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Matrix derivative of quadratic form?

  1. Dec 11, 2014 #1

    perplexabot

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    1. The problem statement, all variables and given/known data
    Find the derivative of f(X).
    f(X) = transpose(a) * X * b

    where:
    X is nxn
    a and b are n x 1
    ai is the i'th element of a
    Xnm is the element in row n and column m
    let transpose(a) = aT
    let transpose(b) = bT

    2. Relevant equations
    I tried using the product rule, which I assume is wrong.
    I know the answer to be a*bT (but I have not the slightest clue how)

    3. The attempt at a solution
    I tried many things, to the point where punching a whole through my screen doesn't really seem like a bad idea anymore.

    My last attempt was to use the product rule along with some matrix properties, here is what I did:
    d(f)/dX = [d(aT*X)/dX]*b + (aT*X)*[d(b)/dX] = [d(aT*X)/dX]*b = (d/dX)[Σai*X1i Σai*X2i ⋅ ⋅ ⋅ Σai*Xni]*b

    I have no idea what to do next. I have a feeling using the product rule doesn't apply to matrices.
    PLEASE HELP ME!!!

    Thanks for reading...
     
  2. jcsd
  3. Dec 12, 2014 #2

    Stephen Tashi

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    As an example take [itex] n = 2 [/itex]

    [itex] a = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} [/itex]

    [itex] b = \begin {pmatrix} b_1 \\ b_2 \end{pmatrix} [/itex]

    [itex] X = \begin{pmatrix} x_{1\ 1} & x_{1\ 2} \\ x_{2\ 1} & x_{2\ 2} \end{pmatrix} [/itex]

    Then [itex] f(X) = a^T X b [/itex] is a single number. ( We could say it is a 1x1 matrix.)



    Then the answer would be [itex] \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} \begin{pmatrix} b_1 & b_2 \end{pmatrix} [/itex] but what kind of multiplication does that represent? It can be worked as ordinary matrix multiplication to produce a 2x2 matrix.

    [itex] ab^t = \begin{pmatrix} a_1b_1 & a_1b_2 \\ a_2b_1 & a_2 b_2 \end{pmatrix} [/itex]

    I don't know the details of your class materials, so I must guess about how "the derivative" of f(X) is defined.

    One guess is that the derivative of [itex] f [/itex] with respect to [itex] X [/itex] is:

    [itex] \begin{pmatrix} \frac{\partial f}{\partial x_{1\ 1}} &\frac{\partial f}{\partial x_{1\ 2}} \\ \frac{\partial f}{\partial x_{2\ 1}} & \frac{\partial f}{\partial x_{2\ 2}} \end{pmatrix}[/itex]

    Is that the definition you use?
     
    Last edited: Dec 12, 2014
  4. Dec 12, 2014 #3

    RUber

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    Looking at the derivative with respect to the first term (1,1), you could use the limit definition to see what happens in the matrix multiplication.
    ## \lim_{h\to 0} \frac{f(X+\begin{pmatrix} h & 0 \\ 0 & 0 \end{pmatrix})-f(X)}{h} = ? ##
     
  5. Dec 12, 2014 #4

    RUber

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    And to take a stab at why the product rule isn't working the way you had it above...
    You are treating b like a constant, where really you have a composition of functions of X. g(X) = Xb, h(X) = aX, so f(X) = h(g(x)). You should use the chain rule instead of the product rule.
     
  6. Dec 12, 2014 #5

    Ray Vickson

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    He should not use any of those things; it is just a straightforward matter, like saying ##(d/dx) (cx) = c## for constant ##c##. In fact,
    [tex] f(X) = \sum_{i=1}^n \sum_{j=1}^n a_i x_{ij} b_j = \sum_{i,j=1}^n c_{ij} x_{ij}, \;\; c_{ij} = a_i b_j [/tex]
     
    Last edited: Dec 12, 2014
  7. Dec 12, 2014 #6

    perplexabot

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    Yes! However I would like to solve it assuming I don't know what the answer is to be.
    I know you are sort of using the definition of a derivative but I don't get why you have a matrix with h in the top left corner.
    I have a couple questions about what you wrote, if I may.

    ##(d/dx) (cx) = x## for constant ##c## should this not be ##(d/dx) (cx) = c## for constant ##c## ?
    For your equation of f(x): [tex] f(X) = \sum_{i=1}^n \sum_{j=1}^n a_i x_{ij} b_j = \sum_{i,j=1}^n c_{ij} x_{ij}, \;\; c_{ij} = a_i b_j [/tex]
    shouldn't the subscripts of x be reversed (ji instead of ij)?
    Also how did the x go away : ( ??

    Thank you so much!
     
  8. Dec 12, 2014 #7

    Ray Vickson

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    Yes, it should have been ##(d/dx) (cx) = c##; I have edited out the error.

    I don't understand the second question: reverse i and j where? What I wrote was ##a^T X b## in expanded form. And, I don't see why you ask why/how the ##x## went away; it didn't---it is still there. Perhaps you wonder where the ##x## went at the end of the displayed equation? Well, when I said ##c_{ij} = a_i b_j##, that was just the definition of ##c_{ij}##. In other words, I wrote the sum with a ##c_{ij}## in it, so I have to define ##c_{ij}## somewhere. Perhaps I should have said " ... where ##c_{ij} = a_i b_j##".
     
  9. Dec 12, 2014 #8

    perplexabot

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    sorry! my last question is wrong. I read your equation as f(X) = aibj, so it is my fault.
    Ok. I think I understand your equation then.

    But what next? Product rule and chain rule? Or do I simply take the derivative of ##c_{ij}x_{ij}## with respect to ##x_{ij}##? If i do the latter procedure, I just get the sum of ##c_{ij}## terms.
    EDIT: Actually I am wrong once again! You don't get the sum of ##c_{ij}##. You get a column vector with each row being a derivative of ##c_{ij}x_{ij}## with respect to an ##x_{ij}##, right?

    Thank you for your patience : )
     
  10. Dec 13, 2014 #9

    perplexabot

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    I finally was able to do this. I was trying to solve it without considering the elements of the matrix, when i think that is not possible. Here is my solution, for anyone that may be interested in the future. Thanks for the help from everyone.

    gotIt.png
     
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