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Positive semidefinite matrix

  1. May 25, 2012 #1
    Hello.

    [itex] [/itex]

    1. The problem statement, all variables and given/known data

    Let [itex]x \in \mathbb R^n[/itex] and [itex]t \in \mathbb R[/itex].


    Prove the following equivalence:
    [itex]\left \| x \right \|_2 \leq t \ \ \Leftrightarrow \ \ \begin{pmatrix} t \cdot I_n & x \\ x^T & t \end{pmatrix} \text{is positive semidefinite }[/itex]

    2. Relevant equations

    [itex]\left \| x \right \|_2 = \sqrt{x_1^2+ ... + x_n^2}[/itex] is the euclidean norm and [itex]I_n [/itex] the identity matrix of dimension n.


    3. The attempt at a solution

    I know that a matrix is positive semidefinite if and only if all eigenvalues of the matrix are [itex]\geq 0[/itex].
    My problem is to calculate the eigenvalues of the given matrix.

    Thank your for your help in advance!

    Bye,
    Brian
     
  2. jcsd
  3. May 25, 2012 #2

    lanedance

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    yeah so I would probably start by trying to find the characteristic equation of the matrix

    as they're only 2 non-zero rows in the first column, hopefully it shoudl simplify a fair bit
     
  4. May 25, 2012 #3

    Ray Vickson

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    It is much easier if you forget about eigenvalues and look directly at the _defintion_ of psd (positive semi-definite). Your matrix [tex] A = \begin{pmatrix} t \cdot I_n & x \\ x^T & t \end{pmatrix}[/tex] is psd if, for all [itex] Y \in \mathbb R^{n+1}[/itex] we have
    [itex] Y^T A Y \geq 0.[/itex] Letting
    [tex] Y = \begin{pmatrix} y \\ z \end{pmatrix}, \; y \in \mathbb{R}^n, \; z \in \mathbb{R},[/tex] the quadratic form [itex]Q(y,z) = Y^T A Y[/itex] is easily computed. For alll [itex] y \in \mathbb{R}^n[/itex] we need [itex] Q \geq 0, [/itex] so considered as an optimization problem in [itex] z[/itex] we can derive a simple necessary and sufficient condition involving [itex] ||x|| \text{ and } t.[/itex]
     
  5. May 26, 2012 #4
    Thanks for your help.

    Now I have calculated [itex]Y^T A Y[/itex]. It is:

    [itex]Y^T A Y = \begin{pmatrix}
    y & z \end{pmatrix} \begin{pmatrix} t \cdot I_n & x \\ x^T & t \end{pmatrix} \begin{pmatrix}
    y\\z \end{pmatrix} = \begin{pmatrix}
    y_1t+zx_1 & \cdots & y_nt+zx_n & y_1x_1+...+y_nx_n+zt
    \end{pmatrix}\begin{pmatrix}
    y_1\\\vdots
    \\ y_n
    \\ z
    \end{pmatrix}= \\
    = \sum_{i=1}^n y_i^2 t + 2z \sum_{i=1}^n y_i x_i. [/itex]

    So I have to find out for which [itex]z[/itex] the inequality [itex]t \cdot \sum_{i=1}^n y_i^2 + 2z \cdot \sum_{i=1}^n y_i x_i \geq 0[/itex] holds?

    I don't know how to find out the solution of the inequality. Please can you help me again?

    Thank you in adavance!

    Bye,

    Brian
     
  6. May 26, 2012 #5

    Ray Vickson

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    Try again. I get [itex] Q(y,z) = Y^T A Y = t ||y||^2 + 2 <x,y> z + t z^2,[/itex] where [itex]<.,.>[/itex] denotes the inner product and [itex] ||\cdot ||[/itex] the usual norm. For any y, Q(y,z) must be ≥ 0, which means that as a function of z it cannot have two distinct roots.

    RGV
     
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