$L^2$ norm of a matrix: Is this statement true?












2












$begingroup$


I am following Nocedal and Wright's Numerical Optimization book for self study. In the Appendix section of the book, the following matrix norms are defined:
enter image description here



They defined the $l2$ norm of the matrix $A$ as the largest eigenvalue of $(A^TA)^{1/2}$.



But I have also seen the following definition:
$||A||_2 =max_{i:n} sqrtlambda_i$ where $lambda_i$ is the i. eigenvalue of the matrix $A^TA$.



(source: http://www.maths.lth.se/na/courses/FMN081/FMN081-06/lecture6.pdf)



I am not sure how these two definitions are equal. $A^TA$ is a symmetric positive definite matrix, hence it has positive eigenvalues. Assume that $lambda_i$ is its largest eigenvalue. $A^TA$ has a unique positive definite square root with the eigenvalues $sqrt{lambda_i}$. Considering only this PD square root matrix, Nocedal's definition is correct. But there can be other square root matrices of $A^TA$ as well, for which different eigenvalues are the largest. And if $A^TA$ has repeating eigenvalues, it will have infinitely many square roots. Hence I think there is an ambiguity in the Nocedal's definition. Am I missing something here? How can be the book's definition correct?






linear-algebra matrices norm matrix-norms







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$endgroup$








  • 3




    $begingroup$
    The function $M mapsto M^{1/2}$ over the set of positive symmetric matrices is usually implicitely defined such that $M^{1/2}$ is also positive. Like $x mapsto sqrt{x}$ is defined as the positive solution of $y=x^2$.
    $endgroup$
    – nicomezi
    Dec 18 '18 at 9:19












  • $begingroup$
    I would not take those formulas as definitions of the $ell_1, ell_2$, and $ell_infty$ matrix norms. There is one single definition that works in all three cases: the operator norm induced by a vector norm $| cdot |$ is defined by $| A | = sup_{x neq 0} | Ax| / | x|$. The formulas listed are then a consequence of this definition.
    $endgroup$
    – littleO
    Dec 18 '18 at 11:55


















2












$begingroup$


I am following Nocedal and Wright's Numerical Optimization book for self study. In the Appendix section of the book, the following matrix norms are defined:
enter image description here



They defined the $l2$ norm of the matrix $A$ as the largest eigenvalue of $(A^TA)^{1/2}$.



But I have also seen the following definition:
$||A||_2 =max_{i:n} sqrtlambda_i$ where $lambda_i$ is the i. eigenvalue of the matrix $A^TA$.



(source: http://www.maths.lth.se/na/courses/FMN081/FMN081-06/lecture6.pdf)



I am not sure how these two definitions are equal. $A^TA$ is a symmetric positive definite matrix, hence it has positive eigenvalues. Assume that $lambda_i$ is its largest eigenvalue. $A^TA$ has a unique positive definite square root with the eigenvalues $sqrt{lambda_i}$. Considering only this PD square root matrix, Nocedal's definition is correct. But there can be other square root matrices of $A^TA$ as well, for which different eigenvalues are the largest. And if $A^TA$ has repeating eigenvalues, it will have infinitely many square roots. Hence I think there is an ambiguity in the Nocedal's definition. Am I missing something here? How can be the book's definition correct?






linear-algebra matrices norm matrix-norms







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The function $M mapsto M^{1/2}$ over the set of positive symmetric matrices is usually implicitely defined such that $M^{1/2}$ is also positive. Like $x mapsto sqrt{x}$ is defined as the positive solution of $y=x^2$.
    $endgroup$
    – nicomezi
    Dec 18 '18 at 9:19












  • $begingroup$
    I would not take those formulas as definitions of the $ell_1, ell_2$, and $ell_infty$ matrix norms. There is one single definition that works in all three cases: the operator norm induced by a vector norm $| cdot |$ is defined by $| A | = sup_{x neq 0} | Ax| / | x|$. The formulas listed are then a consequence of this definition.
    $endgroup$
    – littleO
    Dec 18 '18 at 11:55
















2












2








2





$begingroup$


I am following Nocedal and Wright's Numerical Optimization book for self study. In the Appendix section of the book, the following matrix norms are defined:
enter image description here



They defined the $l2$ norm of the matrix $A$ as the largest eigenvalue of $(A^TA)^{1/2}$.



But I have also seen the following definition:
$||A||_2 =max_{i:n} sqrtlambda_i$ where $lambda_i$ is the i. eigenvalue of the matrix $A^TA$.



(source: http://www.maths.lth.se/na/courses/FMN081/FMN081-06/lecture6.pdf)



I am not sure how these two definitions are equal. $A^TA$ is a symmetric positive definite matrix, hence it has positive eigenvalues. Assume that $lambda_i$ is its largest eigenvalue. $A^TA$ has a unique positive definite square root with the eigenvalues $sqrt{lambda_i}$. Considering only this PD square root matrix, Nocedal's definition is correct. But there can be other square root matrices of $A^TA$ as well, for which different eigenvalues are the largest. And if $A^TA$ has repeating eigenvalues, it will have infinitely many square roots. Hence I think there is an ambiguity in the Nocedal's definition. Am I missing something here? How can be the book's definition correct?






linear-algebra matrices norm matrix-norms







share|cite|improve this question











$endgroup$




I am following Nocedal and Wright's Numerical Optimization book for self study. In the Appendix section of the book, the following matrix norms are defined:
enter image description here



They defined the $l2$ norm of the matrix $A$ as the largest eigenvalue of $(A^TA)^{1/2}$.



But I have also seen the following definition:
$||A||_2 =max_{i:n} sqrtlambda_i$ where $lambda_i$ is the i. eigenvalue of the matrix $A^TA$.



(source: http://www.maths.lth.se/na/courses/FMN081/FMN081-06/lecture6.pdf)



I am not sure how these two definitions are equal. $A^TA$ is a symmetric positive definite matrix, hence it has positive eigenvalues. Assume that $lambda_i$ is its largest eigenvalue. $A^TA$ has a unique positive definite square root with the eigenvalues $sqrt{lambda_i}$. Considering only this PD square root matrix, Nocedal's definition is correct. But there can be other square root matrices of $A^TA$ as well, for which different eigenvalues are the largest. And if $A^TA$ has repeating eigenvalues, it will have infinitely many square roots. Hence I think there is an ambiguity in the Nocedal's definition. Am I missing something here? How can be the book's definition correct?






linear-algebra matrices norm matrix-norms




linear-algebra matrices norm matrix-norms






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edited Dec 18 '18 at 15:52







user593746

















asked Dec 18 '18 at 9:07









Ufuk Can BiciciUfuk Can Bicici

1,24711127




1,24711127








  • 3




    $begingroup$
    The function $M mapsto M^{1/2}$ over the set of positive symmetric matrices is usually implicitely defined such that $M^{1/2}$ is also positive. Like $x mapsto sqrt{x}$ is defined as the positive solution of $y=x^2$.
    $endgroup$
    – nicomezi
    Dec 18 '18 at 9:19












  • $begingroup$
    I would not take those formulas as definitions of the $ell_1, ell_2$, and $ell_infty$ matrix norms. There is one single definition that works in all three cases: the operator norm induced by a vector norm $| cdot |$ is defined by $| A | = sup_{x neq 0} | Ax| / | x|$. The formulas listed are then a consequence of this definition.
    $endgroup$
    – littleO
    Dec 18 '18 at 11:55
















  • 3




    $begingroup$
    The function $M mapsto M^{1/2}$ over the set of positive symmetric matrices is usually implicitely defined such that $M^{1/2}$ is also positive. Like $x mapsto sqrt{x}$ is defined as the positive solution of $y=x^2$.
    $endgroup$
    – nicomezi
    Dec 18 '18 at 9:19












  • $begingroup$
    I would not take those formulas as definitions of the $ell_1, ell_2$, and $ell_infty$ matrix norms. There is one single definition that works in all three cases: the operator norm induced by a vector norm $| cdot |$ is defined by $| A | = sup_{x neq 0} | Ax| / | x|$. The formulas listed are then a consequence of this definition.
    $endgroup$
    – littleO
    Dec 18 '18 at 11:55










3




3




$begingroup$
The function $M mapsto M^{1/2}$ over the set of positive symmetric matrices is usually implicitely defined such that $M^{1/2}$ is also positive. Like $x mapsto sqrt{x}$ is defined as the positive solution of $y=x^2$.
$endgroup$
– nicomezi
Dec 18 '18 at 9:19






$begingroup$
The function $M mapsto M^{1/2}$ over the set of positive symmetric matrices is usually implicitely defined such that $M^{1/2}$ is also positive. Like $x mapsto sqrt{x}$ is defined as the positive solution of $y=x^2$.
$endgroup$
– nicomezi
Dec 18 '18 at 9:19














$begingroup$
I would not take those formulas as definitions of the $ell_1, ell_2$, and $ell_infty$ matrix norms. There is one single definition that works in all three cases: the operator norm induced by a vector norm $| cdot |$ is defined by $| A | = sup_{x neq 0} | Ax| / | x|$. The formulas listed are then a consequence of this definition.
$endgroup$
– littleO
Dec 18 '18 at 11:55






$begingroup$
I would not take those formulas as definitions of the $ell_1, ell_2$, and $ell_infty$ matrix norms. There is one single definition that works in all three cases: the operator norm induced by a vector norm $| cdot |$ is defined by $| A | = sup_{x neq 0} | Ax| / | x|$. The formulas listed are then a consequence of this definition.
$endgroup$
– littleO
Dec 18 '18 at 11:55












1 Answer
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$begingroup$

To avoid any ambiguity in the definition of the square root of a matrix, it is best to start from $ell^2$ norm of a matrix as the induced norm / operator norm coming from the $ell^2$ norm of the vector spaces. So in your case it seems that $Ain mathbb{R}^{mtimes n}$. Then, it holds by the definition of the operator norm



$$
lVert A rVert_2 = lVert A rVert_{ell^2(mathbb{R}^n) to ell^2(mathbb{R}^m)}
= sup_{xin mathbb{R^n}} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}}{lVert x rVert_{ell^2(mathbb{R}^n)}}
$$



By taking the square and expanding the norm to the $ell^2$-scalar product, one arrives at the Rayleigh quotient of $A^T A$



$$
lVert A rVert_2^2 = sup_{xin mathbb{R}^n} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}^2}{lVert x rVert_{ell^2(mathbb{R}^n)}^2} = sup_{x in mathbb{R}^n} frac{ langle x, A^T A xrangle_{ell^2(mathbb{R}^m)}}{langle x , xrangle_{ell^2(mathbb{R}^n)}} = lambda_{max}(A^T A) .
$$






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    $begingroup$

    To avoid any ambiguity in the definition of the square root of a matrix, it is best to start from $ell^2$ norm of a matrix as the induced norm / operator norm coming from the $ell^2$ norm of the vector spaces. So in your case it seems that $Ain mathbb{R}^{mtimes n}$. Then, it holds by the definition of the operator norm



    $$
    lVert A rVert_2 = lVert A rVert_{ell^2(mathbb{R}^n) to ell^2(mathbb{R}^m)}
    = sup_{xin mathbb{R^n}} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}}{lVert x rVert_{ell^2(mathbb{R}^n)}}
    $$



    By taking the square and expanding the norm to the $ell^2$-scalar product, one arrives at the Rayleigh quotient of $A^T A$



    $$
    lVert A rVert_2^2 = sup_{xin mathbb{R}^n} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}^2}{lVert x rVert_{ell^2(mathbb{R}^n)}^2} = sup_{x in mathbb{R}^n} frac{ langle x, A^T A xrangle_{ell^2(mathbb{R}^m)}}{langle x , xrangle_{ell^2(mathbb{R}^n)}} = lambda_{max}(A^T A) .
    $$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      To avoid any ambiguity in the definition of the square root of a matrix, it is best to start from $ell^2$ norm of a matrix as the induced norm / operator norm coming from the $ell^2$ norm of the vector spaces. So in your case it seems that $Ain mathbb{R}^{mtimes n}$. Then, it holds by the definition of the operator norm



      $$
      lVert A rVert_2 = lVert A rVert_{ell^2(mathbb{R}^n) to ell^2(mathbb{R}^m)}
      = sup_{xin mathbb{R^n}} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}}{lVert x rVert_{ell^2(mathbb{R}^n)}}
      $$



      By taking the square and expanding the norm to the $ell^2$-scalar product, one arrives at the Rayleigh quotient of $A^T A$



      $$
      lVert A rVert_2^2 = sup_{xin mathbb{R}^n} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}^2}{lVert x rVert_{ell^2(mathbb{R}^n)}^2} = sup_{x in mathbb{R}^n} frac{ langle x, A^T A xrangle_{ell^2(mathbb{R}^m)}}{langle x , xrangle_{ell^2(mathbb{R}^n)}} = lambda_{max}(A^T A) .
      $$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        To avoid any ambiguity in the definition of the square root of a matrix, it is best to start from $ell^2$ norm of a matrix as the induced norm / operator norm coming from the $ell^2$ norm of the vector spaces. So in your case it seems that $Ain mathbb{R}^{mtimes n}$. Then, it holds by the definition of the operator norm



        $$
        lVert A rVert_2 = lVert A rVert_{ell^2(mathbb{R}^n) to ell^2(mathbb{R}^m)}
        = sup_{xin mathbb{R^n}} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}}{lVert x rVert_{ell^2(mathbb{R}^n)}}
        $$



        By taking the square and expanding the norm to the $ell^2$-scalar product, one arrives at the Rayleigh quotient of $A^T A$



        $$
        lVert A rVert_2^2 = sup_{xin mathbb{R}^n} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}^2}{lVert x rVert_{ell^2(mathbb{R}^n)}^2} = sup_{x in mathbb{R}^n} frac{ langle x, A^T A xrangle_{ell^2(mathbb{R}^m)}}{langle x , xrangle_{ell^2(mathbb{R}^n)}} = lambda_{max}(A^T A) .
        $$






        share|cite|improve this answer











        $endgroup$



        To avoid any ambiguity in the definition of the square root of a matrix, it is best to start from $ell^2$ norm of a matrix as the induced norm / operator norm coming from the $ell^2$ norm of the vector spaces. So in your case it seems that $Ain mathbb{R}^{mtimes n}$. Then, it holds by the definition of the operator norm



        $$
        lVert A rVert_2 = lVert A rVert_{ell^2(mathbb{R}^n) to ell^2(mathbb{R}^m)}
        = sup_{xin mathbb{R^n}} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}}{lVert x rVert_{ell^2(mathbb{R}^n)}}
        $$



        By taking the square and expanding the norm to the $ell^2$-scalar product, one arrives at the Rayleigh quotient of $A^T A$



        $$
        lVert A rVert_2^2 = sup_{xin mathbb{R}^n} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}^2}{lVert x rVert_{ell^2(mathbb{R}^n)}^2} = sup_{x in mathbb{R}^n} frac{ langle x, A^T A xrangle_{ell^2(mathbb{R}^m)}}{langle x , xrangle_{ell^2(mathbb{R}^n)}} = lambda_{max}(A^T A) .
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 18 '18 at 13:04









        littleO

        30.2k648111




        30.2k648111










        answered Dec 18 '18 at 11:37









        André SchlichtingAndré Schlichting

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        1413






























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