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Math Help - Transformation, kernel, image, etc

  1. #1
    MHF Contributor arbolis's Avatar
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    Transformation, kernel, image, etc

    This is an exercise I had on the Linear Algebra final exam I failed. I could only do a part and I'm still stuck for most of it. So I'll write it entirely and I'll ask help only for part 1) until we finish it, then ask for further help.

    -------------------------------------------------------------
    Let \bold B = \{ \alpha_1, \alpha_2, \alpha_3 \} be the ordered basis of \mathbb{R}^3 where \alpha_1=(1,0,1), \alpha_2=(0,1,1), \alpha_3=(0,-1,1).
    Let T : \mathbb{R}^3 \to \mathbb{R}^4 be defined by T(\alpha_1)=(1,-2,1,0), T(\alpha_2)=(2,1,0,-1), T(\alpha_3)=(1,0,1,0).
    1)Give an explicit description of \ker T, calculate its dimension and give a basis.
    2)Give an explicit description of \Im T, calculate its dimension and give a basis.
    3)Calculate [T]_{\bold B}^{\bold B'}, where \bold B' =\{ (1,1,0,0), (0,0,1,1), (1,0,0,4), (0,0,0,1)   \}.
    4)If \alpha=(-3,-1,3), find the coordinates of T(\alpha) with respect to \bold B'.
    -------------------------------------------------------------

    1) My attempt : only thoughts. I must find all the vectors v \in \mathbb{R}^3 such that T(v)=0. I'm at a loss. I don't even know if the transformation is linear. I've absolutely no idea about how to find the kernel of it.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by arbolis View Post
    This is an exercise I had on the Linear Algebra final exam I failed. I could only do a part and I'm still stuck for most of it. So I'll write it entirely and I'll ask help only for part 1) until we finish it, then ask for further help.

    -------------------------------------------------------------
    Let \bold B = \{ \alpha_1, \alpha_2, \alpha_3 \} be the ordered basis of \mathbb{R}^3 where \alpha_1=(1,0,1), \alpha_2=(0,1,1), \alpha_3=(0,-1,1).
    Let T : \mathbb{R}^3 \to \mathbb{R}^4 be defined by T(\alpha_1)=(1,-2,1,0), T(\alpha_2)=(2,1,0,-1), T(\alpha_3)=(1,0,1,0).
    1)Give an explicit description of \ker T, calculate its dimension and give a basis.
    2)Give an explicit description of \Im T, calculate its dimension and give a basis.
    3)Calculate [T]_{\bold B}^{\bold B'}, where \bold B' =\{ (1,1,0,0), (0,0,1,1), (1,0,0,4), (0,0,0,1) \}.
    4)If \alpha=(-3,-1,3), find the coordinates of T(\alpha) with respect to \bold B'.
    -------------------------------------------------------------

    1) My attempt : only thoughts. I must find all the vectors v \in \mathbb{R}^3 such that T(v)=0. I'm at a loss. I don't even know if the transformation is linear. I've absolutely no idea about how to find the kernel of it.
    The question should have stated that T is linear as otherwise the definition does not define its behaviour.

    The kernel or null space of T is the subspace of \mathbb{R}^3 whose image is the zero vector in \mathbb{R}^4. Now let x=u_1\alpha_1+u_2 \alpha_2+u_3\alpha_3 be in the null space of T, then:

     <br />
T(x)=u_1 T(\alpha_1)+u_2 T(\alpha_2)+u_3 T(\alpha_3)={\bf{0}}_{4}<br />

    Write this out in full and you will find that the only solution has u_1=u_2=u_3=0, so the zero vector in \mathbb{R}^3 is the only element in the Kernel of T.

    CB
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    The question should have stated that T is linear as otherwise the definition does not define its behaviour.

    The kernel or null space of T is the subspace of \mathbb{R}^3 whose image is the zero vector in \mathbb{R}^4. Now let x=u_1\alpha_1+u_2 \alpha_2+u_3\alpha_3 be in the null space of T, then:

     <br />
T(x)=u_1 T(\alpha_1)+u_2 T(\alpha_2)+u_3 T(\alpha_3)={\bf{0}}_{4}<br />

    Write this out in full and you will find that the only solution has u_1=u_2=u_3=0, so the zero vector in \mathbb{R}^3 is the only element in the Kernel of T.

    CB
    Thanks so much! It worked. I agree, the professors didn't even specify verbally that the transformation had to be linear.
    So \dim \ker T =0 and it has no basis.


    Part 2) I did it entirely. ( I can write all what I did, but I think it's a waste of time since I did it well)

    Part 3) I've done all wrong. I've written the vectors of the basis \bold B as a linear combination of the vectors of the basis of \bold B', but I can't do that since they do not have the same dimension.
    As of today I still do not know how to proceed.
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  4. #4
    MHF Contributor arbolis's Avatar
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    Using the method I said in my previous, I got that [T]_{\bold B}^{\bold B'}=\begin{bmatrix}  0&1&1 \\ 1&1&2 \\ 1&-1&-1 \\ -4&3&3      \end{bmatrix}. However I got it all wrong. I don't know how to find what they ask me...
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