Results 1 to 7 of 7

Math Help - [SOLVED] Linear Transformations

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    20

    Arrow [SOLVED] Linear Transformations

    Could anyone walk me through these two? I need some help getting going.

    1) Prove that the map T:P_n\rightarrow P_n defined by T\left[p\left(t\right)\right]=\frac{dp}{dt} is a linear transformation and determine if T is one-to-one.

    2) Let T_A:\mathbb{R}^n\rightarrow \mathbb{R}^m. Prove that the range of T_A is a subspace of \mathbb{R}^m.
    Last edited by leverin4; December 6th 2009 at 11:10 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by leverin4 View Post
    Could anyone walk me through these two? I need some help getting going.

    1) Prove that the map T:P_n\rightarrow P_n defined by T\left[p\left(t\right)\right]=\frac{dp}{dt} is a linear transformation and determine if T is one-to-one.
    Take two polynomials p,q\in P_n. Therefore,

    \begin{aligned}T\left(ap+bq\right)&=\frac{\,d}{\,d  t}\left[ap+bq\right]\\ &= \frac{\,d}{\,dt}\left[ap\right]+\frac{\,d}{\,dt}\left[bq\right]\\&=\ldots\end{aligned}

    Can you finish this?

    Note that T is not one to one. Can you think of an easy counterexample?

    2) Let T_A:\mathbb{R}^n\rightarrow \mathbb{R}^m. Prove that the range of T_A is a subspace of \mathbb{R}^m.
    Let w_1,w_2\in\text{range}\!\left(T_A\right). Thus, w_1=T_A\!\left(v_1\right) and w_2=T_A\!\left(v_2\right) for some v_1,v_2\in\mathbb{R}^n.

    Now note that

    \begin{aligned}aw_1+bw_2&=aT_A\!\left(v_1\right)+b  T_A\!\left(v_2\right)\\ &= T_A\!\left(av_1\right)+T_A\!\left(bv_2\right)\\ &= T_A\!\left(av_1+bv_2\right)\end{aligned}

    Since T_A\!\left(av_1+bv_2\right)\in\mathbb{R}^m, we can now say that \text{range}\!\left(T_A\right) is a subspace of \mathbb{R}^m.

    Does this make sense?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    20
    Quote Originally Posted by Chris L T521 View Post
    Take two polynomials p,q\in P_n. Therefore,

    \begin{aligned}T\left(ap+bq\right)&=\frac{\,d}{\,d  t}\left[ap+bq\right]\\ &= \frac{\,d}{\,dt}\left[ap\right]+\frac{\,d}{\,dt}\left[bq\right]\\&=\ldots\end{aligned}

    Can you finish this?

    Note that T is not one to one. Can you think of an easy counterexample?
    I'm a little confused as to where a and b came from.

    If I'm interpreting the problem right, a counter example would be x^3 because then the transformation would lead to 3x^2

    EDIT: Technically, I guess it would be p=t^3 as a counter example. Gotta keep the variables straight.

    Let w_1,w_2\in\text{range}\!\left(T_A\right). Thus, w_1=T_A\!\left(v_1\right) and w_2=T_A\!\left(v_2\right) for some v_1,v_2\in\mathbb{R}^n.

    Now note that

    \begin{aligned}aw_1+bw_2&=aT_A\!\left(v_1\right)+b  T_A\!\left(v_2\right)\\ &= T_A\!\left(av_1\right)+T_A\!\left(bv_2\right)\\ &= T_A\!\left(av_1+bv_2\right)\end{aligned}

    Since T_A\!\left(av_1+bv_2\right)\in\mathbb{R}^m, we can now say that \text{range}\!\left(T_A\right) is a subspace of \mathbb{R}^m.

    Does this make sense?
    Again, I'm confused about a and b. Are they just arbitrary constants? If so, the math makes sense to me, but I don't see how it proves that it's a subspace.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by leverin4 View Post
    I'm a little confused as to where a and b came from.

    If I'm interpreting the problem right, a counter example would be x^3 because then the transformation would lead to 3x^2
    To clarify, a and b are arbitrary constants. Now, to show that injectivity fails, you want to pick polynomials p,q\in P_n:\frac{\,dp}{\,dt}=\frac{\,dq}{\,dt}. Don't make it too hard.

    Again, I'm confused about a and b. Are they just arbitrary constants? If so, the math makes sense to me, but I don't see how it proves that it's a subspace.
    Yes, a and b are arbitrary constants. Recall that if W\subset V and when u,v\in W we have au+bv\in W, then W is a subspace of V.

    In this case we let W=\text{range}\!\left(T_A\right) and V=\mathbb{R}^m.

    Does this clarify things?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2009
    Posts
    20
    Quote Originally Posted by Chris L T521 View Post
    To clarify, a and b are arbitrary constants. Now, to show that injectivity fails, you want to pick polynomials p,q\in P_n:\frac{\,dp}{\,dt}=\frac{\,dq}{\,dt}. Don't make it too hard.
    woops...forgot we were working in n dimensions. So something simple like, in your example in the quote box above, setting

    p: p=t^n
    q: p=t^n+4

    To finish the proof, I just add:

    \begin{aligned}T\left(ap+bq\right)&=\frac{\,d}{\,d  t}\left[ap+bq\right]\\ &= \frac{\,d}{\,dt}\left[ap\right]+\frac{\,d}{\,dt}\left[bq\right]\\&=T(ap)+T(bq)\\&=aT(p)+bT(q)\end{aligned}

    Then I do the same thing with an arbitrary scalar multiple, right?

    Yes, a and b are arbitrary constants. Recall that if W\subset V and when u,v\in W we have au+bv\in W, then W is a subspace of V.

    In this case we let W=\text{range}\!\left(T_A\right) and V=\mathbb{R}^m.

    Does this clarify things?
    This part makes sense now. Thank you.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by leverin4 View Post
    woops...forgot we were working in n dimensions. So something simple like, in your example in the quote box above, setting

    p: p=t^n
    q: p=t^n+4

    To finish the proof, I just add:

    \begin{aligned}T\left(ap+bq\right)&=\frac{\,d}{\,d  t}\left[ap+bq\right]\\ &= \frac{\,d}{\,dt}\left[ap\right]+\frac{\,d}{\,dt}\left[bq\right]\\&=T(ap)+T(bq)\\&=aT(p)+bT(q)\end{aligned}
    The second to last line isn't correct. We can't assume its linear! However, note that \frac{\,d}{\,dt}\left[ap\right]+\frac{\,d}{\,dt}\left[bq\right]=a\frac{\,dp}{\,dt}+b\frac{\,dq}{\,dt}=a T\!\left(p\right)+bT\!\left(q\right).

    Taking T\!\left(ap+bq\right) takes care of the constant muliple part as well. So you only need to check this linear combination, instead of the sum and constant multiple parts separately (in other words, its easier and quicker to do both steps in one step).

    Does this make sense?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Mar 2009
    Posts
    20
    Quote Originally Posted by Chris L T521 View Post
    The second to last line isn't correct. We can't assume its linear! However, note that \frac{\,d}{\,dt}\left[ap\right]+\frac{\,d}{\,dt}\left[bq\right]=a\frac{\,dp}{\,dt}+b\frac{\,dq}{\,dt}=a T\!\left(p\right)+bT\!\left(q\right).

    Taking T\!\left(ap+bq\right) takes care of the constant muliple part as well. So you only need to check this linear combination, instead of the sum and constant multiple parts separately (in other words, its easier and quicker to do both steps in one step).

    Does this make sense?
    ugh, I see my mistake. I'm not a fan of these. Thanks for your help. It's been great.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear Transformations and the General Linear Group
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 26th 2011, 11:50 AM
  2. Basic Linear Algebra - Linear Transformations Help
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: December 7th 2010, 04:59 PM
  3. [SOLVED] Linear Transformations
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: May 31st 2010, 05:18 AM
  4. [SOLVED] Linear transformations
    Posted in the Math Challenge Problems Forum
    Replies: 8
    Last Post: May 13th 2010, 12:50 AM
  5. Replies: 3
    Last Post: June 2nd 2007, 11:08 AM

Search Tags


/mathhelpforum @mathhelpforum