Results 1 to 2 of 2

Thread: Linear Transformation

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    2

    Linear Transformation

    Let T: V -> W be a linear transformation, and let the dimension of V be n, Let {$\displaystyle v_1, v_2, ..., v_m$} be a basis for the kernel of T.

    1) Let V_(m+1), v_(m+2), ..., $\displaystyle v_n$ be vectors in V such that {$\displaystyle v_1$, $\displaystyle v_2$, ..., $\displaystyle v_m$, v_(m+1), ...,$\displaystyle v_n$} in a basis for V. Prove that {Tv_(m+1), Tv_(m+2), ..., $\displaystyle Tv_n$} is a basis for the image of T.

    2) Using 1), prove that the rank of T plus the nullity of T is n.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by cindy View Post
    Let T: V -> W be a linear transformation, and let the dimension of V be n, Let {$\displaystyle v_1, v_2, ..., v_m$} be a basis for the kernel of T.
    It is safe to assume $\displaystyle m\geq 1$.

    1) Let V_(m+1), v_(m+2), ..., $\displaystyle v_n$ be vectors in V such that {$\displaystyle v_1$, $\displaystyle v_2$, ..., $\displaystyle v_m$, v_(m+1), ...,$\displaystyle v_n$} in a basis for V. Prove that {Tv_(m+1), Tv_(m+2), ..., $\displaystyle Tv_n$} is a basis for the image of T.
    Let $\displaystyle \bold{a}$ be in the image of $\displaystyle T$.
    Then $\displaystyle \bold{a} = k_1 \bold{v}_1 + ... + k_m \bold{v}_m + k_{m+1} \bold{v}_{m+1} + ... + k_n \bold{v}_n $.
    Thus, $\displaystyle T\bold{a} = k_1 T\bold{v}_1 + ... + k_m T\bold{v}_m + k_{m+1}T \bold{v}_{m+1} + ... + k_n T\bold{v}_n = k_{m+1}T \bold{v}_{m+1} + ... + k_n T\bold{v}_n$.
    Thus, $\displaystyle T\bold{v}_{m+1}, ... T\bold{v}_n$ spams the image of $\displaystyle T$.

    You need to show $\displaystyle T\bold{v}_{m+1}, ... T\bold{v}_n$ is linearly independent.
    If $\displaystyle a_{m+1}T\bold{v}_{m+1} + ... + a_n T\bold{v}_n = \bold{0}$
    Then $\displaystyle T(a_{m+1}\bold{v}_{m+1} + ... + a_n \bold{v}_n) = \bold{0}$.
    Thus, $\displaystyle a_{m+1}\bold{v}_{m+1} + ... + a_n \bold{v}_n \in \ker T$.

    We can therefore write, $\displaystyle a_{m+1}\bold{v}_{m+1} + ... + a_n \bold{v}_n = a_1\bold{v}_1 + ... + a_2 \bold{v}_m$.
    This implies $\displaystyle a_1=a_2= ... 0$ because those vectors form a basis.
    ---
    Then #2 follows because $\displaystyle m + (n-m) = n$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Aug 1st 2011, 10:00 PM
  2. Example of a linear transformation T:R^4 --> R^3?
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Apr 5th 2011, 07:04 PM
  3. linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 28th 2009, 06:40 AM
  4. Linear Algebra.Linear Transformation.Help
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Mar 5th 2009, 01:14 PM
  5. Linear Transformation
    Posted in the Pre-Calculus Forum
    Replies: 10
    Last Post: May 25th 2008, 12:14 AM

Search Tags


/mathhelpforum @mathhelpforum