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Linear Transformations

  1. May 26, 2015 #1
    Assume the mapping T: P2 -> P2 defined by:
    T(a0 + a1t+a2t2) = 3a0 + (5a0 - 2a1)t + (4a1 + a2)t2
    is linear.Find the matrix representation of T relative to the basis B = {1,t,t2}

    My book says to first compute the images of the basis vector. This is the point where I'm stuck at because I'm not sure how the books arrives at the images:
    T(b1) = T(1) = 3+5t
    T(b2) = T(t) = -2t+4t2
    T(b3) = T(t2) = t2

    Where are these results coming from?
    I don't understand where 1 is supposed to go to solve for T(1). I guess its the notation that is throwing me off. Usually when solving for a transformation, it has something such as T(x) = x^2, and you solve the transformation by substituting the value of the input for x. But now my input is 1 for an entire expression (a0 + a1t+a2t2)
     
  2. jcsd
  3. May 26, 2015 #2

    Svein

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    Start with what you know. If you want T(1), look at your definition. In order to get T(1), put a0=1, a1=0 and a2=0. Then the definition says T(1)= 3⋅1 + (5⋅1 - 2⋅0)t + (4⋅0 + 0)t2. In the same way, to find T(t) , put a0=0, a1=1 and a2=0. The rest is left as an exercise...
     
  4. May 26, 2015 #3

    Fredrik

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    I'm not a fan of calling the functions 1, t and t2 (these are notations for numbers, not functions). I would define functions ##e_0, e_1, e_2## by
    \begin{align*}
    &e_0(x)=1\\
    &e_1(x)=x\\
    &e_2(x)=x^2
    \end{align*} for all real numbers x. Then T is defined by ##T(a_0e_0+a_1e_1+a_2e_2)=3a_0e_0+(5a_0-2a_1)e_1+(4a_1+a_2)e_2## for all real numbers ##a_1,a_2,a_3##. Now let's do what Svein did, in my notation:
    $$T(e_0)=T(1e_0+0e_1+0e_2)=3\cdot 1 e_0+(5\cdot 1-2\cdot 0)e_1+(4\cdot 0+0)e_2=3e_0+5e_1.$$
     
  5. May 26, 2015 #4
    How do you know which ax corresponds to the input? For T(1), you set a0 =1 and for T(t) you set a1 = 1. I don't think it could be the order in which it appears in B, because B is just a set of vectors and the order shouldn't matter.
     
  6. May 26, 2015 #5
    So what makes
    e0(x)=1 and not e0(x)=x? If the order of the vectors in the Basis changed, how would I know that e0(x)=1? . Also, why are the others always zero?
     
  7. May 26, 2015 #6

    Fredrik

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    Yes, strictly speaking, it's ambiguous to talk about the components of a vector in a specific basis. We should always be talking about the components of a vector with respect to an ordered basis like the triple ##(b_1,b_2,b_3)## rather than the components of a vector with respect to the basis ##\{b_1,b_2,b_3\}##. Unfortunately people are sloppy with the language. But they're at least being sloppy in a consistent way. When they talk about the components of a vector with respect to ##\{b_1,b_2,b_3\}##, they always mean with respect to ##(b_1,b_2,b_3)##, and never with respect to e.g. ##(b_3,b_1,b_2)##.

    The way I did it is just a convention. You could number the functions differently if you want to.

    Now you're probably thinking "wait a minute, the formula for the number on row i, column j of the matrix depends on the order of the basis vectors, so each choice of how to order them could give me a different matrix". This would be a correct observation. A linear operator and a basis don't uniquely determine a matrix. A linear operator and an ordered basis on the other hand...

    In this problem, it's safe to assume that you should find the matrix of T with respect to the ordered basis ##(b_1,b_2,b_3)## (i.e. my ##(e_0,e_1,e_2)##).

    I'm not sure what others you're referring to.
     
    Last edited: May 26, 2015
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