1 norm $||_1$, of non square matrix












1












$begingroup$


Does $1$ norm exist for non-square matrices? By $1$ norm I mean



$d (x,y)=sum_{i=1}^{n} |x^i-y^i|, x=(x_1,dots, x_n), y=(y_1,dots, y_n)$



Suppose $A$ is $mtimes n, (mne n)$ matrix what can we say about $|A|_1$? Also, can we say $|A|_1=|A^T|_1= |A^TA|_1=|AA^T|_1$? and $|AB|_1le |A|_1|B|_1$ Thanks for helping.










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$endgroup$












  • $begingroup$
    What property of the norm are you uncertain about $d$ satisfying?
    $endgroup$
    – mathworker21
    Jan 7 at 10:46










  • $begingroup$
    @OscarLanzi I don't understand what are you talking about, if possible give an example and explain.
    $endgroup$
    – Markov
    Jan 7 at 12:05










  • $begingroup$
    Cannot do so in a comment. If this fails to be clear I am forced to delete.
    $endgroup$
    – Oscar Lanzi
    Jan 7 at 12:17
















1












$begingroup$


Does $1$ norm exist for non-square matrices? By $1$ norm I mean



$d (x,y)=sum_{i=1}^{n} |x^i-y^i|, x=(x_1,dots, x_n), y=(y_1,dots, y_n)$



Suppose $A$ is $mtimes n, (mne n)$ matrix what can we say about $|A|_1$? Also, can we say $|A|_1=|A^T|_1= |A^TA|_1=|AA^T|_1$? and $|AB|_1le |A|_1|B|_1$ Thanks for helping.










share|cite|improve this question









$endgroup$












  • $begingroup$
    What property of the norm are you uncertain about $d$ satisfying?
    $endgroup$
    – mathworker21
    Jan 7 at 10:46










  • $begingroup$
    @OscarLanzi I don't understand what are you talking about, if possible give an example and explain.
    $endgroup$
    – Markov
    Jan 7 at 12:05










  • $begingroup$
    Cannot do so in a comment. If this fails to be clear I am forced to delete.
    $endgroup$
    – Oscar Lanzi
    Jan 7 at 12:17














1












1








1





$begingroup$


Does $1$ norm exist for non-square matrices? By $1$ norm I mean



$d (x,y)=sum_{i=1}^{n} |x^i-y^i|, x=(x_1,dots, x_n), y=(y_1,dots, y_n)$



Suppose $A$ is $mtimes n, (mne n)$ matrix what can we say about $|A|_1$? Also, can we say $|A|_1=|A^T|_1= |A^TA|_1=|AA^T|_1$? and $|AB|_1le |A|_1|B|_1$ Thanks for helping.










share|cite|improve this question









$endgroup$




Does $1$ norm exist for non-square matrices? By $1$ norm I mean



$d (x,y)=sum_{i=1}^{n} |x^i-y^i|, x=(x_1,dots, x_n), y=(y_1,dots, y_n)$



Suppose $A$ is $mtimes n, (mne n)$ matrix what can we say about $|A|_1$? Also, can we say $|A|_1=|A^T|_1= |A^TA|_1=|AA^T|_1$? and $|AB|_1le |A|_1|B|_1$ Thanks for helping.







matrix-norms






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share|cite|improve this question










asked Jan 7 at 10:35









MarkovMarkov

17.3k1059180




17.3k1059180












  • $begingroup$
    What property of the norm are you uncertain about $d$ satisfying?
    $endgroup$
    – mathworker21
    Jan 7 at 10:46










  • $begingroup$
    @OscarLanzi I don't understand what are you talking about, if possible give an example and explain.
    $endgroup$
    – Markov
    Jan 7 at 12:05










  • $begingroup$
    Cannot do so in a comment. If this fails to be clear I am forced to delete.
    $endgroup$
    – Oscar Lanzi
    Jan 7 at 12:17


















  • $begingroup$
    What property of the norm are you uncertain about $d$ satisfying?
    $endgroup$
    – mathworker21
    Jan 7 at 10:46










  • $begingroup$
    @OscarLanzi I don't understand what are you talking about, if possible give an example and explain.
    $endgroup$
    – Markov
    Jan 7 at 12:05










  • $begingroup$
    Cannot do so in a comment. If this fails to be clear I am forced to delete.
    $endgroup$
    – Oscar Lanzi
    Jan 7 at 12:17
















$begingroup$
What property of the norm are you uncertain about $d$ satisfying?
$endgroup$
– mathworker21
Jan 7 at 10:46




$begingroup$
What property of the norm are you uncertain about $d$ satisfying?
$endgroup$
– mathworker21
Jan 7 at 10:46












$begingroup$
@OscarLanzi I don't understand what are you talking about, if possible give an example and explain.
$endgroup$
– Markov
Jan 7 at 12:05




$begingroup$
@OscarLanzi I don't understand what are you talking about, if possible give an example and explain.
$endgroup$
– Markov
Jan 7 at 12:05












$begingroup$
Cannot do so in a comment. If this fails to be clear I am forced to delete.
$endgroup$
– Oscar Lanzi
Jan 7 at 12:17




$begingroup$
Cannot do so in a comment. If this fails to be clear I am forced to delete.
$endgroup$
– Oscar Lanzi
Jan 7 at 12:17










1 Answer
1






active

oldest

votes


















3












$begingroup$

There is no problem to define an "entrywise" matrix norm defined as follows:
$$
|A|_1=|vec(A)|_1=sum_{i,j}|A_{i,j}|
$$

see wikipedia, Matrix norms. It is a norm and the induced distance is (if $A$, $B$ have same dimensions):
$$
d(A,B)=sum_{i,j}|A_{i,j}-B_{i,j}|
$$

This norm is a sub-multiplicative norm (see here):
$$
|AB|_1leq|A|_1|B|_1
$$

but, attention, in general:
$$
|A^tA|_1neq|AA^t|_1
$$



Another example of "entrywise" matrix norm often encountered in practice is the Frobenius norm. This norm is defined as follows:



$$
|A|_F = sqrt{text{tr}(A^tA)}=sqrt{sum_{i,j}|A_{i,j}|^2}
$$

Frobenius norm also fulfills the sub-multiplicative property (Cauchy-Schwarz in action, see
here):
$$
|AB|_Fleq |A|_F|B|_F
$$



Compared to the previous case $|.|_1$, like we have $text{tr}(A^tA)=text{tr}(AA^t)$, Frobenius norm also fulfills the property:



$$
|AA^t|_F=|A^tA|_F
$$





To be complete one must also say a word about Matrix norms induced by vector norms. One can define:
$$
|A|_1=sup_{xneq 0}frac{|Ax|_1}{|x|_1}
$$

(attention despite identical the notation, this norm is different from the previously defined $|vec(.)|_1 $, here we have $|A|_1=max_j sum_i|A_{i,j}|$)



An immediate generalization, valid for any $1leq p leq infty$, is:
$$
|A|_p=sup_{xneq 0}frac{|Ax|_p}{|x|_p}
$$



Such norms are called Matrix norms induced by vector norms, they automatically fulfill the sub-multiplicative property:
$$
|AB|_p leq |A|_p |B|_p
$$



Attention: however in general,
$$
|AA^t|_p neq |A^tA|_p
$$





Also note this important fact, as we are in finite dimension all these previously defined matrix norms are equivalent in the sense that for any two matrix norm $|.|_alpha$ and $|.|_beta$ it exists $r$ and $s$ such that:
$$
forall A, r|A|_alphaleq |A|_beta leq s|A|_alpha
$$

in peculiar if a sequence $nrightarrow (A)_n$ is convergent for a given matrix norm it is also convergent for all the other norms.






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    1 Answer
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    active

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    active

    oldest

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    3












    $begingroup$

    There is no problem to define an "entrywise" matrix norm defined as follows:
    $$
    |A|_1=|vec(A)|_1=sum_{i,j}|A_{i,j}|
    $$

    see wikipedia, Matrix norms. It is a norm and the induced distance is (if $A$, $B$ have same dimensions):
    $$
    d(A,B)=sum_{i,j}|A_{i,j}-B_{i,j}|
    $$

    This norm is a sub-multiplicative norm (see here):
    $$
    |AB|_1leq|A|_1|B|_1
    $$

    but, attention, in general:
    $$
    |A^tA|_1neq|AA^t|_1
    $$



    Another example of "entrywise" matrix norm often encountered in practice is the Frobenius norm. This norm is defined as follows:



    $$
    |A|_F = sqrt{text{tr}(A^tA)}=sqrt{sum_{i,j}|A_{i,j}|^2}
    $$

    Frobenius norm also fulfills the sub-multiplicative property (Cauchy-Schwarz in action, see
    here):
    $$
    |AB|_Fleq |A|_F|B|_F
    $$



    Compared to the previous case $|.|_1$, like we have $text{tr}(A^tA)=text{tr}(AA^t)$, Frobenius norm also fulfills the property:



    $$
    |AA^t|_F=|A^tA|_F
    $$





    To be complete one must also say a word about Matrix norms induced by vector norms. One can define:
    $$
    |A|_1=sup_{xneq 0}frac{|Ax|_1}{|x|_1}
    $$

    (attention despite identical the notation, this norm is different from the previously defined $|vec(.)|_1 $, here we have $|A|_1=max_j sum_i|A_{i,j}|$)



    An immediate generalization, valid for any $1leq p leq infty$, is:
    $$
    |A|_p=sup_{xneq 0}frac{|Ax|_p}{|x|_p}
    $$



    Such norms are called Matrix norms induced by vector norms, they automatically fulfill the sub-multiplicative property:
    $$
    |AB|_p leq |A|_p |B|_p
    $$



    Attention: however in general,
    $$
    |AA^t|_p neq |A^tA|_p
    $$





    Also note this important fact, as we are in finite dimension all these previously defined matrix norms are equivalent in the sense that for any two matrix norm $|.|_alpha$ and $|.|_beta$ it exists $r$ and $s$ such that:
    $$
    forall A, r|A|_alphaleq |A|_beta leq s|A|_alpha
    $$

    in peculiar if a sequence $nrightarrow (A)_n$ is convergent for a given matrix norm it is also convergent for all the other norms.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      There is no problem to define an "entrywise" matrix norm defined as follows:
      $$
      |A|_1=|vec(A)|_1=sum_{i,j}|A_{i,j}|
      $$

      see wikipedia, Matrix norms. It is a norm and the induced distance is (if $A$, $B$ have same dimensions):
      $$
      d(A,B)=sum_{i,j}|A_{i,j}-B_{i,j}|
      $$

      This norm is a sub-multiplicative norm (see here):
      $$
      |AB|_1leq|A|_1|B|_1
      $$

      but, attention, in general:
      $$
      |A^tA|_1neq|AA^t|_1
      $$



      Another example of "entrywise" matrix norm often encountered in practice is the Frobenius norm. This norm is defined as follows:



      $$
      |A|_F = sqrt{text{tr}(A^tA)}=sqrt{sum_{i,j}|A_{i,j}|^2}
      $$

      Frobenius norm also fulfills the sub-multiplicative property (Cauchy-Schwarz in action, see
      here):
      $$
      |AB|_Fleq |A|_F|B|_F
      $$



      Compared to the previous case $|.|_1$, like we have $text{tr}(A^tA)=text{tr}(AA^t)$, Frobenius norm also fulfills the property:



      $$
      |AA^t|_F=|A^tA|_F
      $$





      To be complete one must also say a word about Matrix norms induced by vector norms. One can define:
      $$
      |A|_1=sup_{xneq 0}frac{|Ax|_1}{|x|_1}
      $$

      (attention despite identical the notation, this norm is different from the previously defined $|vec(.)|_1 $, here we have $|A|_1=max_j sum_i|A_{i,j}|$)



      An immediate generalization, valid for any $1leq p leq infty$, is:
      $$
      |A|_p=sup_{xneq 0}frac{|Ax|_p}{|x|_p}
      $$



      Such norms are called Matrix norms induced by vector norms, they automatically fulfill the sub-multiplicative property:
      $$
      |AB|_p leq |A|_p |B|_p
      $$



      Attention: however in general,
      $$
      |AA^t|_p neq |A^tA|_p
      $$





      Also note this important fact, as we are in finite dimension all these previously defined matrix norms are equivalent in the sense that for any two matrix norm $|.|_alpha$ and $|.|_beta$ it exists $r$ and $s$ such that:
      $$
      forall A, r|A|_alphaleq |A|_beta leq s|A|_alpha
      $$

      in peculiar if a sequence $nrightarrow (A)_n$ is convergent for a given matrix norm it is also convergent for all the other norms.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        There is no problem to define an "entrywise" matrix norm defined as follows:
        $$
        |A|_1=|vec(A)|_1=sum_{i,j}|A_{i,j}|
        $$

        see wikipedia, Matrix norms. It is a norm and the induced distance is (if $A$, $B$ have same dimensions):
        $$
        d(A,B)=sum_{i,j}|A_{i,j}-B_{i,j}|
        $$

        This norm is a sub-multiplicative norm (see here):
        $$
        |AB|_1leq|A|_1|B|_1
        $$

        but, attention, in general:
        $$
        |A^tA|_1neq|AA^t|_1
        $$



        Another example of "entrywise" matrix norm often encountered in practice is the Frobenius norm. This norm is defined as follows:



        $$
        |A|_F = sqrt{text{tr}(A^tA)}=sqrt{sum_{i,j}|A_{i,j}|^2}
        $$

        Frobenius norm also fulfills the sub-multiplicative property (Cauchy-Schwarz in action, see
        here):
        $$
        |AB|_Fleq |A|_F|B|_F
        $$



        Compared to the previous case $|.|_1$, like we have $text{tr}(A^tA)=text{tr}(AA^t)$, Frobenius norm also fulfills the property:



        $$
        |AA^t|_F=|A^tA|_F
        $$





        To be complete one must also say a word about Matrix norms induced by vector norms. One can define:
        $$
        |A|_1=sup_{xneq 0}frac{|Ax|_1}{|x|_1}
        $$

        (attention despite identical the notation, this norm is different from the previously defined $|vec(.)|_1 $, here we have $|A|_1=max_j sum_i|A_{i,j}|$)



        An immediate generalization, valid for any $1leq p leq infty$, is:
        $$
        |A|_p=sup_{xneq 0}frac{|Ax|_p}{|x|_p}
        $$



        Such norms are called Matrix norms induced by vector norms, they automatically fulfill the sub-multiplicative property:
        $$
        |AB|_p leq |A|_p |B|_p
        $$



        Attention: however in general,
        $$
        |AA^t|_p neq |A^tA|_p
        $$





        Also note this important fact, as we are in finite dimension all these previously defined matrix norms are equivalent in the sense that for any two matrix norm $|.|_alpha$ and $|.|_beta$ it exists $r$ and $s$ such that:
        $$
        forall A, r|A|_alphaleq |A|_beta leq s|A|_alpha
        $$

        in peculiar if a sequence $nrightarrow (A)_n$ is convergent for a given matrix norm it is also convergent for all the other norms.






        share|cite|improve this answer











        $endgroup$



        There is no problem to define an "entrywise" matrix norm defined as follows:
        $$
        |A|_1=|vec(A)|_1=sum_{i,j}|A_{i,j}|
        $$

        see wikipedia, Matrix norms. It is a norm and the induced distance is (if $A$, $B$ have same dimensions):
        $$
        d(A,B)=sum_{i,j}|A_{i,j}-B_{i,j}|
        $$

        This norm is a sub-multiplicative norm (see here):
        $$
        |AB|_1leq|A|_1|B|_1
        $$

        but, attention, in general:
        $$
        |A^tA|_1neq|AA^t|_1
        $$



        Another example of "entrywise" matrix norm often encountered in practice is the Frobenius norm. This norm is defined as follows:



        $$
        |A|_F = sqrt{text{tr}(A^tA)}=sqrt{sum_{i,j}|A_{i,j}|^2}
        $$

        Frobenius norm also fulfills the sub-multiplicative property (Cauchy-Schwarz in action, see
        here):
        $$
        |AB|_Fleq |A|_F|B|_F
        $$



        Compared to the previous case $|.|_1$, like we have $text{tr}(A^tA)=text{tr}(AA^t)$, Frobenius norm also fulfills the property:



        $$
        |AA^t|_F=|A^tA|_F
        $$





        To be complete one must also say a word about Matrix norms induced by vector norms. One can define:
        $$
        |A|_1=sup_{xneq 0}frac{|Ax|_1}{|x|_1}
        $$

        (attention despite identical the notation, this norm is different from the previously defined $|vec(.)|_1 $, here we have $|A|_1=max_j sum_i|A_{i,j}|$)



        An immediate generalization, valid for any $1leq p leq infty$, is:
        $$
        |A|_p=sup_{xneq 0}frac{|Ax|_p}{|x|_p}
        $$



        Such norms are called Matrix norms induced by vector norms, they automatically fulfill the sub-multiplicative property:
        $$
        |AB|_p leq |A|_p |B|_p
        $$



        Attention: however in general,
        $$
        |AA^t|_p neq |A^tA|_p
        $$





        Also note this important fact, as we are in finite dimension all these previously defined matrix norms are equivalent in the sense that for any two matrix norm $|.|_alpha$ and $|.|_beta$ it exists $r$ and $s$ such that:
        $$
        forall A, r|A|_alphaleq |A|_beta leq s|A|_alpha
        $$

        in peculiar if a sequence $nrightarrow (A)_n$ is convergent for a given matrix norm it is also convergent for all the other norms.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 7 at 12:07

























        answered Jan 7 at 10:45









        Picaud VincentPicaud Vincent

        1,52439




        1,52439






























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