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Linear Maps/Coordinates/Basis

  1. Dec 7, 2007 #1
    Hi I have a final comming up soon so I was looking at an older final and saw this question

    the prof did it already but I dont understand how...here it is

    You are given that L : R2 −> R2 is a linear map for which B =
    {(1, 1), (1, 2)} is a basis of R2 consisting of eigenvectors of L with corresponding
    eigenvalues 1/2 and 1 respectively – i.e. L(1, 1) = 1/2(1, 1) and
    L(1, 2) = (1, 2).

    c) Determine the coordinates of the vector (3, 5) with respect to the basis
    B and determine the value of L^n (3, 5) as n tends to infinity

    Ok to find coordinates of that vector 3,5 its easy enough

    [1 1 | 3 ]
    [1 2 | 5 ]

    and row reduce it to get

    [1 0 | 1 ]
    [0 1 | 2 ]

    so (3,5)B = (1,2)

    to determine L(3,5) I got 1/2(1,1) + 2(1,2)

    now the prof said to determine L^n (3,5) that equals (1/2)^n(1,1) + 2(1,2)

    why didnt he put the 2^n why do you put the 1/2^n only and NOT the 2 to the power of n??

    I dont get how to get L^n (3,5)
  2. jcsd
  3. Dec 8, 2007 #2


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    You've learned diagonalization of square matrices right? So once you see that that there are 2 eigenvalues associated with that 2x2 matrix you know that it is diagonalizable. The 2x2 matrix concerned is the standard matrix associated with linear mapping L. Let's call it A. Once you've worked out the invertible matrix P associated with diagonalizing A, then all you need do is to find the diagonal matrix D, then multiply raise its diagonal entries to the power of n. You notice that one of the diagonal entries is 1/2 and the other one is 1. 1^n = 1 for all n, but for (1/2)^n and n->infinity, you'll get the limit as 0. Once you get that, multiply calculate P(D^n)P^-1 and then multiply this matrix on the right by (3,5), and that's your answer. Should be (2,4), as given by your prof.

    I don't get your professor's method above. How is the distributive property valid for iterated linear mappings?
  4. Dec 8, 2007 #3


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    Staff Emeritus
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    If x= au+ bv, where a, b are scalars and u, v are vectors, then for linear transformation L, Lx= a Lu+ b Lv. In particular, if Lu= (1/2)u and Lv= v, then Lx= a(1/2)u+ bv, L2x= a(1/2) Lu+ b Lv= a(1/2)2u+ b v, L3x= a(1/2)3+ b Lv, etc.

    To answer your question, "why didnt he put the 2^n why do you put the 1/2^n only and NOT the 2 to the power of n??", what you really should be asking is "why is the 1/2 to the nth power", not "why is the 2 not to the nth power"!

    The "1/2" is raised to a power because it is an eigenvalue of L- each time you apply L you get another "1/2". The "2" is NOT an eigenvalue of L, it is the coefficient of (1, 2) in the linear combination for (3, 5)- it just goes along for the ride: Ln(au+bv)= aLnu+ bLnv. Then eigenvalue corresponding to (1, 2) is 1. If you insist upon "consistencey", you could write Ln(3,5)= Ln(1(1,1)+ 2(1,2))= (1)(1/2)n(1, 1)+ (2)(1n)(1, 2). Of course, 1n= 1!

    Your professor told you exactly how:
    Even if you were not sure why that was true you ought to be able to use it!
    Ln(3, 5)= (1/2)^n(1, 1)+ 2(1, 2)= ((1/2)n+ 2, (1/2)n+ 4).

    As defnnnder said, since the limit, as n goes to infinity, of (1/2)n is 0, the limit of Ln, as n goes to infinity, is (2, 4).

    ?? Since the "distributive property" is true for a linear transformation (by definition of "linear transformation") it is trivially true for iterated linear mappings:

    L2(au+ bv)= L(L(au+bv))= L(aLu+ bLv)= a L(Lu)+ b L(Lv)= aL2u+ bL2v. The general case could be proved by induction.
  5. Dec 8, 2007 #4
    so (3,5)B = (1,2)

    the b coordinate of 3,5 is (1,2)

    but to calculate L(3,5)

    don't I need to know

    1) the matrix representation of linear map with respect to the basis

    2) then multiply that by the B coordinates like the following


    [0......1][2] =

    [ 1/2]
    [ 2 ]

    therefore L(3,5) = 1/2*(1,1) + 2*(1,2)

    so if the transformation is applied again the EIGENVALUE will be applied to each?

    L^2(3,5) = (1/2 (1/2*(1,1)) + (1(2*(1,2))

    Last edited: Dec 8, 2007
  6. Dec 8, 2007 #5


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    No, you don't need to know any of those things. You only need to know that (3,5)= (1,1)+ 2(1,2) and that L(1,1)= 1/2, L(1,2)= 1. L(3,5)= L(1,1)+ 2L(1,2)= 1/2(1,1)+ 2(1)(1, 2)= (1/2, 1/2)+ (2, 4)= (5/2, 9/2). And Ln(3,5)= Ln(1,1)+ 2Ln(1, 2)= (1/2)n(1,1)+ 2(1, 2).

    (It's not a bad exercise, though!)

    Well, I wouldn]t say "applied" to each- it is multiplied by each. You know that L(1,1)= 1/2(1,1) so L2(1,1)= L(L(1,1))= L((1/2)(1,1)= (1/2)L(1,1) (by the L(av)= aL(v) property of linear transformations) = (1/2)(1/2)(1, 1)= (1/4)(1,1).
  7. Dec 8, 2007 #6
    alright pretty cool

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