Results 1 to 3 of 3

Math Help - Euclidian norm

  1. #1
    Junior Member
    Joined
    Sep 2009
    From
    Johannesburg, South Africa
    Posts
    71

    Euclidian norm

    We have the Euclidean norm (or l_2-norm) for a vector on the vector space R^m ; \parallel{v}\parallel_2 = \sqrt{\sum_{j=1}^m\mid{v_j}\mid^2}. Given this vector norm on R^m the induced matrix norm is; \parallel{A}\parallel_2 = \sqrt{max. eigenvalue\ of\ A^T{A}}. Let \\ U\ \in\ R^{m\times{m}} be orthogonal ( U^T{U}=I). \\ I want to show that 1) \parallel{Uv}\parallel{{_2^2}}=\parallel{v}\parall  el{_2^2} for all  v\in\ R^m and 2) that \parallel{UA}\parallel_2 = \parallel{AU}\parallel_2 for any  A \in\ R^{m\times{m}}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by bram kierkels View Post
    We have the Euclidean norm (or l_2-norm) for a vector on the vector space R^m ; \parallel{v}\parallel_2 = \sqrt{\sum_{j=1}^m\mid{v_j}\mid^2}. Given this vector norm on R^m the induced matrix norm is; \parallel{A}\parallel_2 = \sqrt{max. eigenvalue\ of\ A^T{A}}. Let \\ U\ \in\ R^{m\times{m}} be orthogonal ( U^T{U}=I). \\ I want to show that 1) \parallel{Uv}\parallel{{_2^2}}=\parallel{v}\parall  el{_2^2} for all  v\in\ R^m and 2) that \parallel{UA}\parallel_2 = \parallel{AU}\parallel_2 for any  A \in\ R^{m\times{m}}
    the second part of your problem is trivial because (UA)^TUA=A^TU^TUA=A^TA and (AU)^TAU=U^TA^TAU=U^{-1}A^TAU and we know that similar matrices have the same eigenvalues.

    for the first part, suppose e_j, \ 1 \leq j \leq m, is the j-th column of U. since U^{-1}=U^T, we have e_i \cdot e_j = \delta_{ij}, where \delta_{ij} is the Kronecker's delta. now suppose v=\begin{bmatrix}v_1 & v_2 & . & . & . & v_m \end{bmatrix}^T. then:

    ||Uv||_2^2=Uv \cdot Uv=\left(\sum_{i=1}^m v_ie_i \right) \cdot \left(\sum_{j=1}^m v_je_j \right)=\sum_{1\leq i,j \leq m}v_iv_j \delta_{ij}=\sum_{j=1}^mv_j^2=||v||_2^2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    From
    Johannesburg, South Africa
    Posts
    71
    Quote Originally Posted by NonCommAlg View Post
    the second part of your problem is trivial because (UA)^TUA=A^TU^TUA=A^TA and (AU)^TAU=U^TA^TAU=U^{-1}A^TAU and we know that similar matrices have the same eigenvalues.

    for the first part, suppose e_j, \ 1 \leq j \leq m, is the j-th column of U. since U^{-1}=U^T, we have e_i \cdot e_j = \delta_{ij}, where \delta_{ij} is the Kronecker's delta. now suppose v=\begin{bmatrix}v_1 & v_2 & . & . & . & v_m \end{bmatrix}^T. then:

    ||Uv||_2^2=Uv \cdot Uv=\left(\sum_{i=1}^m v_ie_i \right) \cdot \left(\sum_{j=1}^m v_je_j \right)=\sum_{1\leq i,j \leq m}v_iv_j \delta_{ij}=\sum_{j=1}^mv_j^2=||v||_2^2.
    .

    Thanks a lot,
    I have one more question about this norm;
    Given a vector norm ||\cdot|| on R^m the induced matrix norm for m x m matrices A is defined by
    ||A||\ =\ max_{v\neq{0}}\frac{||Av||}{||v||}
    That is , ||A|| is the smallest number \alpha such that  ||Av||\leq\alpha||v||\ \forall\ v \in R^m
    So given the Euclidian-norm for a vector as written in the first question why is the induced matrix norm ||A||_2=\sqrt{max.\ eigenvalue A^TA}
    In my book it is given but I donīt see it directly, i think it has something to do with the fact that A^TA is symmetric.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. vector with euclidian norm of 1
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 4th 2011, 07:46 AM
  2. Field norm and ideal norm
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: July 22nd 2010, 08:47 PM
  3. Replies: 3
    Last Post: July 13th 2010, 06:37 PM
  4. Replies: 2
    Last Post: November 7th 2009, 12:13 PM
  5. Vector Norm and Matrix Norm
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 18th 2008, 10:49 AM

Search Tags


/mathhelpforum @mathhelpforum