Results 1 to 6 of 6

Math Help - Linear transformation proof problem

  1. #1
    Member
    Joined
    Jan 2010
    Posts
    232

    Linear transformation proof problem

    Let T:R^n\rightarrow R^m be a linear transformation.
    If v_1,v_2,...,v_k are vectors in R^n such that the set \{T(v_1),T(v_2),...,T(v_k)\} is linearly independent in R^m, prove that the set \{v_1,v_2,...,v_k\} is linearly independent.
    If n=m and, for every linearly independent set \{w_1,w_2,...,w_k\} in R^n, the set \{T(w_1),T(w_2),...,T(w_k)\} is always linearly independent, prove that there is an inverse transformation S:R^n\rightarrow R^n. (i.e. S\circ T, T\circ S are both the identity transformation).

    This one is pretty tricky, needless to say.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2010
    Posts
    422
    Suppose the set \{v_i\} were not linearly independent so there are a^i not all zero s.t. a^iv_i=0. Apply T to this equation to reach a contradiction.

    For the second, try showing that \mathrm{ker}\; T = 0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2010
    Posts
    232
    Quote Originally Posted by maddas View Post
    Suppose the set \{v_i\} were not linearly independent so there are a^i not all zero s.t. a^iv_i=0. Apply T to this equation to reach a contradiction.

    For the second, try showing that \mathrm{ker}\; T = 0.
    We don't have the Kelvin function available, so that wouldn't work.

    Also, though I could probably write out all of the first part, I'm tied up with other Algebra-related material, so do you think you could show a bit more detail?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Feb 2010
    Posts
    422
    Kelvin function? I meant the kernel lol.

    1) If a^iv_i=0 then T(a^iv_i)=T(0) meaning a^iT(v_i)=0. But the T(v_i) are linearly independent so the a^i are identically zero.

    2) Let v_i be a basis for the vector space. Let x=x^iv_i be a vector in the kernel of T. Then T(x)=x^iT(v_i)=0. But because T(v_i) are linearly independent, this implies that the x^i are identically zero so x=0. So the kernel of T is zero. It follows that T is injective. Now show it is surjective and use these to deduce it is invertible.

    P.S. I use the Einstein summation convention. I'm going to get in trouble if I don't get back to work now lol. Good luck!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2010
    Posts
    232
    Quote Originally Posted by maddas View Post
    Kelvin function? I meant the kernel lol.

    1) If a^iv_i=0 then T(a^iv_i)=T(0) meaning a^iT(v_i)=0. But the T(v_i) are linearly independent so the a^i are identically zero.

    2) Let v_i be a basis for the vector space. Let x=x^iv_i be a vector in the kernel of T. Then T(x)=x^iT(v_i)=0. But because T(v_i) are linearly independent, this implies that the x^i are identically zero so x=0. So the kernel of T is zero. It follows that T is injective. Now show it is surjective and use these to deduce it is invertible.

    P.S. I use the Einstein summation convention. I'm going to get in trouble if I don't get back to work now lol. Good luck!
    Thanks for your help on the first half, but as for the second half, we don't have kernels available. I got the name mixed up last post.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Feb 2010
    Posts
    422
    There is no reason not to use kernels. I'm not sure what you mean by them not being available..... The kernel of a map is the set of vectors it sends to zero. There is a beautiful relationship between the kernel and the image of a map called the rank-nullity theorem.

    ...but in any case, this particular proof doesn't use them in any essential way so I'll reformulate it without them:

    Lemma. If T is linear and Tv=0\implies v=0 then T is injective.

    Proof. For suppose not. Then there are distinct x and y s.t. Tx=Ty. But then T(x-y)=0, which implies that x-y=0, or x=y, contrary to our assumption.

    Then to prove your 2) Take an x=x^iv_i and suppose that Tx=0. Then Tx=x^iTv_i=0 so by linear independence of the Tv_i, the x^i are all zero, meaning x=0. Therefore if Tx=0, then x=0, and we may apply the lemma to see that T is injective. Surjectiveness is easy and so T is invertible.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. linear transformation problem
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: April 1st 2010, 04:13 AM
  2. Proof for a basis of a linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: February 8th 2010, 06:00 AM
  3. Bijective Linear Transformation Proof
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 17th 2009, 09:05 PM
  4. Help with linear transformation / subspace proof
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: July 28th 2009, 11:04 AM
  5. Proof about eigenvalues/vectors of a linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: June 28th 2009, 06:21 AM

Search Tags


/mathhelpforum @mathhelpforum