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    Can I get help on this?

    Not sure where to start. Everything resulted in failure. Not too sure what to do.

    I thoughts I'd say Tr=r where r is a column matrix abc plus lamda(def) but I'm not sure how to deal with that
    And I should use the fact that it goes through origin I take it? So r=0 for some lamda?

    No 8.


    I know the rules. A simple text advise response will help. Dont need working done. Thanks.

    TeeEm
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    What have you done to the beautiful book?!
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    (Original post by Kvothe the arcane)
    Can I get help on this?

    Not sure where to start. Everything resulted in failure. Not too sure what to do.

    I thoughts I'd say Tr=r where r is a column matrix abc plus lamda(def) but I'm not sure how to deal with that
    And I should use the fact that it goes through origin I take it? So r=0 for some lamda?

    No 8.


    I know the rules. A simple text advise response will help. Dont need working done. Thanks.

    @teeem
    find eigenvalues
    the one with lamda = 1 is the one
    find its eigenvector

    that is the direction of the line

    which essentially the same as multiplying by vector (x,y,z) should give you (x,y,z)
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    Not sure how to give a hint without telling you directly.

    consider if every point is mapped onto itself, you get something like this ax+by+cz=x' (x'=translated coordinate, but if it's invariant, it's the same...so), you'll have other equations for y' and z'.

    Good luck,
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    (Original post by Kvothe the arcane)
    Can I get help on this?

    Not sure where to start. Everything resulted in failure. Not too sure what to do.

    I thoughts I'd say Tr=r where r is a column matrix abc plus lamda(def) but I'm not sure how to deal with that
    And I should use the fact that it goes through origin I take it? So r=0 for some lamda?

    No 8.


    I know the rules. A simple text advise response will help. Dont need working done. Thanks.

    TeeEm
    At this point, have you covered eigenvectors?
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    (Original post by Slowbro93)
    At this point, have you covered eigenvectors?
    Its the next mini section. Many moons ago.
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    (Original post by TeeEm)
    find eigenvalues
    the one with lamda = 1 is the one
    find its eigenvector

    that is the direction of the line

    which essentially the same as multiplying by vector (x,y,z) should give you (x,y,z)
    Oh, so this is meant to make you discover eigenvectors without being taught. Thanks.
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    (Original post by Kvothe the arcane)
    Its the next mini section. Many moons ago.
    Okay, this question is leading up to the topic. If for example you had  \bold{y} = m \bold{x} + \bold{c} where x, y, and c are vectors, then for it to go through the origin you'd need c to be 0.

    Now take this to this example, you can start with a general vector say (x, y, z) and when you apply the transformation T, you should also get (x, y, z) back. How can you solve for x, y and z?
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    (Original post by Kvothe the arcane)
    Oh, so this is meant to make you discover eigenvectors without being taught. Thanks.
    in that case multiply the matrix by vector (x,y,z) and set it to (x,y,z).

    solving this is actually finding an eigenvector with lamda 1
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    TeeEm Slowbro93 Argylesocksrox thanks

    I did originally say that r=\begin{pmatrix} d \\ e \\f \end{pmatrix} + \lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix}

    And some lines later I got a=-b; a=c but it wasn't obvious to me how to proceed.

    The final answer is r=\lambda \begin{pmatrix}1 \\ -1 \\ 1 \end{pmatrix}

    I suppose even without substituting a number I'd end up with r=\begin{pmatrix} 0 \\ 0 \\0 \end{pmatrix} + \lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix} =  \begin{pmatrix}a \\ -a \\ a \end{pmatrix} \implies  \begin{pmatrix}a \\ b \\c \end{pmatrix}= \dfrac{1}{\lambda} \begin{pmatrix}a \\ -a \\ a \end{pmatrix}
    \therefore r=\mu \begin{pmatrix}a \\ -a \\ a \end{pmatrix} and I suppose the answer follows as a can be anything and 1's the simplest number.
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    (Original post by Kvothe the arcane)
    TeeEm Slowbro93 Argylesocksrox thanks

    I did originally say that r=\begin{pmatrix} d \\ e \\f \end{pmatrix} + \lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix}

    And some lines later I got a=-b; a=c but it wasn't obvious to me how to proceed.

    The final answer is r=\lambda \begin{pmatrix}1 \\ -1 \\ 1 \end{pmatrix}

    I suppose even without substituting a number I'd end up with r=\begin{pmatrix} 0 \\ 0 \\0 \end{pmatrix} + \lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \lambda \begin{pmatrix}a \\ -a \\ a \end{pmatrix} \Rightarrow \lambda \begin{pmatrix}a \\ b \\c \end{pmatrix}= \dfrac{1}{\lambda} \begin{pmatrix}a \\ -a \\ a \end{pmatrix}
    \therefore r=\mu \begin{pmatrix}a \\ -a \\ a \end{pmatrix} and I suppose the answer follows as a can be anything and 1's the simplest number.
    no worries
 
 
 
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