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Diagonal Scaling of a 2x2 Positive Definite Matrix

  1. Nov 16, 2015 #1
    Given a Positive Definite Matrix ## A \in {\mathbb{R}}^{2 \times 2} ## given by:

    $$ A = \begin{bmatrix}
    {A}_{11} & {A}_{12} \\
    {A}_{12} & {A}_{22}
    \end{bmatrix} $$

    And a Matrix ## B ## Given by:

    $$ B = \begin{bmatrix}
    \frac{1}{\sqrt{{A}_{11}}} & 0 \\
    0 & \frac{1}{\sqrt{{A}_{22}}}
    \end{bmatrix} $$

    Now, defining the Diagonal Scaling of ## A ## given by ## C = B A B ##.
    One could see the main diagonal elements of ## C ## are all ## 1 ##.
    Actually ## C ## is given by:

    $$ C = B A B = \begin{bmatrix}
    1 & \frac{{A}_{12}}{\sqrt{ {A}_{11} {A}_{22} }} \\
    \frac{{A}_{12}}{\sqrt{ {A}_{11} {A}_{22} }} & 1
    \end{bmatrix} $$

    This scaling, intuitively, makes the Matrix closer to the identity Matrix (Smaller off diagonal values, 1 on the main diagonal) and hence improve the Condition Number.
    Yet I couldn't prove it.

    How could one prove ## \kappa \left( C \right) = \kappa \left( B A B \right) \leq \kappa \left( A \right) ##?
     
  2. jcsd
  3. Nov 16, 2015 #2

    fresh_42

    Staff: Mentor

    Can you explain to me what a 'Condition Number' and what ## \kappa \left( C \right) ## is?
     
  4. Nov 16, 2015 #3
    The Condition Number is the division of the largest Eigen Value by the Smallest (Under L2 Norm):

    https://en.wikipedia.org/wiki/Condition_number

    The symbol ## \kappa \left( A \right) ## stands for the condition number of ## A ##.

    Thank You.
     
  5. Nov 17, 2015 #4

    fresh_42

    Staff: Mentor

    The eigenvalues of (2,2)-matrices can be explicitly calculated and therefore the condition numbers. The resulting expression depends on the entries of elements of ##A##. After two dozens of steps or so I think I have proven it with a brute-force attack.
     
    Last edited: Nov 17, 2015
  6. Nov 17, 2015 #5
    Yea, this is what I tried.
    I was after something more elegant.
     
  7. Nov 17, 2015 #6

    fresh_42

    Staff: Mentor

    I don't know whether it hepls or I correctly transformed it. However, defining

    ## T = trace (A) , D = det (A)## and ##det(B) = µ ## or ## µ^2 = det (C) \cdot det (A^{-1})##

    I've been able to reduce the inequation to

    ## \bar{λ}^2 ≤ μ^2 λ^2 ## with the larger eigenvalues ## \bar{λ} ## of ##C## and ##λ## of ##A##

    by using that fact the determinant of a (2,2)-matrix is the product of its eigenvalues and therefore eliminating one of them. Still a little work to do but keeping ## T, D, μ ## as long as possible and taking ##trace (C) = 2## it should be somehow shorter than brute-force.
     
  8. Nov 17, 2015 #7
    Hi,
    Could you show your steps?

    Thank You.
     
  9. Nov 17, 2015 #8

    fresh_42

    Staff: Mentor

    If I understood the conditional number correct then with eigenvalues ##λ_1 ≤ λ_2## we get:

    ##κ(A) = λ_2 λ^{-1}_1##.

    The characteristic polynomial of ##A## is ## det (A - x \cdot 1) = (x - λ_1)(x - λ_2) = x^2 - x \cdot T + D ## and ## D = λ_1 λ_2 ##. Similar is true for ##K(C)##.

    Therefore ##D \cdot κ(A) = λ^{2}_2## and ##\bar{D} \cdot κ(C) = \bar{λ}^{2}_2## if the barred constants are those of ##C##.

    Now I've calculated ##\bar{D} = det (B) det(A) det (B) = det^2 (B) \cdot D = μ^2 D##. Dropping now the indices of ##λ## and dividing by ##D## the inequality reads

    ## \bar{λ}^2 ≤ μ^2 λ^2##. Note that also ##λ_1 + λ_2 = trace (A)## is true. Same with ##B## or ##C##. I didn't use this but maybe it shortens even more by using it to prove the rest.
     
  10. Nov 17, 2015 #9
    You're saying it is equivalent to show your eigen values inequality to the statement above?
     
  11. Nov 17, 2015 #10

    fresh_42

    Staff: Mentor

    Yes, as long as the maximal eigenvalues aren't 0. Just substitute ##λ_1## by ##D \cdot λ^{-1}_2##. Same with ##C = BAB## and finally define ##μ = det(B)##, i.e. ##\bar{D} = \bar{λ}_1 \bar{λ}_2 = det(C) = det^2 (B) \cdot det (A) = μ^2 D##. Note: I did not check your matrix multiplications. And don't try to take square roots. The squares above help a lot.
     
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