambiguous. It also could mean that they want the change-of-coordinates basis from E to B (ie, if you had a vector in E and wanted to convert it to B). You could also do this by hand. Or you could use the handy shortcut that the inverse of the matrix from B to E is the matrix from E to B.
     
  14. Aug 7, 2010 #13
    I tried it by hand and got the property to work out to the I matrix both ways

    Also, for 4.3 I got x_1 = [0 1]^T and x_2 = [0 3]^T
    If you get a chance could you please check if those are correct? Thank you
     
  15. Aug 7, 2010 #14
    OK, the first one can't be right at first glance, because the magnitude of x1 in B is 2, but the magnitude of [0 1] is 1.

    For the second one, they gave you the representation in the E basis (both elements are positive) and asked for the B basis. Now your vector [0 3] says that our vector is 3b2. But the first component of b2 is negative. So 3b2 also has a negative first component.
     
  16. Aug 7, 2010 #15
    what I did was I took [sqrt(2) sqrt(2)]^T and did
    x1 = sqrt(2)*b1 + sqrt(2) * b2

    is this not the right way to do it?
     
  17. Aug 7, 2010 #16
    [tex] \sqrt{2} b_1 = \sqrt{2} \left(\frac{1}{\sqrt{2}} e_1 + \frac{1}{\sqrt{2}} e_2 \right)= e_1 + e_2 [/tex]
    [tex] \sqrt{2} b_2 = \sqrt{2} \left(\frac{-1}{\sqrt{2}} e_1 + \frac{1}{\sqrt{2}} e_2 \right) = -e_1 + e_2 [/tex]
    [tex] e_1 + e_2 - e_1 + e_2 = 2 e_2 [/tex]
     
  18. Aug 7, 2010 #17
    Oh ok, so x1 would be [0,2]^T
    but i'm still confused about x2. I did it that way and still got the same answer. I don't really understand what you said about the 2nd one in post 14
     
  19. Aug 7, 2010 #18
    For x2, they are giving you the vector in the standard basis and asking for its representation in the B basis. So you have to do the transformation backwards.
     
  20. Aug 7, 2010 #19
    Oh! I don't know how I missed that, I read the problem so many times. Sorry. So the vector would actually be [3 0]^T
     
  21. Aug 7, 2010 #20
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