Is the row space of a matrix A a subspace A? If so, what are the objects of A?

Multi tool use
Multi tool use












1












$begingroup$


I have problems with an assertion that I read in a definition of the row space.
I hope somebody can help me :)



This part is clear:
Let A be a mxn-matrix, with rows $ r_{1},...,r_{m} in K^{n} $ The set of all possible linear combinations of $ r_{1},...,r_{m} $ is the row space of A.
Also the row space is a subspace of $K^{n}$.



My Problem:



In one definition that I have read it is stated that the row space of A is also a subspace of A. So the concrete Matrix A has to be a vector space. What are the objects of A?



Can somebody give me an interpretation of what it exactly means that the row space of A is a subspace of A?



PS: I'm not used to write about math in english, please ask if something doesn't makes sense to you.






matrices vector-spaces terminology







share|cite|improve this question











$endgroup$












  • $begingroup$
    The phrase "subspace of $A$" is simply nonsense. Does the reference use exactly those words? If not you should tell us exactly what it does say - possibly you're misinterpreting or mistranslating sommething.
    $endgroup$
    – David C. Ullrich
    Jan 9 at 15:24










  • $begingroup$
    sry, I pressed "enter" too soon. The comment I wanted to post: I'm pretty sure that I translated "subspace of A" correctly but it seems like they(my source) made a mistake. I'm already satisfied with the answers (and your comment). Thank you :)
    $endgroup$
    – CherryBlossom1878
    Jan 9 at 19:38


















1












$begingroup$


I have problems with an assertion that I read in a definition of the row space.
I hope somebody can help me :)



This part is clear:
Let A be a mxn-matrix, with rows $ r_{1},...,r_{m} in K^{n} $ The set of all possible linear combinations of $ r_{1},...,r_{m} $ is the row space of A.
Also the row space is a subspace of $K^{n}$.



My Problem:



In one definition that I have read it is stated that the row space of A is also a subspace of A. So the concrete Matrix A has to be a vector space. What are the objects of A?



Can somebody give me an interpretation of what it exactly means that the row space of A is a subspace of A?



PS: I'm not used to write about math in english, please ask if something doesn't makes sense to you.






matrices vector-spaces terminology







share|cite|improve this question











$endgroup$












  • $begingroup$
    The phrase "subspace of $A$" is simply nonsense. Does the reference use exactly those words? If not you should tell us exactly what it does say - possibly you're misinterpreting or mistranslating sommething.
    $endgroup$
    – David C. Ullrich
    Jan 9 at 15:24










  • $begingroup$
    sry, I pressed "enter" too soon. The comment I wanted to post: I'm pretty sure that I translated "subspace of A" correctly but it seems like they(my source) made a mistake. I'm already satisfied with the answers (and your comment). Thank you :)
    $endgroup$
    – CherryBlossom1878
    Jan 9 at 19:38
















1












1








1





$begingroup$


I have problems with an assertion that I read in a definition of the row space.
I hope somebody can help me :)



This part is clear:
Let A be a mxn-matrix, with rows $ r_{1},...,r_{m} in K^{n} $ The set of all possible linear combinations of $ r_{1},...,r_{m} $ is the row space of A.
Also the row space is a subspace of $K^{n}$.



My Problem:



In one definition that I have read it is stated that the row space of A is also a subspace of A. So the concrete Matrix A has to be a vector space. What are the objects of A?



Can somebody give me an interpretation of what it exactly means that the row space of A is a subspace of A?



PS: I'm not used to write about math in english, please ask if something doesn't makes sense to you.






matrices vector-spaces terminology







share|cite|improve this question











$endgroup$




I have problems with an assertion that I read in a definition of the row space.
I hope somebody can help me :)



This part is clear:
Let A be a mxn-matrix, with rows $ r_{1},...,r_{m} in K^{n} $ The set of all possible linear combinations of $ r_{1},...,r_{m} $ is the row space of A.
Also the row space is a subspace of $K^{n}$.



My Problem:



In one definition that I have read it is stated that the row space of A is also a subspace of A. So the concrete Matrix A has to be a vector space. What are the objects of A?



Can somebody give me an interpretation of what it exactly means that the row space of A is a subspace of A?



PS: I'm not used to write about math in english, please ask if something doesn't makes sense to you.






matrices vector-spaces terminology




matrices vector-spaces terminology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 14:16









José Carlos Santos

155k22124227




155k22124227










asked Jan 9 at 14:00









CherryBlossom1878CherryBlossom1878

83




83












  • $begingroup$
    The phrase "subspace of $A$" is simply nonsense. Does the reference use exactly those words? If not you should tell us exactly what it does say - possibly you're misinterpreting or mistranslating sommething.
    $endgroup$
    – David C. Ullrich
    Jan 9 at 15:24










  • $begingroup$
    sry, I pressed "enter" too soon. The comment I wanted to post: I'm pretty sure that I translated "subspace of A" correctly but it seems like they(my source) made a mistake. I'm already satisfied with the answers (and your comment). Thank you :)
    $endgroup$
    – CherryBlossom1878
    Jan 9 at 19:38




















  • $begingroup$
    The phrase "subspace of $A$" is simply nonsense. Does the reference use exactly those words? If not you should tell us exactly what it does say - possibly you're misinterpreting or mistranslating sommething.
    $endgroup$
    – David C. Ullrich
    Jan 9 at 15:24










  • $begingroup$
    sry, I pressed "enter" too soon. The comment I wanted to post: I'm pretty sure that I translated "subspace of A" correctly but it seems like they(my source) made a mistake. I'm already satisfied with the answers (and your comment). Thank you :)
    $endgroup$
    – CherryBlossom1878
    Jan 9 at 19:38


















$begingroup$
The phrase "subspace of $A$" is simply nonsense. Does the reference use exactly those words? If not you should tell us exactly what it does say - possibly you're misinterpreting or mistranslating sommething.
$endgroup$
– David C. Ullrich
Jan 9 at 15:24




$begingroup$
The phrase "subspace of $A$" is simply nonsense. Does the reference use exactly those words? If not you should tell us exactly what it does say - possibly you're misinterpreting or mistranslating sommething.
$endgroup$
– David C. Ullrich
Jan 9 at 15:24












$begingroup$
sry, I pressed "enter" too soon. The comment I wanted to post: I'm pretty sure that I translated "subspace of A" correctly but it seems like they(my source) made a mistake. I'm already satisfied with the answers (and your comment). Thank you :)
$endgroup$
– CherryBlossom1878
Jan 9 at 19:38






$begingroup$
sry, I pressed "enter" too soon. The comment I wanted to post: I'm pretty sure that I translated "subspace of A" correctly but it seems like they(my source) made a mistake. I'm already satisfied with the answers (and your comment). Thank you :)
$endgroup$
– CherryBlossom1878
Jan 9 at 19:38












2 Answers
2






active

oldest

votes


















0












$begingroup$

A matrix is not a vector space. Generally, it represents a linear transformation, from one vector space to another.



Now, since the number of columns always equals the dimension of the domain space, we can view the row space as a subspace of $V$ (where $T:Vto W$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer :) If $f:Vrightarrow W $ is a linear transformation and I search for the transformation matrix A of f. Let $ B_{1}={v_{1},...,v_{n}}$ be the basis of V and $ B_{2}={w_{1},...,w_{m}} $ of W, then $f(v_{1})=w=a_{11}w_{1}+...+a_{m1}w_{m}$ with $a_{11},..,a_{m1}$ representing the elements of the first column of A. So if we continue we get n columns which is the dimension of the domain space V, but why should (for example) always $(a_{11},...,a_{1n})^{T}in V$ be true ?
    $endgroup$
    – CherryBlossom1878
    Jan 9 at 20:37












  • $begingroup$
    It's always an $n$-tuple, of elements from the field, so it can be viewed as such.
    $endgroup$
    – Chris Custer
    Jan 9 at 20:41










  • $begingroup$
    Why is it always an n-tuple of elements from the field? If you aren't in the mood to explain it, I would also be happy about a link to a good explanation :)
    $endgroup$
    – CherryBlossom1878
    Jan 9 at 20:54










  • $begingroup$
    I was just thinking I might refer you to Paul Halmos' book Finite Dimensional Vector Spaces. The reason though, that it's an $n$-tuple, where $n=operatorname{dim}V$, is simply that matrix multiplication won't work otherwise. Try a few examples.
    $endgroup$
    – Chris Custer
    Jan 9 at 21:07










  • $begingroup$
    I think that I understand it now, thank you again :) I will check if the book is availible in the library from my university. I think it would be good for my development to also read english literature about math.
    $endgroup$
    – CherryBlossom1878
    Jan 9 at 21:45



















0












$begingroup$

No, a matrix is not a vector space. Asserting that the row space of a matrix is a subspace of that matrix is just an abuse of language.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    A matrix is not a vector space. Generally, it represents a linear transformation, from one vector space to another.



    Now, since the number of columns always equals the dimension of the domain space, we can view the row space as a subspace of $V$ (where $T:Vto W$).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for your answer :) If $f:Vrightarrow W $ is a linear transformation and I search for the transformation matrix A of f. Let $ B_{1}={v_{1},...,v_{n}}$ be the basis of V and $ B_{2}={w_{1},...,w_{m}} $ of W, then $f(v_{1})=w=a_{11}w_{1}+...+a_{m1}w_{m}$ with $a_{11},..,a_{m1}$ representing the elements of the first column of A. So if we continue we get n columns which is the dimension of the domain space V, but why should (for example) always $(a_{11},...,a_{1n})^{T}in V$ be true ?
      $endgroup$
      – CherryBlossom1878
      Jan 9 at 20:37












    • $begingroup$
      It's always an $n$-tuple, of elements from the field, so it can be viewed as such.
      $endgroup$
      – Chris Custer
      Jan 9 at 20:41










    • $begingroup$
      Why is it always an n-tuple of elements from the field? If you aren't in the mood to explain it, I would also be happy about a link to a good explanation :)
      $endgroup$
      – CherryBlossom1878
      Jan 9 at 20:54










    • $begingroup$
      I was just thinking I might refer you to Paul Halmos' book Finite Dimensional Vector Spaces. The reason though, that it's an $n$-tuple, where $n=operatorname{dim}V$, is simply that matrix multiplication won't work otherwise. Try a few examples.
      $endgroup$
      – Chris Custer
      Jan 9 at 21:07










    • $begingroup$
      I think that I understand it now, thank you again :) I will check if the book is availible in the library from my university. I think it would be good for my development to also read english literature about math.
      $endgroup$
      – CherryBlossom1878
      Jan 9 at 21:45
















    0












    $begingroup$

    A matrix is not a vector space. Generally, it represents a linear transformation, from one vector space to another.



    Now, since the number of columns always equals the dimension of the domain space, we can view the row space as a subspace of $V$ (where $T:Vto W$).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for your answer :) If $f:Vrightarrow W $ is a linear transformation and I search for the transformation matrix A of f. Let $ B_{1}={v_{1},...,v_{n}}$ be the basis of V and $ B_{2}={w_{1},...,w_{m}} $ of W, then $f(v_{1})=w=a_{11}w_{1}+...+a_{m1}w_{m}$ with $a_{11},..,a_{m1}$ representing the elements of the first column of A. So if we continue we get n columns which is the dimension of the domain space V, but why should (for example) always $(a_{11},...,a_{1n})^{T}in V$ be true ?
      $endgroup$
      – CherryBlossom1878
      Jan 9 at 20:37












    • $begingroup$
      It's always an $n$-tuple, of elements from the field, so it can be viewed as such.
      $endgroup$
      – Chris Custer
      Jan 9 at 20:41










    • $begingroup$
      Why is it always an n-tuple of elements from the field? If you aren't in the mood to explain it, I would also be happy about a link to a good explanation :)
      $endgroup$
      – CherryBlossom1878
      Jan 9 at 20:54










    • $begingroup$
      I was just thinking I might refer you to Paul Halmos' book Finite Dimensional Vector Spaces. The reason though, that it's an $n$-tuple, where $n=operatorname{dim}V$, is simply that matrix multiplication won't work otherwise. Try a few examples.
      $endgroup$
      – Chris Custer
      Jan 9 at 21:07










    • $begingroup$
      I think that I understand it now, thank you again :) I will check if the book is availible in the library from my university. I think it would be good for my development to also read english literature about math.
      $endgroup$
      – CherryBlossom1878
      Jan 9 at 21:45














    0












    0








    0





    $begingroup$

    A matrix is not a vector space. Generally, it represents a linear transformation, from one vector space to another.



    Now, since the number of columns always equals the dimension of the domain space, we can view the row space as a subspace of $V$ (where $T:Vto W$).






    share|cite|improve this answer









    $endgroup$



    A matrix is not a vector space. Generally, it represents a linear transformation, from one vector space to another.



    Now, since the number of columns always equals the dimension of the domain space, we can view the row space as a subspace of $V$ (where $T:Vto W$).







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 9 at 14:42









    Chris CusterChris Custer

    11.4k3824




    11.4k3824












    • $begingroup$
      Thank you for your answer :) If $f:Vrightarrow W $ is a linear transformation and I search for the transformation matrix A of f. Let $ B_{1}={v_{1},...,v_{n}}$ be the basis of V and $ B_{2}={w_{1},...,w_{m}} $ of W, then $f(v_{1})=w=a_{11}w_{1}+...+a_{m1}w_{m}$ with $a_{11},..,a_{m1}$ representing the elements of the first column of A. So if we continue we get n columns which is the dimension of the domain space V, but why should (for example) always $(a_{11},...,a_{1n})^{T}in V$ be true ?
      $endgroup$
      – CherryBlossom1878
      Jan 9 at 20:37












    • $begingroup$
      It's always an $n$-tuple, of elements from the field, so it can be viewed as such.
      $endgroup$
      – Chris Custer
      Jan 9 at 20:41










    • $begingroup$
      Why is it always an n-tuple of elements from the field? If you aren't in the mood to explain it, I would also be happy about a link to a good explanation :)
      $endgroup$
      – CherryBlossom1878
      Jan 9 at 20:54










    • $begingroup$
      I was just thinking I might refer you to Paul Halmos' book Finite Dimensional Vector Spaces. The reason though, that it's an $n$-tuple, where $n=operatorname{dim}V$, is simply that matrix multiplication won't work otherwise. Try a few examples.
      $endgroup$
      – Chris Custer
      Jan 9 at 21:07










    • $begingroup$
      I think that I understand it now, thank you again :) I will check if the book is availible in the library from my university. I think it would be good for my development to also read english literature about math.
      $endgroup$
      – CherryBlossom1878
      Jan 9 at 21:45


















    • $begingroup$
      Thank you for your answer :) If $f:Vrightarrow W $ is a linear transformation and I search for the transformation matrix A of f. Let $ B_{1}={v_{1},...,v_{n}}$ be the basis of V and $ B_{2}={w_{1},...,w_{m}} $ of W, then $f(v_{1})=w=a_{11}w_{1}+...+a_{m1}w_{m}$ with $a_{11},..,a_{m1}$ representing the elements of the first column of A. So if we continue we get n columns which is the dimension of the domain space V, but why should (for example) always $(a_{11},...,a_{1n})^{T}in V$ be true ?
      $endgroup$
      – CherryBlossom1878
      Jan 9 at 20:37












    • $begingroup$
      It's always an $n$-tuple, of elements from the field, so it can be viewed as such.
      $endgroup$
      – Chris Custer
      Jan 9 at 20:41










    • $begingroup$
      Why is it always an n-tuple of elements from the field? If you aren't in the mood to explain it, I would also be happy about a link to a good explanation :)
      $endgroup$
      – CherryBlossom1878
      Jan 9 at 20:54










    • $begingroup$
      I was just thinking I might refer you to Paul Halmos' book Finite Dimensional Vector Spaces. The reason though, that it's an $n$-tuple, where $n=operatorname{dim}V$, is simply that matrix multiplication won't work otherwise. Try a few examples.
      $endgroup$
      – Chris Custer
      Jan 9 at 21:07










    • $begingroup$
      I think that I understand it now, thank you again :) I will check if the book is availible in the library from my university. I think it would be good for my development to also read english literature about math.
      $endgroup$
      – CherryBlossom1878
      Jan 9 at 21:45
















    $begingroup$
    Thank you for your answer :) If $f:Vrightarrow W $ is a linear transformation and I search for the transformation matrix A of f. Let $ B_{1}={v_{1},...,v_{n}}$ be the basis of V and $ B_{2}={w_{1},...,w_{m}} $ of W, then $f(v_{1})=w=a_{11}w_{1}+...+a_{m1}w_{m}$ with $a_{11},..,a_{m1}$ representing the elements of the first column of A. So if we continue we get n columns which is the dimension of the domain space V, but why should (for example) always $(a_{11},...,a_{1n})^{T}in V$ be true ?
    $endgroup$
    – CherryBlossom1878
    Jan 9 at 20:37






    $begingroup$
    Thank you for your answer :) If $f:Vrightarrow W $ is a linear transformation and I search for the transformation matrix A of f. Let $ B_{1}={v_{1},...,v_{n}}$ be the basis of V and $ B_{2}={w_{1},...,w_{m}} $ of W, then $f(v_{1})=w=a_{11}w_{1}+...+a_{m1}w_{m}$ with $a_{11},..,a_{m1}$ representing the elements of the first column of A. So if we continue we get n columns which is the dimension of the domain space V, but why should (for example) always $(a_{11},...,a_{1n})^{T}in V$ be true ?
    $endgroup$
    – CherryBlossom1878
    Jan 9 at 20:37














    $begingroup$
    It's always an $n$-tuple, of elements from the field, so it can be viewed as such.
    $endgroup$
    – Chris Custer
    Jan 9 at 20:41




    $begingroup$
    It's always an $n$-tuple, of elements from the field, so it can be viewed as such.
    $endgroup$
    – Chris Custer
    Jan 9 at 20:41












    $begingroup$
    Why is it always an n-tuple of elements from the field? If you aren't in the mood to explain it, I would also be happy about a link to a good explanation :)
    $endgroup$
    – CherryBlossom1878
    Jan 9 at 20:54




    $begingroup$
    Why is it always an n-tuple of elements from the field? If you aren't in the mood to explain it, I would also be happy about a link to a good explanation :)
    $endgroup$
    – CherryBlossom1878
    Jan 9 at 20:54












    $begingroup$
    I was just thinking I might refer you to Paul Halmos' book Finite Dimensional Vector Spaces. The reason though, that it's an $n$-tuple, where $n=operatorname{dim}V$, is simply that matrix multiplication won't work otherwise. Try a few examples.
    $endgroup$
    – Chris Custer
    Jan 9 at 21:07




    $begingroup$
    I was just thinking I might refer you to Paul Halmos' book Finite Dimensional Vector Spaces. The reason though, that it's an $n$-tuple, where $n=operatorname{dim}V$, is simply that matrix multiplication won't work otherwise. Try a few examples.
    $endgroup$
    – Chris Custer
    Jan 9 at 21:07












    $begingroup$
    I think that I understand it now, thank you again :) I will check if the book is availible in the library from my university. I think it would be good for my development to also read english literature about math.
    $endgroup$
    – CherryBlossom1878
    Jan 9 at 21:45




    $begingroup$
    I think that I understand it now, thank you again :) I will check if the book is availible in the library from my university. I think it would be good for my development to also read english literature about math.
    $endgroup$
    – CherryBlossom1878
    Jan 9 at 21:45











    0












    $begingroup$

    No, a matrix is not a vector space. Asserting that the row space of a matrix is a subspace of that matrix is just an abuse of language.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      No, a matrix is not a vector space. Asserting that the row space of a matrix is a subspace of that matrix is just an abuse of language.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        No, a matrix is not a vector space. Asserting that the row space of a matrix is a subspace of that matrix is just an abuse of language.






        share|cite|improve this answer









        $endgroup$



        No, a matrix is not a vector space. Asserting that the row space of a matrix is a subspace of that matrix is just an abuse of language.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 14:15









        José Carlos SantosJosé Carlos Santos

        155k22124227




        155k22124227






























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            The Binding of Isaac: Rebirth/Afterbirth

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            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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            menu    Wanted List (rus) Main Projects    FAMI Dumping project    Pirate cart's ID numbers list    My carts collection    Unpublished NES cheats    >All titles     Numbered titles     A titles     B titles     C titles     D titles     E titles     F titles     G titles     H titles     I titles     J titles     K titles     L titles     M titles     N titles     O titles     P titles     Q titles     R titles     S titles     T titles     U titles     V titles     W titles     X titles     Y titles     Z titles     Unique cheats     Buried secrets     Fixed generic cheats     Generic cheats     Secret passwords     Password generators     List of games with no cheats found    Unpublished Game Boy cheats    Unpublished Game Boy Color cheats Misc Stuff    Pirate Market    BONZA    FORTUNA    Decrypted SmyNES ROMs    NES inside (rus)    FCEUMM project Blog    cah4e3.wordpress.com Mail me:    cah4e3 at gmail dot c...
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            Dobbiaco

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            Dobbiaco comune ( IT ) Dobbiaco ( DE ) Toblach Dobbiaco Vecchia Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Guido Bocher (lista civica Indipendenti-Unabhängige) dal 10-5-2015 (2º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Coordinate: 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Altitudine 1 213 m s.l.m. Superficie 125,42 km² Abitanti 3 356 [2] (31-7-2017) Densità 26,76 ab./km² Frazioni Carbonin ( Schluderbach ), Santa Maria ( Aufkirchen ), Valle San Silvestro ( Wahlen ) Comuni confinanti Auronzo di Cadore (BL), Braies, Cortina d'Ampezzo (BL), Innervillgraten (AT-7), San Candido, Sesto, Valle di Casies, Villabassa Altre informazioni Cod. postale 39034 Prefisso 0474 Fuso...
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