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Math Help - introduction To Linear Transformations (sec. 1.8 Linear Algebra book by Lay)

  1. #1
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    introduction To Linear Transformations (sec. 1.8 Linear Algebra book by Lay)

    I have three homework problems that I got stuck on while, trying to solve. I'd really appreciate it if anyone could help me with these. Here they are

    For # 14 & 16 use a rectangular coordinate system to plot u= [vertical: 5 2]
    , v=[vert: -2 4], and their images under the given transformation T. (Make a large sketch for each exercise.) Describe geometrically what T does to each vector x in R^2.
    #14 T(x)= [.5 0 ] [x1] horiz.
    [0 -.5] [x2]horiz.
    # 16 T(x)= [0 1] [x1] horiz.
    [1 0] [x2] horiz.

    (T of each exercices is just one matrix of two columns and two rows)

    #34 Let T: R^n----> R^m be a linear transformation. Show that if T maps two linearly independent vectors onto a linearly dependent set, then the equation T(x)=0 has a notrivial solution.
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  2. #2
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    Quote Originally Posted by googoogaga View Post
    #34 Let T: R^n----> R^m be a linear transformation. Show that if T maps two linearly independent vectors onto a linearly dependent set, then the equation T(x)=0 has a notrivial solution.
    It is necessary and sufficient to show that \ker T \not = \{ \bold{0} \}.

    Let \bold{u},\bold{v} \in \mathbb{R}^n be these two linearly independent vectors.

    We are given that T(\bold{u}),T(\bold{v}) are linearly dependent.

    Thus, there exists k_1,k_2 \in \mathbb{R} so that k_1T(\bold{u})+k_2T(\bold{v}) = \bold{0} and k_1,k_2 are not both zero.

    Hence, T(k_1\bold{u}+k_2\bold{v}) = \bold{0} since T is a linear transformation.

    But, k_1\bold{u}+k_2\bold{v} \not = \bold{0} because k_1,k_2 are not both zero and \bold{u},\bold{v} are linearly independent.

    Hence, there exists a non-zero element vector \bold{w} = k_1\bold{u}+k_2\bold{v} such that T(\bold{w}) = \bold{0}.

    Thus, \ker T\not = \{\bold{0}\}.

    ---
    Note: This can be generalized to n\geq 2 linearly independent vectors getting mapped into a linearly dependent set.
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  3. #3
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    Quote Originally Posted by googoogaga View Post
    I have three homework problems that I got stuck on while, trying to solve. I'd really appreciate it if anyone could help me with these. Here they are

    For # 14 & 16 use a rectangular coordinate system to plot u= [vertical: 5 2]
    , v=[vert: -2 4], and their images under the given transformation T. (Make a large sketch for each exercise.) Describe geometrically what T does to each vector x in R^2.
    #14 T(x)= [.5 0 ] [x1] horiz.
    [0 -.5] [x2]horiz.
    # 16 T(x)= [0 1] [x1] horiz.
    [1 0] [x2] horiz.

    (T of each exercices is just one matrix of two columns and two rows)
    14.)

    I'll describe this graphically.

    You have the vector v at (-1, 4); then T(v) is at (-1, 2) (both in the direction of the origin.

    You have the vector u at (5, 2); then T(u) is at (2.5, 1) (both in the direction of the origin.

    Thus, there's a CONTRACTION of a factor of 1/2.

    16.) It's easy to see that there's a reflection through the line x_2 = x_1 (or y = x).

    Vector v is at (-2, 4) and is in the direction of T(v) at (4, -2).

    Vector u is at (5, 2) and is in the direction of T(u) at (2, 5).

    Draw a dotted line x_2 = x_1 (or y = x) and you can see the reflection.
    Last edited by AfterShock; June 2nd 2007 at 11:14 AM.
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  4. #4
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    Quote Originally Posted by googoogaga View Post
    #34 Let T: R^n----> R^m be a linear transformation. Show that if T maps two linearly independent vectors onto a linearly dependent set, then the equation T(x)=0 has a notrivial solution.
    Suppose that {u,v} are linearly independent in R^n and we have T(u) and T(v) are linearly dependent. There exists weights c_1, c_2 (both of them not zero), such that:

    c_1*T(u) + c_2*T(v) = 0

    Since T is linear, T(c_1*u + c_2*v) = 0. Thus,

    x= c_1*u+ c_2*v will satisfy T(x) = 0. And, we know that it's impossible for x to be the zero vector since that'd imply that a nontrivial linear combination of u and v is zero, and we know that is impossible since u and v are linearly independent.

    Therefore, we can conlude that the equation T(x) = 0 has a nontrivial solution.
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