$L^2$ norm of a matrix: Is this statement true?
$begingroup$
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:
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
asked Dec 18 '18 at 9:07
Ufuk Can BiciciUfuk Can Bicici
1,24711127
$endgroup$
add a comment |
$begingroup$
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:
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
asked Dec 18 '18 at 9:07
Ufuk Can BiciciUfuk Can Bicici
1,24711127
$endgroup$
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
add a comment |
$begingroup$
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:
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
asked Dec 18 '18 at 9:07
Ufuk Can BiciciUfuk Can Bicici
1,24711127
$endgroup$
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:
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
asked Dec 18 '18 at 9:07
Ufuk Can BiciciUfuk Can Bicici
1,24711127
asked Dec 18 '18 at 9:07
Ufuk Can BiciciUfuk Can Bicici
1,24711127
edited Dec 18 '18 at 15:52
user593746
asked Dec 18 '18 at 9:07
Ufuk Can BiciciUfuk Can Bicici
1,24711127
asked Dec 18 '18 at 9:07
Ufuk Can BiciciUfuk Can Bicici
1,24711127
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
add a comment |
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
add a comment |
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 $A^T A$
$$
lVert A rVert_2^2 = sup_{xin mathbb{R}^n} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}^2}{lVert x rVert_{ell^2(mathbb{R}^n)}^2} = sup_{x in mathbb{R}^n} frac{ langle x, A^T A xrangle_{ell^2(mathbb{R}^m)}}{langle x , xrangle_{ell^2(mathbb{R}^n)}} = lambda_{max}(A^T A) .
$$
edited Dec 18 '18 at 13:04
littleO
30.2k648111
answered Dec 18 '18 at 11:37
André SchlichtingAndré Schlichting
1413
$endgroup$
add a comment |
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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 $A^T A$
$$
lVert A rVert_2^2 = sup_{xin mathbb{R}^n} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}^2}{lVert x rVert_{ell^2(mathbb{R}^n)}^2} = sup_{x in mathbb{R}^n} frac{ langle x, A^T A xrangle_{ell^2(mathbb{R}^m)}}{langle x , xrangle_{ell^2(mathbb{R}^n)}} = lambda_{max}(A^T A) .
$$
edited Dec 18 '18 at 13:04
littleO
30.2k648111
answered Dec 18 '18 at 11:37
André SchlichtingAndré Schlichting
1413
$endgroup$
add a comment |
$begingroup$
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 $A^T A$
$$
lVert A rVert_2^2 = sup_{xin mathbb{R}^n} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}^2}{lVert x rVert_{ell^2(mathbb{R}^n)}^2} = sup_{x in mathbb{R}^n} frac{ langle x, A^T A xrangle_{ell^2(mathbb{R}^m)}}{langle x , xrangle_{ell^2(mathbb{R}^n)}} = lambda_{max}(A^T A) .
$$
edited Dec 18 '18 at 13:04
littleO
30.2k648111
answered Dec 18 '18 at 11:37
André SchlichtingAndré Schlichting
1413
$endgroup$
add a comment |
$begingroup$
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 $A^T A$
$$
lVert A rVert_2^2 = sup_{xin mathbb{R}^n} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}^2}{lVert x rVert_{ell^2(mathbb{R}^n)}^2} = sup_{x in mathbb{R}^n} frac{ langle x, A^T A xrangle_{ell^2(mathbb{R}^m)}}{langle x , xrangle_{ell^2(mathbb{R}^n)}} = lambda_{max}(A^T A) .
$$
edited Dec 18 '18 at 13:04
littleO
30.2k648111
answered Dec 18 '18 at 11:37
André SchlichtingAndré Schlichting
1413
$endgroup$
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 $A^T A$
$$
lVert A rVert_2^2 = sup_{xin mathbb{R}^n} frac{ lVert A x rVert_{ell^2(mathbb{R}^m)}^2}{lVert x rVert_{ell^2(mathbb{R}^n)}^2} = sup_{x in mathbb{R}^n} frac{ langle x, A^T A xrangle_{ell^2(mathbb{R}^m)}}{langle x , xrangle_{ell^2(mathbb{R}^n)}} = lambda_{max}(A^T A) .
$$
edited Dec 18 '18 at 13:04
littleO
30.2k648111
answered Dec 18 '18 at 11:37
André SchlichtingAndré Schlichting
1413
edited Dec 18 '18 at 13:04
littleO
30.2k648111
edited Dec 18 '18 at 13:04
littleO
30.2k648111
edited Dec 18 '18 at 13:04
littleO
30.2k648111
30.2k648111
answered Dec 18 '18 at 11:37
André SchlichtingAndré Schlichting
1413
answered Dec 18 '18 at 11:37
André SchlichtingAndré Schlichting
1413
answered Dec 18 '18 at 11:37
André SchlichtingAndré Schlichting
1413
1413
add a comment |
add a comment |
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$begingroup$
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