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Linear algebra and choosing training vector

  1. Aug 18, 2015 #1

    perplexabot

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    Hey all. So I have been reading this article and have a question I would like to ask. I will be referring to this article extensively so it would be kind of you to open it: http://www.ee.ucr.edu/~yhua/MILCOM_2013_Reprint.pdf

    I believe reading the article is not required to answer my questions (i did personally read it tho up until where I am stuck), but I will refer to the equation numbers in the article instead of typing them out here.

    So, on page 5 of the pdf, there is a matrix, G2, or equation number (22). I am basically trying to figure out how they constructed this matrix! Specifically the left two partitions.

    • Equation (15) gives the definition of G, and shows that G depends on u(k).
    • Equation (12) defines u(k) to be [u1T | u2T | ##\bar{g}## T | 1].
    • The two lines after equation (12) define u1T and u2T
    • Equation (5) defines ##\bar{g}## T, giT, and grT
    • Now that we have all the definitions stated we can go back to equation (22), or G2. It is stated in the two lines above equation (22) that m = 2 (m is the column size of ##\bar{g}## T). Right?
    IT IS AT THIS POINT WHERE MY CONFUSION BEGINS
    • Now if ##\bar{g}## T contains only two elements that means, according to equation (5), giT, and grT would be scalars, right?
    • Continuing with this logic and going to the definition of u2T, the difference of the Kronecker product of two scalars, would be zero! would it not? Hence, u2T = 0
    • According to the previous bullet point, this would make the second column/partition (from the left) of equation (22) equal to 0, which contradicts what is shown in equation (22)
    • Doing the same with u1T and the first column/partition (from the left) of equation (22) would also yield a different answer.
    That is my logic on this. I know I am doing something wrong, but I am not sure what it is. Please help me out here as I have been trying to understand this for a quite a bit. Thank you for read and your time : )
     
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  3. Aug 19, 2015 #2

    perplexabot

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    Hey! I don't know if anyone cares but I believe I got it! It was right in front of my face the whole time. So the G2 I was trying to find was for a real system (not complex), this required the use of equation (21)!!! I considered that earlier but what I thought was that since equation (21) had only 3 partitions while G2 had 4 partitions that equation doesn't work. It turns out tho that the first partition of equation (21) provided three elements when solved. These three elements covered the first two partitions of G2.

    Damn, if feels good to figure something out after days of trial and error. Sorry to take your time and thank you for reading. I am still open to comments of course : )
     
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