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    Hiya I was wondering if someone could help me with question 2ii) in the picture.

    Note  ||z||_\infty = \displaystyle\mathrm{max}_{j=1,.  .,N} |z_j|

    I proved 2i) basically using the triangle inequality.

    For 2ii) I was thinking that

    if we choose  d   \mathrm{such \ that}  d_i = 1  \mathrm{for \ all}    1 \leq i \leq N

    then

     ||Bd||_\infty = \displaystyle\mathrm{max}_{i=1,.  .,N} \displaystyle\sum_{j=1}^N |b_i_j|

    I'm pretty sure this is true anyway. Then surely since it is true (which I think it is) then I can say that the inequality I am trying to prove in 2ii) holds. This feels really really cheap though and I don't think it is correct, could anyone point me in the right direction?
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    (Original post by TheBhramaBull)
    if we choose  d   \mathrm{such \ that}  d_i = 1  \mathrm{for \ all}    1 \leq i \leq N

    then

     ||Bd||_\infty = \displaystyle\mathrm{max}_{i=1,.  .,N} \displaystyle\sum_{j=1}^N |b_i_j|

    I'm pretty sure this is true anyway.
    Let B= \begin{pmatrix} 1 & -1 \\-1 & 1\end{pmatrix}

    Now tell me it's true.
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    (Original post by ghostwalker)
    Let B= \begin{pmatrix} 1 & -1 \\-1 & 1\end{pmatrix}

    Now tell me it's true.

    Point taken haha that is so obvious too. Do you have any advice on question 2ii) then?
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    (Original post by TheBhramaBull)
    Point taken haha that is so obvious too. Do you have any advice on question 2ii) then?
    Consider the particular i, that gives the maximum for the RHS, then construct your d
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    (Original post by ghostwalker)
    Consider the particular i, that gives the maximum for the RHS, then construct your d
    I can only think to choose d such that d_i = -1 if b_ij < 0, d_i = 1 if b_ij greater than or equal to 0. But we cannot choose d we need to find it so im stuck.

    I don't understand, which particular i gives the maximum for the RHS, why is there a particular i that gives the maximum for the RHS?
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    (Original post by TheBhramaBull)
    I can only think to choose d such that d_i = -1 if b_ij < 0, d_i = 1 if b_ij greater than or equal to 0. But we cannot choose d we need to find it so im stuck.

    I don't understand, which particular i gives the maximum for the RHS, why is there a particular i that gives the maximum for the RHS?
    For a given B one of the values for i must make the RHS a maximum.

    There isn't one value of i for all Bs, but for any given B, there must be at least one value of i, that makes the RHS a maximum.


    Same with the d; there isn't going to be one value for d that works for all Bs, but you can construct one for any given B
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    (Original post by ghostwalker)
    For a given B one of the values for i must make the RHS a maximum.

    There isn't one value of i for all Bs, but for any given B, there must be at least one value of i, that makes the RHS a maximum.


    Same with the d; there isn't going to be one value for d that works for all Bs, but you can construct one for any given B
    But then does that mean that I am allowed to use the answer:

    choose d such that d_i = -1 if the sum from j=1 to N of b_ij < 0, or d_i = 1 if the sum of b_ij from j = 1 to N is greater than or equal to 0.

    Sorry but that's what I am understanding from your previous post. I do appreciate the help, thanks a lot.
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    (Original post by TheBhramaBull)
    But then does that mean that I am allowed to use the answer:

    choose d such that d_i = -1 if the sum from j=1 to N of b_ij < 0, or d_i = 1 if the sum of b_ij from j = 1 to N is greater than or equal to 0.

    Sorry but that's what I am understanding from your previous post. I do appreciate the help, thanks a lot.
    Apply your thought process to the example I gave; does it work?

    Also I'm not clear whether you saying this for all i. You need to be precise. Look at my previous post.
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    (Original post by ghostwalker)
    Apply your thought process to the example I gave; does it work?

    Also I'm not clear whether you saying this for all i. You need to be precise. Look at my previous post.
    Okay so how about choose d such that for

    i = 1,2,...,N if \displaystyle\sum_{j=1}^N b_i_j &lt; 0 then set d_i = -1 , if \displaystyle\sum_{j=1}^N b_i_j \geq 0 then set d_i = 1

    Or is this not sufficient. It's basically the same as what I said before, although I feel it is more precise, and should seem a bit clearer to you.
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    (Original post by TheBhramaBull)
    Okay so how about choose d such that for

    i = 1,2,...,N if \displaystyle\sum_{j=1}^N b_i_j &lt; 0 then set d_i = -1 , if \displaystyle\sum_{j=1}^N b_i_j \geq 0 then set d_i = 1

    Or is this not sufficient. It's basically the same as what I said before, although I feel it is more precise, and should seem a bit clearer to you.
    OK, it's clearer what you're saying, but I don't think you've followed what I was saying in my previous posts.

    2 things:

    1) check that value of d using the example I gave in my first post. Does it work?

    2) You don't need to be concerned with all the values that "i" can take, only one that makes the RHS a maximum. That's what you have to construct your "d" against, and it's not necessarily going to be all "1"s or all "-1"s.
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    (Original post by ghostwalker)
    OK, it's clearer what you're saying, but I don't think you've followed what I was saying in my previous posts.

    2 things:

    1) check that value of d using the example I gave in my first post. Does it work?

    2) You don't need to be concerned with all the values that "i" can take, only one that makes the RHS a maximum. That's what you have to construct your "d" against, and it's not necessarily going to be all "1"s or all "-1"s.

    Okay so

    1. By my calculations its does work using your first example where by my rules we would have d = (1,1) and ||Bd||\infity = 0 and so the inequality holds.

    2. I still am struggling with this, I don't know how one can say for any square matrix B, only one value in d makes the RHS a maximum, and for this specific d_i I don't see how one could decide whether it should be 1 or -1.
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    (Original post by TheBhramaBull)
    Okay so

    1. By my calculations its does work using your first example where by my rules we would have d = (1,1) and ||Bd||\infity = 0 and so the inequality holds.
    LHS = 0, and RHS = 2. And the last time I checked 0 \ge 2 was false.

    2. I still am struggling with this, I don't know how one can say for any square matrix B, only one value in d makes the RHS a maximum, and for this specific d_i I don't see how one could decide whether it should be 1 or -1.
    I really don't know what else to say; perhaps someone else may be able to explain it better.

    For a given B, there is at least one i, that makes the RHS a maximum. For that value of i, you will need to find "d", i.e. all the d_i, that make the inequality true (there are multiple values of d that will work, you just need to find one.

    It is unfortunate that the same suffix notation is being used for both sides of the inequality which may be confusing things. Perhaps call them d_k on the LHS, and you need to find all the ones that make up a specific d.
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    (Original post by ghostwalker)
    LHS = 0, and RHS = 2. And the last time I checked 0 \ge 2 was false.



    I really don't know what else to say; perhaps someone else may be able to explain it better.

    For a given B, there is at least one i, that makes the RHS a maximum. For that value of i, you will need to find "d", i.e. all the d_i, that make the inequality true (there are multiple values of d that will work, you just need to find one.

    It is unfortunate that the same suffix notation is being used for both sides of the inequality which may be confusing things. Perhaps call them d_k on the LHS, and you need to find all the ones that make up a specific d.
    1. Can you explain this please? I have RHS = |-1 + 1| = 0
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    (Original post by TheBhramaBull)
    1. Can you explain this please? I have RHS = |-1 + 1| = 0
    This is equation labelled (4) in your attachment.


    RHS = max {i=1,2} sum {j=1,2} |b_i,j|

    |b_ij| = 1 for all entries in the example. Hence summing two of them gives you "2".
 
 
 
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