How to check if a determined set generates a (2x2) Matrix












0












$begingroup$


Having a determined set: $(a) ={left[ {begin{array}{ccccc}1 & 1\1 & 1\end{array} } right] left[ {begin{array}{ccccc}0 & 1\0 & 1\end{array} } right] left[ {begin{array}{ccccc}1 & 0\1 & 0\end{array} } right] left[ {begin{array}{ccccc}0 & 2\0 & 2\end{array} } right] }$



How can I check if it generates $M_{2*2}$?



$M_{2*2}$ represents any space with a 2 by 2 matrix, right?



I know the answer is that the set does not generate the space, but why not, even with only one 2 by 2 matrix making the set, shouldn't it generate, even though not completely, a part of the $M_{2*2}$?






linear-algebra matrices vector-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    By generating, you mean using linear combinations? Then notice that for all generated matrices the two rows are identical.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 26 '18 at 13:53










  • $begingroup$
    @Jean-ClaudeArbaut Yes! The problem is, if $M_{2*2}$ is just a space with 2 by 2 matrices, even if the set doesn't generate all of the space, it still generates it, doesn't it? Thank you for the precious help!
    $endgroup$
    – Miguel Ferreira
    Dec 26 '18 at 14:05










  • $begingroup$
    "even if the set doesn't generate all of the space, it still generates it" How does this make any sense? It does not generate the space of $2times2$ matrices, period. You can find what is generated: it's exactly the space $E$ of $2times2$ matrices with equal rows. Note the 2nd and 3rd are linearly independent and generate $E$, the other two are linear combinations.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 26 '18 at 14:11












  • $begingroup$
    @Jean-ClaudeArbaut Thank you!
    $endgroup$
    – Miguel Ferreira
    Dec 26 '18 at 14:27
















0












$begingroup$


Having a determined set: $(a) ={left[ {begin{array}{ccccc}1 & 1\1 & 1\end{array} } right] left[ {begin{array}{ccccc}0 & 1\0 & 1\end{array} } right] left[ {begin{array}{ccccc}1 & 0\1 & 0\end{array} } right] left[ {begin{array}{ccccc}0 & 2\0 & 2\end{array} } right] }$



How can I check if it generates $M_{2*2}$?



$M_{2*2}$ represents any space with a 2 by 2 matrix, right?



I know the answer is that the set does not generate the space, but why not, even with only one 2 by 2 matrix making the set, shouldn't it generate, even though not completely, a part of the $M_{2*2}$?






linear-algebra matrices vector-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    By generating, you mean using linear combinations? Then notice that for all generated matrices the two rows are identical.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 26 '18 at 13:53










  • $begingroup$
    @Jean-ClaudeArbaut Yes! The problem is, if $M_{2*2}$ is just a space with 2 by 2 matrices, even if the set doesn't generate all of the space, it still generates it, doesn't it? Thank you for the precious help!
    $endgroup$
    – Miguel Ferreira
    Dec 26 '18 at 14:05










  • $begingroup$
    "even if the set doesn't generate all of the space, it still generates it" How does this make any sense? It does not generate the space of $2times2$ matrices, period. You can find what is generated: it's exactly the space $E$ of $2times2$ matrices with equal rows. Note the 2nd and 3rd are linearly independent and generate $E$, the other two are linear combinations.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 26 '18 at 14:11












  • $begingroup$
    @Jean-ClaudeArbaut Thank you!
    $endgroup$
    – Miguel Ferreira
    Dec 26 '18 at 14:27














0












0








0





$begingroup$


Having a determined set: $(a) ={left[ {begin{array}{ccccc}1 & 1\1 & 1\end{array} } right] left[ {begin{array}{ccccc}0 & 1\0 & 1\end{array} } right] left[ {begin{array}{ccccc}1 & 0\1 & 0\end{array} } right] left[ {begin{array}{ccccc}0 & 2\0 & 2\end{array} } right] }$



How can I check if it generates $M_{2*2}$?



$M_{2*2}$ represents any space with a 2 by 2 matrix, right?



I know the answer is that the set does not generate the space, but why not, even with only one 2 by 2 matrix making the set, shouldn't it generate, even though not completely, a part of the $M_{2*2}$?






linear-algebra matrices vector-spaces







share|cite|improve this question











$endgroup$




Having a determined set: $(a) ={left[ {begin{array}{ccccc}1 & 1\1 & 1\end{array} } right] left[ {begin{array}{ccccc}0 & 1\0 & 1\end{array} } right] left[ {begin{array}{ccccc}1 & 0\1 & 0\end{array} } right] left[ {begin{array}{ccccc}0 & 2\0 & 2\end{array} } right] }$



How can I check if it generates $M_{2*2}$?



$M_{2*2}$ represents any space with a 2 by 2 matrix, right?



I know the answer is that the set does not generate the space, but why not, even with only one 2 by 2 matrix making the set, shouldn't it generate, even though not completely, a part of the $M_{2*2}$?






linear-algebra matrices vector-spaces




linear-algebra matrices vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 14:38









José Carlos Santos

163k22131234




163k22131234










asked Dec 26 '18 at 13:48









Miguel FerreiraMiguel Ferreira

864




864








  • 1




    $begingroup$
    By generating, you mean using linear combinations? Then notice that for all generated matrices the two rows are identical.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 26 '18 at 13:53










  • $begingroup$
    @Jean-ClaudeArbaut Yes! The problem is, if $M_{2*2}$ is just a space with 2 by 2 matrices, even if the set doesn't generate all of the space, it still generates it, doesn't it? Thank you for the precious help!
    $endgroup$
    – Miguel Ferreira
    Dec 26 '18 at 14:05










  • $begingroup$
    "even if the set doesn't generate all of the space, it still generates it" How does this make any sense? It does not generate the space of $2times2$ matrices, period. You can find what is generated: it's exactly the space $E$ of $2times2$ matrices with equal rows. Note the 2nd and 3rd are linearly independent and generate $E$, the other two are linear combinations.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 26 '18 at 14:11












  • $begingroup$
    @Jean-ClaudeArbaut Thank you!
    $endgroup$
    – Miguel Ferreira
    Dec 26 '18 at 14:27














  • 1




    $begingroup$
    By generating, you mean using linear combinations? Then notice that for all generated matrices the two rows are identical.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 26 '18 at 13:53










  • $begingroup$
    @Jean-ClaudeArbaut Yes! The problem is, if $M_{2*2}$ is just a space with 2 by 2 matrices, even if the set doesn't generate all of the space, it still generates it, doesn't it? Thank you for the precious help!
    $endgroup$
    – Miguel Ferreira
    Dec 26 '18 at 14:05










  • $begingroup$
    "even if the set doesn't generate all of the space, it still generates it" How does this make any sense? It does not generate the space of $2times2$ matrices, period. You can find what is generated: it's exactly the space $E$ of $2times2$ matrices with equal rows. Note the 2nd and 3rd are linearly independent and generate $E$, the other two are linear combinations.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 26 '18 at 14:11












  • $begingroup$
    @Jean-ClaudeArbaut Thank you!
    $endgroup$
    – Miguel Ferreira
    Dec 26 '18 at 14:27








1




1




$begingroup$
By generating, you mean using linear combinations? Then notice that for all generated matrices the two rows are identical.
$endgroup$
– Jean-Claude Arbaut
Dec 26 '18 at 13:53




$begingroup$
By generating, you mean using linear combinations? Then notice that for all generated matrices the two rows are identical.
$endgroup$
– Jean-Claude Arbaut
Dec 26 '18 at 13:53












$begingroup$
@Jean-ClaudeArbaut Yes! The problem is, if $M_{2*2}$ is just a space with 2 by 2 matrices, even if the set doesn't generate all of the space, it still generates it, doesn't it? Thank you for the precious help!
$endgroup$
– Miguel Ferreira
Dec 26 '18 at 14:05




$begingroup$
@Jean-ClaudeArbaut Yes! The problem is, if $M_{2*2}$ is just a space with 2 by 2 matrices, even if the set doesn't generate all of the space, it still generates it, doesn't it? Thank you for the precious help!
$endgroup$
– Miguel Ferreira
Dec 26 '18 at 14:05












$begingroup$
"even if the set doesn't generate all of the space, it still generates it" How does this make any sense? It does not generate the space of $2times2$ matrices, period. You can find what is generated: it's exactly the space $E$ of $2times2$ matrices with equal rows. Note the 2nd and 3rd are linearly independent and generate $E$, the other two are linear combinations.
$endgroup$
– Jean-Claude Arbaut
Dec 26 '18 at 14:11






$begingroup$
"even if the set doesn't generate all of the space, it still generates it" How does this make any sense? It does not generate the space of $2times2$ matrices, period. You can find what is generated: it's exactly the space $E$ of $2times2$ matrices with equal rows. Note the 2nd and 3rd are linearly independent and generate $E$, the other two are linear combinations.
$endgroup$
– Jean-Claude Arbaut
Dec 26 '18 at 14:11














$begingroup$
@Jean-ClaudeArbaut Thank you!
$endgroup$
– Miguel Ferreira
Dec 26 '18 at 14:27




$begingroup$
@Jean-ClaudeArbaut Thank you!
$endgroup$
– Miguel Ferreira
Dec 26 '18 at 14:27










2 Answers
2






active

oldest

votes


















1












$begingroup$

Well, the space $M_{2times 2}$ has dimension $4$, you have $4$ matrices, and the second and the fourth are clearly l.d.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes! Sure! But if we remove the second and the fourth matrices, will the set still not generate a space $M_{2*2}$? I mean, not the entire space, sure, but some of it? Thank you for the help!
    $endgroup$
    – Miguel Ferreira
    Dec 26 '18 at 14:03










  • $begingroup$
    @MiguelFerreira If you remove matrices, you obviously can't generate a larger space.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 26 '18 at 14:12



















1












$begingroup$

Your set has $4$ matrices, but the fourth one is twice the second on. So, forget the fourth one. Now, the first one is the sum of the second and the third ones. So, forget the first one too. So, what you're after is ths space$$operatorname{span}left{begin{bmatrix}0&1\0&1end{bmatrix},begin{bmatrix}1&0\1&0end{bmatrix}right}.$$This is the space$$left{begin{bmatrix}a&b\a&bend{bmatrix},middle|,a,binmathbb Rright},$$which clearly is not $M_{2,2}(mathbb{R})$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Well, the space $M_{2times 2}$ has dimension $4$, you have $4$ matrices, and the second and the fourth are clearly l.d.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Yes! Sure! But if we remove the second and the fourth matrices, will the set still not generate a space $M_{2*2}$? I mean, not the entire space, sure, but some of it? Thank you for the help!
      $endgroup$
      – Miguel Ferreira
      Dec 26 '18 at 14:03










    • $begingroup$
      @MiguelFerreira If you remove matrices, you obviously can't generate a larger space.
      $endgroup$
      – Jean-Claude Arbaut
      Dec 26 '18 at 14:12
















    1












    $begingroup$

    Well, the space $M_{2times 2}$ has dimension $4$, you have $4$ matrices, and the second and the fourth are clearly l.d.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Yes! Sure! But if we remove the second and the fourth matrices, will the set still not generate a space $M_{2*2}$? I mean, not the entire space, sure, but some of it? Thank you for the help!
      $endgroup$
      – Miguel Ferreira
      Dec 26 '18 at 14:03










    • $begingroup$
      @MiguelFerreira If you remove matrices, you obviously can't generate a larger space.
      $endgroup$
      – Jean-Claude Arbaut
      Dec 26 '18 at 14:12














    1












    1








    1





    $begingroup$

    Well, the space $M_{2times 2}$ has dimension $4$, you have $4$ matrices, and the second and the fourth are clearly l.d.






    share|cite|improve this answer









    $endgroup$



    Well, the space $M_{2times 2}$ has dimension $4$, you have $4$ matrices, and the second and the fourth are clearly l.d.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 26 '18 at 13:53









    ajotatxeajotatxe

    53.8k23990




    53.8k23990












    • $begingroup$
      Yes! Sure! But if we remove the second and the fourth matrices, will the set still not generate a space $M_{2*2}$? I mean, not the entire space, sure, but some of it? Thank you for the help!
      $endgroup$
      – Miguel Ferreira
      Dec 26 '18 at 14:03










    • $begingroup$
      @MiguelFerreira If you remove matrices, you obviously can't generate a larger space.
      $endgroup$
      – Jean-Claude Arbaut
      Dec 26 '18 at 14:12


















    • $begingroup$
      Yes! Sure! But if we remove the second and the fourth matrices, will the set still not generate a space $M_{2*2}$? I mean, not the entire space, sure, but some of it? Thank you for the help!
      $endgroup$
      – Miguel Ferreira
      Dec 26 '18 at 14:03










    • $begingroup$
      @MiguelFerreira If you remove matrices, you obviously can't generate a larger space.
      $endgroup$
      – Jean-Claude Arbaut
      Dec 26 '18 at 14:12
















    $begingroup$
    Yes! Sure! But if we remove the second and the fourth matrices, will the set still not generate a space $M_{2*2}$? I mean, not the entire space, sure, but some of it? Thank you for the help!
    $endgroup$
    – Miguel Ferreira
    Dec 26 '18 at 14:03




    $begingroup$
    Yes! Sure! But if we remove the second and the fourth matrices, will the set still not generate a space $M_{2*2}$? I mean, not the entire space, sure, but some of it? Thank you for the help!
    $endgroup$
    – Miguel Ferreira
    Dec 26 '18 at 14:03












    $begingroup$
    @MiguelFerreira If you remove matrices, you obviously can't generate a larger space.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 26 '18 at 14:12




    $begingroup$
    @MiguelFerreira If you remove matrices, you obviously can't generate a larger space.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 26 '18 at 14:12











    1












    $begingroup$

    Your set has $4$ matrices, but the fourth one is twice the second on. So, forget the fourth one. Now, the first one is the sum of the second and the third ones. So, forget the first one too. So, what you're after is ths space$$operatorname{span}left{begin{bmatrix}0&1\0&1end{bmatrix},begin{bmatrix}1&0\1&0end{bmatrix}right}.$$This is the space$$left{begin{bmatrix}a&b\a&bend{bmatrix},middle|,a,binmathbb Rright},$$which clearly is not $M_{2,2}(mathbb{R})$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Your set has $4$ matrices, but the fourth one is twice the second on. So, forget the fourth one. Now, the first one is the sum of the second and the third ones. So, forget the first one too. So, what you're after is ths space$$operatorname{span}left{begin{bmatrix}0&1\0&1end{bmatrix},begin{bmatrix}1&0\1&0end{bmatrix}right}.$$This is the space$$left{begin{bmatrix}a&b\a&bend{bmatrix},middle|,a,binmathbb Rright},$$which clearly is not $M_{2,2}(mathbb{R})$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Your set has $4$ matrices, but the fourth one is twice the second on. So, forget the fourth one. Now, the first one is the sum of the second and the third ones. So, forget the first one too. So, what you're after is ths space$$operatorname{span}left{begin{bmatrix}0&1\0&1end{bmatrix},begin{bmatrix}1&0\1&0end{bmatrix}right}.$$This is the space$$left{begin{bmatrix}a&b\a&bend{bmatrix},middle|,a,binmathbb Rright},$$which clearly is not $M_{2,2}(mathbb{R})$.






        share|cite|improve this answer









        $endgroup$



        Your set has $4$ matrices, but the fourth one is twice the second on. So, forget the fourth one. Now, the first one is the sum of the second and the third ones. So, forget the first one too. So, what you're after is ths space$$operatorname{span}left{begin{bmatrix}0&1\0&1end{bmatrix},begin{bmatrix}1&0\1&0end{bmatrix}right}.$$This is the space$$left{begin{bmatrix}a&b\a&bend{bmatrix},middle|,a,binmathbb Rright},$$which clearly is not $M_{2,2}(mathbb{R})$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 26 '18 at 14:30









        José Carlos SantosJosé Carlos Santos

        163k22131234




        163k22131234






























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