Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Matrices - An apparent contradiction?

  1. Apr 27, 2007 #1
    1. The problem statement, all variables and given/known data
    Consider T:V->V
    where the basis vectors in V are {v1, v2}

    T(v1) = a(v1) + b(v2)
    T(v2) = c(v1) + d(v2)

    With T = the square matrix
    a c
    b d

    Now let v1=(1,1) v2=(1,-1)

    So T= the squre matrix
    1 2
    1 1

    But T(1,1)=(3,2) which does not equal (2,0)
    T(1,-1)=(-1,0) which does not equal (3,1)

    The theory does not fit this simple example! What has gone wrong?
    Last edited: Apr 27, 2007
  2. jcsd
  3. Apr 27, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Up 'till now you've made the following assumptions: V is 2 dimensional and we choose the two linearly independent vectors v1 and v2 as a basis. There exists an isomorphism (linear bijection w/ a linear inverse) between V and R² induced by the fact that {v1, v2} is a basis for V and so, since every vector in V can be written as v = k1*v1 + k2*v2 for real numbers k1, k2. So you made the association v<-->(k1, k2). This implies that v1<-->(1,0) and v2<-->(0,1) and so for the linear transformation T who has matrix representation

    a c
    b d,

    we have T(v1) = a*v1 + b*v2 and T(v2)=c*v1 + d*v2 because when working in R² through the bijection, we calculate that T((1,0))=(a,b), T((0,1))=(c,d).

    Here what you've done is you've said ok let's see a concrete example and set V = R², v1=(1,1), v2=(1,-1) and T=

    1 2
    1 1.

    But your mistake was in expecting T(v1) = a*v1 + b*v2 and T(v2)=c*v1 + d*v2 to hold. Remember that these formulas were a result of the fact that we had made a bijection from the abstract vector space V to R² and had associated v1 with (1,0) and v2 with (0,1). Now we're already in R², and though you could still make the association (1,1)<-->(1,0) and (1,-1)<-->(0,1), that would just amount to changing the basis.
  4. Apr 27, 2007 #3
    I'm not sure if I understand you well but are saying the original formula, T(v1) = a(v1) + b(v2), T(v2) = c(v1) + d(v2) only applies to the case where v1 and v2 are the standard basis in whatever space V is?

    Is it in the definition? I was thinking of doing a change of basis hence the different T matrix the second time. But it didn't work if following the formula as I have shown.
  5. Apr 27, 2007 #4


    User Avatar
    Homework Helper
    Gold Member

    What quasar987 is saying is this. If your basis is [tex] B= \{(1,1),(1,-1) \}[/tex], then any [tex] v \in V , v=(v1,v2)[/tex] (where (v1,v2) is in the standard basis (1,0),(0,1)), can be written as [tex] (v1,v2) = a_1 (1,1) + a_2 (1,-1) [/tex].

    Solving for [tex]a_1[/tex] and [tex]a_2[/tex], you get

    [tex]a_2 = \frac{v_1-v_2}{2}[/tex]

    This means that any vector [tex] (v_1,v_2)[/tex] in the standard basis [tex]\{(0,1),(1,0) \}[/tex], will be [tex]((v_1+v_2)/2,(v_1-v_2/2))[/tex] with respect to the basis [tex] B= \{(1,1),(1,-1) \}[/tex].

    Note that your transformation matrix is relative to the basis [tex] B= \{(1,1),(1,-1) \}[/tex]. Therefore, for any vector [tex] v \in V[/tex], [tex] [T]_B [v]_B = [T(v)]_B[/tex].

    As quasar pointed out, your mistake was using v in the standard basis, even though your transformation matrix is relative to the basis B.
    Last edited: Apr 28, 2007
  6. Apr 28, 2007 #5


    User Avatar
    Science Advisor

    I'm going to use (a, b) to mean a vector written in the "usual basis", (1, 0), and (0, 1), and <a, b> to mean a vector written in your new basis, (1, 1) and (1, -1). (And notice that those are written in the "usual" basis!)

    In your new basis, (1, 1) and (1, -1) become <1, 0> and <0, 1> precisely because the are you basis vectors! (1, 1)= 1(1,1)+ 0(1,1)= <1, 0> and (1, -1)= 0(1, 1)+ 1(1, -1)= <0, 1>. Whatever basis you use, your basis vectors are always <1, 0> and <0, 1> when written in that basis! That's the whole point of the method for finding the matrix as written in that basis!
    Multiplying your matrix times <1, 0> and < 0, 1> will give <1, 1>= 1(1, 1)+ 1(1, -1)= (2, 0) and <2, 1>= 2(1,1)+ 1(1, -1)= (3, 1) just as it should.
  7. Apr 29, 2007 #6
    I kind of see but the map is V->V and there is only one set of basis of vectors in V which I have set as v1=(1,1) v2=(1,-1)

    So I like to think of transforming T(v1) to a vector in V written as a linear combo of v1 and v2.

    But you are suggesting T(<1,0>)=<1,1> and T(<0,1>)=<2,1>

    <1,0> is not a basis vector in V. So why transform it?
  8. Apr 29, 2007 #7
    The problem you're having is that you're confusing the basis vectors with their coordinate vectors. The transformation itself functions on vectors written in the standard basis. But the matrix functions on coordinate vectors in terms of whatever basis you used to construct it.

    For example, when you take the matrix you got times (a,b), what you're actually doing is performing the transformation of a*(1,1) + b*(1,-1), which is obviously going to be different than the transformation of a*(1,0) + b*(0,1). Not only that, when you take this transformation, you're going to get an answer in terms of the second, "destination" basis (which happens to be the same as the first in this case) you used to construct the matrix.

    Now, to adopt the notation that was used earlier (i.e., (a,b) indicates a coordinate vector in terms of the standard basis and <a,b> indicates a coordinate vector in terms of the basis you gave) what you're doing in terms of transformations when you perform the multiplication by the matrix is T(<1,1>) = T(1*(1,1) + 1*(1,-1)) = T(1,1) + T(1,-1) = (2,0) + (3,1) = (5, 1) = 3*(1,1) + 2*(1,-1) = <3,2>.

    I hope that made sense.
  9. Apr 29, 2007 #8
    I kind of follow your reasoning but don't see how you done the actual transformation or arithmetic. T(1,1)=(3,2) and T(1,-1)=(-1,0). So...
    T(<1,1>) = T(1*(1,1) + 1*(1,-1)) = T(1,1) + T(1,-1) = (3,2) + (-1,0) = (2, 2) = 2*(1,1) + 0*(1,-1) = <2,0>.

    T(<1,-1>)=T(1*(1,1) - 1*(1,-1)) = T(1,1) - T(1,-1) = (3,2) - (-1,0) = (4,2) = 3*(1,1) + 1*(1,-1) = <3,1>.

    Exactly as I wanted. The point is they should be in the new basis written in <,> so this notation has proved to be very useful. Is the important thing to always write the vectors in the new basis wrt to the standard basis. Transform it in that form then convert back wrt coordinate vectors in the new basis.
    Last edited: Apr 29, 2007
  10. Apr 30, 2007 #9
    Or I could do it all in one go by moving to the standard basis and multiply by a transformation matrix PTP^-1 where P is the change of coordinates matrix with columns as the basis vectors. In my case it turned out to be A=
    5/2 -1/2
    1/2 -1/2

    So I could convert <1,0> to (1,1) then multiply by A and convert back to the new basis by multiplying the result by P^-1
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook