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Change of basis

  1. Mar 4, 2008 #1
    find the change of basis matrix PC<->B from the given ordered basis B to the given ordered basis C of the vector space V:

    V=R^2; B={(-5,-3),(4,28)}; C={(6,2),(1,-1)}

    3. The attempt at a solution

    I'm having a hard time grasping the concept of changing basis. Can someone please explain it to me? my books explanation seems rather abstract and hard to follow.
  2. jcsd
  3. Mar 4, 2008 #2


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    Can you write an arbitrary vector [itex]\vec{v}=<v_x,v_y>[/itex] in terms of either basis?
  4. Mar 4, 2008 #3


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    The change of basis matrix is a matrix that when multiplied by a column B-coordinate vector gives the same vector represented as a column C coordinate vector.

    (a b)(-5) = (6)
    (c d)(-3) (2)

    (a b)(4 ) = (1)
    (c d)(28) (-1)

    Do the multiplication and solve the relevant systems to find a,b,c,d.
  5. Mar 4, 2008 #4


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    First of all, to "change bases", you have to have two bases! Let's use P2, the set of quadratic polynomials, ax2+ bx+ c, as an example (we can add two polynomials and multiply a polynomial by a real number so that is a vector space over the real numbers). The "standard" basis is x2, x, and 1. As long as we understand we are using those "vectors" as basis, in that order, we can use <a, b, c> to represent ax2+ bx+ c: the "x2, x, and 1 are "understood".

    But (x- 1)2= x2- 2x+ 1, x- 1, and 1 is also a basis. Since the "standard" basis for P2 has 3 vectors, it has dimension 3. This new set also contains 3 "vectors" so as long as they are independent . To show they are independent, look at p(x-1)2+ q(x-1)+ r= 0 (the 0 function: 0 for all x). Since that is 0 for all x, taking x= 1gives r= 0. If that is 0 for all x, it is a constant and so its derivative, 2p(x-1)+ q= 0 for all x. Again taking x= 1, we get q= 0. That is, the original equation was really p(x-1)2= 0 and, taking x= 2 now, p= 0. Since p= q= r= 0, those "vectors" are independent and so a basis.

    If I wanted to write ax2+ bx+ c in that new basis, I would have to find p, q, r so that p(x- 1)2+ q(x-1)+ r= ax2+ bx+ c for all x. Do pretty much what we did before: taking x= 1, r= a+ b+ c. Taking the derivative of each side,
    2p(x-1)+ q= 2ax+ b, again for all x. Taking x= 1 again, q= 2a+ b. Taking the second derivative of both sides, 2p= 2a so p= a. (Multiplying p(x-1)2 and q(x-1) out, collecting coefficients of like and setting corresponding coefficients on both sides equal would give the same thing: p= a, q= 2a+ b, and r= a+ b+ c.) That is, the polynomial represented by <a, b, c> in the first basis (that is, ax2+ bx+ c) is represented by (a, 2a+ b, a+ b+ c) (which is a(x- 1)2+ (2a+ b)(x-1)+ (a+b+c)) in the new basis. If you multiply those out you should see that you get the same thing.

    The "change of basis" matrix must convert the <a, b, c> of a vector written in one basis into the components for the same vector as written in a new basis. In the example above, multiplying the basis by <a, b, c> must give <a, 2a+ b, a+ b+ c> . In particular, it must change <1, 0, 0> to <1, 2, 1> (a= 1, b= 0, c= 0 here), change <0, 1, 0> to <0, 1, 1> (a= 0, b= 1, c= 0), and change <0, 0, 1> to <0, 0, 1> (a= 0, b= 0, c= 1). The reason I like to use <1, 0, 0>, <0, 1, 0>, and <0, 0, 1> is because it is easy to see that a matrix multiplied by <1, 0, 0> just gives the numbers in the first column, multiplied by <0, 1, 0> gives the numbers in the second column, multiplied by <0, 0, 1> gives the numbers in the third column. In other words, the "change of basis" matrix must have <1, 2, 1>, <0, 1, 0> and <0, 0, 1> as the first, second and third columns: it is
    [tex]\left(\begin{array}{ccc} 1 & 2 & 1 \\0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)[/tex]
    Last edited by a moderator: Mar 4, 2008
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