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Math Help - Linear Transformation

  1. #1
    Junior Member pearlyc's Avatar
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    Linear Transformation

    Hey guys,

    I was revising Linear Transformation and I came accross a couple of doubts here and there. Hope you guys would be able to explain to me for me to understand it better (:

    1. I know how to attempt questions with two variables, eg. S(x,y) = (2x-y, x+y) to prove that they are linear transformations. What is the method to attempt those with three variables involved? eg. T(x,y,z) = (0, 2x+y) or F(x,y,z) = {\color{white}.} \quad \left[ \begin{array}{ccc}y & z\\ -x & 0 \end{array}\right]

    2. What is the way to see the transformation patterns ? eg. {\color{white}.} \quad \left[ \begin{array}{ccc}0 & 1\\ 1 & 0 \end{array}\right] is the reflection of y = x. Say, how do we determine what is the transformation for a matrix like {\color{white}.} \quad \left[ \begin{array}{ccc}0 & 0\\ 1 & 0 \end{array}\right] or {\color{white}.} \quad \left[ \begin{array}{ccc}1 & 0\\ a & 1 \end{array}\right], etc.

    3. Whats the meaning of surjective & injective?

    4. I was taught how to tackle most of the questions in matrix form. When it comes to polynomial, I'd go nuts figuring how to attempt the question Especially when it comes to change of basis questions. Like what is [T]s and things like that when from matrix you can find it as the matrix representation.

    eg. T[p(x)] = p(2x+1), B = {1, x, x^2}

    T (a_0 + a_1x + a_2x^2) = a_0 + a_1(2x+1) + a_2(2x+1)^2

    Find [T]B.

    Thanks for your timeeee!
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by pearlyc View Post
    1. I know how to attempt questions with two variables, eg. S(x,y) = (2x-y, x+y) to prove that they are linear transformations. What is the method to attempt those with three variables involved? eg. T(x,y,z) = (0, 2x+y) or F(x,y,z) = {\color{white}.} \quad \left[ \begin{array}{ccc}y & z\\ -x & 0 \end{array}\right]

    How different is a three variable one?

    If T(x,y,z) = (0, 2x+y) is a linear transformation, then you can find a matrix A such that this A transforms (x,y,z) to (0,2x+y). That is

    A(x,y,z)^T = (0,2x+y)^T
    So the matrix should map a 3 x 1 vector to a 2 x 1 vector. This means A must be 2 x 3.

    So assume A to be a general 2 x 3 matrix and solve for it to get A as

    A = \begin{pmatrix} 0 & 0 & 0 \\ 2 & 1 & 0 \end{pmatrix}
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  3. #3
    Lord of certain Rings
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    Quote Originally Posted by pearlyc View Post
    Hey guys,

    2. What is the way to see the transformation patterns ? eg. {\color{white}.} \quad \left[ \begin{array}{ccc}0 & 1\\ 1 & 0 \end{array}\right] is the reflection of y = x. Say, how do we determine what is the transformation for a matrix like {\color{white}.} \quad \left[ \begin{array}{ccc}0 & 0\\ 1 & 0 \end{array}\right] or {\color{white}.} \quad \left[ \begin{array}{ccc}1 & 0\\ a & 1 \end{array}\right], etc.
    This is simple.. Assume (x,y) to be the mapping vector and see what the mapped vector is

    For \quad \left[ \begin{array}{ccc}0 & 1\\ 1 & 0 \end{array}\right]
    <br />
 \quad \left[ \begin{array}{ccc}0 & 1\\ 1 & 0 \end{array}\right] \cdot \begin{pmatrix} x\\ y\end{pmatrix} = \begin{pmatrix} y\\ x\end{pmatrix}

    So this transformation maps all (x,y) to (y,x). Thus its the reflection

    For \quad \left[ \begin{array}{ccc}1 & 0\\ a & 1 \end{array}\right]

    Do the same thing again to see that it maps (x,y) to (x,ax+y)...


    3. Whats the meaning of surjective & injective?
    Surjective = ONTO mapping (if you know this term)
    Injective = ONE-ONE mapping (if you know this term)
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by pearlyc View Post
    4. I was taught how to tackle most of the questions in matrix form. When it comes to polynomial, I'd go nuts figuring how to attempt the question Especially when it comes to change of basis questions. Like what is [T]s and things like that when from matrix you can find it as the matrix representation.

    eg. T[p(x)] = p(2x+1), B = {1, x, x^2}

    T (a_0 + a_1x + a_2x^2) = a_0 + a_1(2x+1) + a_2(2x+1)^2

    Find [T]B.
    Dont worry about it just because its a polynomial. After all a_0 + a_1x + a_2x^2 can be viewed as the tuple (a_0, a_1, a_2) with B as basis. With this look it is just another \mathbb{R}^3 vector, isnt it?

    So the transformation looks like T[p(x)] = p(2x+1) and we want to find the matrix representation for this, right?

    T (a_0 + a_1x + a_2x^2) = a_0 + a_1(2x+1) + a_2(2x+1)^2
    a_0 + a_1(2x+1) + a_2(2x+1)^2 = a_0 + 2a_1x+a_1 + 4a_2 x^2 + 4 a_2 x +a_2
    a_0 + 2a_1x+a_1 + 4a_2 x^2 + 4 a_2 x +a_2 = (a_0 + a_1 + 4a_2) +(2a_1 + 4a_2)x + (4a_2)x^2

    So a_0 + a_1x + a_2x^2 gets mapped to (a_0 + a_1 + 4a_2) +(2a_1 + 4a_2)x + (4a_2)x^2.

    In terms of triples, its the same as saying (a_0, a_1, a_2) maps to
    (a_0 + a_1 + 4a_2, 2a_1 + 4a_2 , 4a_2)

    So now this problem is the same as the first question

    Just a\find a 3 x 3 matrix that maps (a_0, a_1, a_2) maps to
    (a_0 + a_1 + 4a_2, 2a_1 + 4a_2 , 4a_2).
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  5. #5
    Junior Member pearlyc's Avatar
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    Quote Originally Posted by Isomorphism View Post
    How different is a three variable one?

    If T(x,y,z) = (0, 2x+y) is a linear transformation, then you can find a matrix A such that this A transforms (x,y,z) to (0,2x+y). That is

    A(x,y,z)^T = (0,2x+y)^T
    So the matrix should map a 3 x 1 vector to a 2 x 1 vector. This means A must be 2 x 3.

    So assume A to be a general 2 x 3 matrix and solve for it to get A as

    A = \begin{pmatrix} 0 & 0 & 0 \\ 2 & 1 & 0 \end{pmatrix}
    How do I do the whole proving the two properties :

    T(u+v) = T(u) + T(v) and T(au) = aT(u)?
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  6. #6
    Junior Member pearlyc's Avatar
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    Quote Originally Posted by Isomorphism View Post
    Dont worry about it just because its a polynomial. After all a_0 + a_1x + a_2x^2 can be viewed as the tuple (a_0, a_1, a_2) with B as basis. With this look it is just another \mathbb{R}^3 vector, isnt it?

    So the transformation looks like T[p(x)] = p(2x+1) and we want to find the matrix representation for this, right?

    T (a_0 + a_1x + a_2x^2) = a_0 + a_1(2x+1) + a_2(2x+1)^2
    a_0 + a_1(2x+1) + a_2(2x+1)^2 = a_0 + 2a_1x+a_1 + 4a_2 x^2 + 4 a_2 x +a_2
    a_0 + 2a_1x+a_1 + 4a_2 x^2 + 4 a_2 x +a_2 = (a_0 + a_1 + 4a_2) +(2a_1 + 4a_2)x + (4a_2)x^2

    So a_0 + a_1x + a_2x^2 gets mapped to (a_0 + a_1 + 4a_2) +(2a_1 + 4a_2)x + (4a_2)x^2.

    In terms of triples, its the same as saying (a_0, a_1, a_2) maps to
    (a_0 + a_1 + 4a_2, 2a_1 + 4a_2 , 4a_2)

    So now this problem is the same as the first question

    Just a\find a 3 x 3 matrix that maps (a_0, a_1, a_2) maps to
    (a_0 + a_1 + 4a_2, 2a_1 + 4a_2 , 4a_2).
    Does this mean the 3 x 3 matrix for that is
    (1 1 4
    0 2 4
    0 0 4) ?
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  7. #7
    Lord of certain Rings
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    Quote Originally Posted by pearlyc View Post
    Does this mean the 3 x 3 matrix for that is
    (1 1 4
    0 2 4
    0 0 4) ?
    Bingo
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  8. #8
    Junior Member pearlyc's Avatar
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    Quote Originally Posted by pearlyc View Post
    How do I do the whole proving the two properties :

    T(u+v) = T(u) + T(v) and T(au) = aT(u)?
    How about this one?

    And the 3 x 3 matrix from the post before this is [T]s right? (:

    Thanks a lot for your help, really.
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  9. #9
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    Quote Originally Posted by pearlyc View Post
    How about this one?
    Once you find a matrix A such that T(x) = Ax, there is nothing to prove since you must have proved the map Ax is linear for any matrix A.

    But if you are still adamant, you can try the axioms directly.

    And the 3 x 3 matrix from the post before this is [T]s right?
    Yes
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  10. #10
    Junior Member pearlyc's Avatar
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    Thanks a lot.

    Okay so if you have [T]s, to find [T]b you'll need Pb->s and Ps->b right? How do we find this from a polynomial function? Let's take the question above as an example?

    How do you prove whether a polynomial function is a linear transformation then?

    I am still so weak in this whole polynomial business

    Thanks for your help though!

    For example,

    T[p(x)] = xp(x)
    T(a + bx + cx^2) = ax + bx^2 + cx^3


    Show that T is a linear transformation.
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  11. #11
    Lord of certain Rings
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    Quote Originally Posted by pearlyc View Post
    Thanks a lot.

    Okay so if you have [T]s, to find [T]b you'll need Pb->s and Ps->b right? How do we find this from a polynomial function? Let's take the question above as an example?

    How do you prove whether a polynomial function is a linear transformation then?

    I am still so weak in this whole polynomial business

    Thanks for your help though!

    For example,

    T[p(x)] = xp(x)
    T(a + bx + cx^2) = ax + bx^2 + cx^3

    Show that T is a linear transformation.
    Basically the same trick works again... This time you will have to try it out. I will give you a hint:

    a + bx + cx^2 is equivalent to (a,b,c) in terms of triples and ax + bx^2 + cx^3 is equivalent to (0,a,b,c). So you want to map the triple (a,b,c) to (0,a,b,c). So find a matrix A such that it maps (a,b,c) to (0,a,b,c). As I have already told you before, once you find a matrix, its immediately clear that the map is linear.
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