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I have a question about linear transformations... watch

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    Define T: P_2 --> R^2 by T(p) = \begin{bmatrix} p(0) \\p(1) \end{bmatrix}. For instance, if p(t)=3+5t+7t^2, then T(p) = \begin{bmatrix} 3 \\15 \end{bmatrix}

    Find a polynomial p in P_2 that spans the kernel of T, and describe the range of T.

    The answer is:

    Any quadratic polynomial that vanishes at 0 and 1 must be a multiple of p(t) = t(t-1). The range of T is R^2...I know how to find a polynomial that spans the kernel, but I don't know how to find the range...I don't know why its R^2

    Thank you in advance.
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    take any element, (x,y) in R^2, then (x,y) is in the range of T if we can find p s.t. p(0) = x, p(1) = y. Note that P(t) = yt + x(1-t) is such a polynomial for any (x,y) (assuming that P_2 is any polynomial with degree 2 or less, if it's degree of exactly 2, add an t^2 term and then fiddle the other terms to make it work)

    Hence, every (x,y) is in the range, so R^2 is a subset of the range. Its also obvious that the range is a subset of R^2 and hence they are equal.
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    (Original post by JoMo1)
    take any element, (x,y) in R^2, then (x,y) is in the range of T if we can find p s.t. p(0) = x, p(1) = y. Note that P(t) = yt + x(1-t) is such a polynomial for any (x,y) (assuming that P_2 is any polynomial with degree 2 or less, if it's degree of exactly 2, add an t^2 term and then fiddle the other terms to make it work)

    Hence, every (x,y) is in the range, so R^2 is a subset of the range. Its also obvious that the range is a subset of R^2 and hence they are equal.
    I'm sorry, but I'm not sure if I understood this...I asked my professor a similar question:

    Define a linear transformation T: P_2 --> R^2 by
    T(p) = \begin{bmatrix} p(0) \\p(0) \end{bmatrix}. Find polynomials p_1 and p_2 in P_2 that span the kernel of T, and describe the range of T.

    We can say p_1 = x and p_2 = x^2 ...my professor said...if we look at the polynomial ax^2 + bx + c, then at t=0, we will have
    T(p) = \begin{bmatrix} c \\c \end{bmatrix}. So the span is \begin{bmatrix} 1 \\1 \end{bmatrix}. Is it ok if you try to explain it in a similar way so that it is easier for me? (if possible) By the way, I did not understand the professor's explanation very well (I thought I did, but I got confused when I looked at the question again)...
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    Well generally explaining it to you in a way you've already not understood makes very little sense, so instead here's a more general explanation:

    The range of a function is all the values it can generate. In the example your professor used, it can generate points in R^2 of the form (c,c), which are spanned (if looked at as a vector space) by (1,1). This means that anything in the range of T can be created by a linear combination of (1,1)s, i.e. c*(1,1).

    In the other example you've figured out or been told that the range is R^2. The sensible way to prove that is to show that the range is a subset of R^2 and vice versa. So, first we take anything in R^2 and prove it's in the range of T. To do this we need to show there's a p s.t. T(p) generates each point in R^2(which is what I showed in my first post). The range also has to be a subset of R^2 as T maps into R^2 by definition. Hence we have equality.
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    (Original post by JoMo1)
    Well generally explaining it to you in a way you've already not understood makes very little sense, so instead here's a more general explanation:

    The range of a function is all the values it can generate. In the example your professor used, it can generate points in R^2 of the form (c,c), which are spanned (if looked at as a vector space) by (1,1). This means that anything in the range of T can be created by a linear combination of (1,1)s, i.e. c*(1,1).

    In the other example you've figured out or been told that the range is R^2. The sensible way to prove that is to show that the range is a subset of R^2 and vice versa. So, first we take anything in R^2 and prove it's in the range of T. To do this we need to show there's a p s.t. T(p) generates each point in R^2(which is what I showed in my first post). The range also has to be a subset of R^2 as T maps into R^2 by definition. Hence we have equality.
    Thank you for clarifying. Actually, I didn't mean that I had to prove that the range was R^2...I know the answer, because its the answer of an odd question in our textbook...so what if you didn't know that the range was R^2...how would you find out then? In other words, I'm not being asked to prove that the range is R^2; I'm being asked to find the range.
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    It completely depends on the question. In the problem you asked it was rather obvious: for a quadratic ax^2 + bx + x, p(0) = c, p(1) = a+b+c. It's pretty easy to see from this that we can get any values for p(0) and p(1) so it spans the whole of R^2.

    There is no set formula for doing this, you just have to look at it and see what's going on.
 
 
 
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