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Coordinates of vector to the base

  1. Jan 3, 2005 #1
    Hi all, I don't know how to solve this:

    The vector u has the coordinates [itex]( u_1, u_2, u_3 ) [/itex] to the base [itex] a = { ( 1,0,0), (1,1,0), (1,1,1) }[/itex] and it has the coordinates [itex]( u^{'}_{1}, u^{'}_{2}, u^{'}_{3} ) [/itex] to the base b. We also know:

    u_1 = 2u^{'}_1 - u^{'}_2 + u^{'}_3

    u_2 = u^{'}_1 + 3u^{'}_2 - u^{'}_3

    u_3 = u^{'}_1 + u^{'}_2 - 4u^{'}_3

    Find out the base b !

    I have no idea how to solve it...I tried it but I couldn't find a way without adding unknown variables to the equations.

    Could someone help please?
  2. jcsd
  3. Jan 3, 2005 #2


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    In base, A, u has the coordinates [itex]( u_1, u_2, u_3 )[/itex]. This means, explicitely, that u is

    [tex]\vec{u}=u_1(1,0,0)+u_2(1,1,0)+u_3(1,1,1) \ \ (*)[/tex]

    In base, B, u has the coordinates [itex]( u'_1, u'_2, u'_3 )[/itex]. This means, explicitely, that u is

    [tex]\vec{u}=u'_1 \vec{b_1}+u'_2\vec{b_2}+u'_3\vec{b_3} \ \ (**)[/tex]

    where the [itex]\vec{b_i}[/itex] are the vectors of base B, which is what we want to find.

    Using the relations between the coordinates of base A and B, we can rewrite (*) has




    So according to (**),

    [tex]\vec{b_1}=(4,2,1), \vec{b_2}=(...), \vec{b_3}=(...)[/tex]
  4. Jan 5, 2005 #3


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    the basis vectors b1, b2, and b3, are exactly the vectors that have coordinates
    (u1', u2',u3') = (1,0,0), (0,1,0) and (0,0,1) in the basis B.

    So to find the first vector say b1, all you have to do is plug in (u1', u2',u3') = (1,0,0), into those equations you are given, find u1, u2, and u3, and multiply those by the basis vectors in A.

    I.e. you get u1 = 2, u2 = 1, u3 = 1, which means that b1 = 2(1,0,0) + 1(1,1,0) + 1(1,1,1) = (4,2,1), etc,,,, i.e. do the same thing to find b2, then b3.

    computational questions like this are not what i love about linear algebra. linear algebra is a beautiful subject. this kind of thing has about as much to do with that subject as multiplying big numbers has to do with number theory. of course it is useful in a pinch to be able to calculate.

    Lets call the original copy of R^n simply V. Then giving a basis is simply giving an isomorphism to another copy of R^n, that assigns differfent coordinates to each vector. For example, the basis A assigns the coordinates (1,0,0), (0,1,0), and (0,0,1) to the three vectors in A. I.e. it is given by a map V-->R^n that takes (1,0,0) to (1,0,0), and takes (1,1,0) to (0,1,0), and takes (1,1,1) to (0,0,1). It is not clear what the matrix of this map is.

    To know the matrix of a map we need to know what it does to the standard basis. Thus writing down the matrix made from the basis A itself, i.e. with (1,0,0) in the first column, (1,1,0) in the second, and (1,1,1) in the third defines instead the map, R^n-->V, inverse to the coordinate system defined by A.

    Similarly, the matrix whose columns contain the desired basis for B would define the inverse map R^n-->V of the coordinate system defined by B. If we can find this matrix its columns will solve our probloem.

    Now what have we been given in the equations expressing u in terms of u'?

    Well those equations give the A coordinates as a function of the B coordinates. So this is the map R^n-->V-->R^n composed of first the inverse of the B coordinate map, followed by the A coordinate system, i.e. by the B basis matrix followed by the inverse of the A - basis matrix.

    i.e. the matrix of coefficients of the given equations is the matrix A^(-1)B where A is the matrix whose columns are the A basis, and where B is the matrix whose columns are the B basis. Since we have the A basis we have the matrix A, and we are also given the matrix A^(-1)B as the matrix of coefficients. Thus all we need to do to get the amtrix B is multiply

    A by A^(-1)B. I.e. put the vectors of the basis A into the columns of a matrix called A, and multiply it by the matrix of coefficients of those equations expressing the u's in terms of the u's.

    I realize this is incomprehensible, but it is clear if one can draw pictures with arrows in them showing the direction of the maps.
    Last edited: Jan 5, 2005
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