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Math Help - One-to-one parameterization

  1. #1
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    One-to-one parameterization

    Let A= <br /> <br />
\left[\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] <br />

    Give a one-to-one parameterization of the ker(A)

    Would it just be \left[\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}\right]
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  2. #2
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    How about (t,s,-(s+t)) s,t \in \mathbb{R}
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  3. #3
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    Thanks Jose, may I ask how you got that answer??
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  4. #4
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    Notice that x \in kerA iff Ax=0. If x=(x_1,x_2,x_3) then we get three equations that are exactly the same x_1+x_2+x_3=0. So given any values for x_1 and x_2 x_3 is completely determined (i.e. kerA has dimension 2).
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  5. #5
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    If you multiply a general vector (on the right side!) v=(x,y,z)' with the given matrix,
    you will end up with the vector (x+y+z,x+y+z,x+y+z)' <-- (transposed).

    Now if A reprisents a linear map h:V->W. Then ker(A) corresponds to the
    subspace which image under h is 0.

    Now what that means is what does v=(x,y,z) have to be so that when you multiply it
    with A you get 0?
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  6. #6
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    Now what that means is what does v=(x,y,z) have to be so that when you multiply it
    with A you get 0?
    I don't know, what method should I use? I am used to row reducing but in this case it looks like the bottom two rows will become 0 0 0.
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  7. #7
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    Quote Originally Posted by Zocken View Post
    I don't know, what method should I use? I am used to row reducing but in this case it looks like the bottom two rows will become 0 0 0.
    That just means you have 2 free variables. That is the reason you
    can parametrise the solution.

    Do you understand that, this subspace is the kernel of A?
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  8. #8
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    One last question...Let b=Ae1. Find a one to one parameterization of the solution set Ax=b.

    Would this makes sense using the above answer?

    <br /> <br />
e1+t\left[\begin{matrix} 1 \\ 0 \\ -1 \end{matrix}\right]+s\left[\begin{matrix} 0 \\ 1 \\ -1 \end{matrix}\right]<br />
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  9. #9
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    Yes, that is correct.

    The solution describes a plane with normal vector (1,1,1) going through the point e1.

    So the function is not one-to-one , because all the points on that plane will go to b.
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