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I Can this matrix equation be solved?

  1. May 30, 2017 #1
    Hello, if I have some given vector [itex]c \in R^n[/itex], then I want to find solutions [itex]X \in R^{n\times n}[/itex] to the following equation:
    [tex]
    X C X^T = C
    [/tex]
    where [itex]C = c c^T[/itex]. Certainly [itex]X = I[/itex] is a solution, but I'm looking for any nontrivial solutions. We can also assume [itex]X[/itex] is invertible if that helps.

    This equation seems related to the Sylvester equation. If the solution is nontrivial, I would appreciate if someone could point me in the direction of any work which has been done on this type of equation.
     
  2. jcsd
  3. May 30, 2017 #2

    fresh_42

    Staff: Mentor

    Assume you have chosen a basis such that ##c=(1,0,\ldots,0)^T##. How many solutions do you get?
     
  4. May 30, 2017 #3

    WWGD

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    Isn't C then a scalar? But then ## CXX^{T} ## is a matrix and C is a scalar. Am I missing something?
     
  5. May 30, 2017 #4

    fresh_42

    Staff: Mentor

    Depends on how you write vectors, as a column or as a row: one gives a scalar and the other a rank one matrix. Either case has many solutions.
     
  6. May 30, 2017 #5

    WWGD

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    But how can an ## n \times n ## matrix equal a ## 1 \times 1 ## unless n=1? EDIT: Do you mean we extend the scalar C into a matrix (C,0,0,..,0)?
     
  7. May 30, 2017 #6

    fresh_42

    Staff: Mentor

    You mean if ##C## is a scalar? Then ##XCX^T=C=CI## by the usual identification of the two ones in algebras with unit. Otherwise ##\begin{bmatrix}c_1\\ \vdots \\c_n\end{bmatrix} \cdot \begin{bmatrix}c_1 & \ldots & c_n \end{bmatrix}## is a ##n \times n## matrix of rank ##1##.
     
  8. May 30, 2017 #7

    WWGD

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    I mean, which case do you assume then, that C is ## n \times n ## or a "scalar matrix"?
     
  9. May 31, 2017 #8
    C is an n-by-n matrix of rank 1
     
  10. May 31, 2017 #9

    WWGD

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    Thanks, kind of reminds me of least-square problems, solved by ortho projections.
     
  11. May 31, 2017 #10

    StoneTemplePython

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    It's not clear to me what kind of "solution" OP is looking for since this problem is very much underdetermined. There are some interesting underlying relations here between singular vectors of ##\mathbf X## and the structure of ##\mathbf C##, but that's about all I can add.

    note: if ##\mathbf c = \mathbf 0##, any ##\mathbf X## is allowed.
    - - - -
    Now assume ##\mathbf c \neq \mathbf 0##. Thus ##\mathbf C = \mathbf{cc}^T## is a rank one matrix.

    We can do singular value decomposition on ##\mathbf X## such that ##\mathbf X = \mathbf{U \Sigma V}^T##

    For ##\mathbf{X C X}^T = \mathbf{X c} \mathbf c^T \mathbf X^T= \mathbf{cc}^T##, we know that ##\mathbf c## is given by one (or more) of the right singular vectors of ##\mathbf X## that has a singular value of 1 ##^{**}##. Why? Using the trace operation we see

    ##trace\big(\mathbf{cc}^T\big) = \mathbf c ^T \mathbf c = trace\big(\mathbf{X c} \mathbf c^T \mathbf X^T\big) ##

    ##\mathbf c ^T \mathbf c = trace\big(\mathbf{X c} \mathbf c^T \mathbf X^T\big) = trace\big(\mathbf c^T \mathbf X^T \mathbf{Xc}\big) = \mathbf c^T \mathbf X^T \mathbf{Xc} = \mathbf c^T \mathbf V \mathbf \Sigma^T \mathbf U^T \mathbf {U \Sigma V}^T \mathbf c = \mathbf c^T \mathbf V \mathbf{\Sigma}^2 \mathbf{V}^T \mathbf c = \mathbf y^T \mathbf{\Sigma}^2 \mathbf{y} = \mathbf y^T \mathbf y##

    where ##\mathbf{V}^T \mathbf c := \mathbf y##, and of course ##\big \Vert \mathbf c \big \Vert_2^2 = \big \Vert \mathbf{V}^T \mathbf c \big \Vert_2^2 = \big \Vert \mathbf y \big \Vert_2^2##

    - - - - - - - -
    ##^{**}## to be more accurate: it is possible that that the squared singular values that non-zero elements of ##\mathbf y## gets scaled by are not one-- the exact constraint we need to observe is

    ## \mathbf y^T \mathbf{\Sigma}^2 \mathbf{y} = \sigma_1 ^2 y_1^2 + \sigma_2^2 y_2^2 +.... + \sigma_n ^2 y_n^2 = y_1^2 + y_2^2 +.... + y_n^2 =\mathbf y^T \mathbf y##
    - - - - - - - -

    There's also a lot of flexibility in determining ##\mathbf U##. If ##\mathbf X## is full rank, you can also tease out the constraint that, ##\mathbf c ^T \mathbf c = \mathbf c^T \mathbf U \mathbf{\Sigma}^{-2} \mathbf{U}^T \mathbf c = \mathbf w^T \mathbf{\Sigma}^{-2} \mathbf w = \mathbf w^T \mathbf w##, where ##\mathbf{U}^T \mathbf c := \mathbf w##.

    Losing some generality, here's one particularly easy approach for creating a legal ##\mathbf X##: write out ##\mathbf C = \mathbf {PDP}^T##, where ##\mathbf D## is the zero matrix except the top left entry is given by ##trace\big(\mathbf C\big)##. From here ensure that all values of ##y_k = 0## for ##k \geq 2##, and that ##\sigma_1 = 1##, then set ##\mathbf U := \mathbf P##.

    Again, this whole thing is a very much underdetermined problem so there's a lot of different ways to approach this.
     
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