A set of vectors in $R^2$ for which $x + y$ stays in the set but $frac12x$ may be outside.












3












$begingroup$


These questions refer to subspace.



One requirement can be met while the other fails. Show this by finding

(a) A set of vectors in $R^2$ for which $x + y$ stays in the set but $frac12x$ may be outside.

(b) A set of vectors in $R^2$ (other than two quarter-planes) for which every $cx$ stays in the set but $x + y$ may be outside.



these are the answers my textbook gives,

(a) The vectors with integer components allow addition, but not multiplication by $frac12$

This is saying that you can't multiply a vector by a non-integer? why not?



(b) Remove the x axis from the xy plane (but leave the origin). Multiplication by any c is allowed but not all vector additions.

If you remove the x-axis then you are only left with vectors on the y-axis, so shouldn't it still be closed under addition?






linear-algebra







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    These questions refer to subspace.



    One requirement can be met while the other fails. Show this by finding

    (a) A set of vectors in $R^2$ for which $x + y$ stays in the set but $frac12x$ may be outside.

    (b) A set of vectors in $R^2$ (other than two quarter-planes) for which every $cx$ stays in the set but $x + y$ may be outside.



    these are the answers my textbook gives,

    (a) The vectors with integer components allow addition, but not multiplication by $frac12$

    This is saying that you can't multiply a vector by a non-integer? why not?



    (b) Remove the x axis from the xy plane (but leave the origin). Multiplication by any c is allowed but not all vector additions.

    If you remove the x-axis then you are only left with vectors on the y-axis, so shouldn't it still be closed under addition?






    linear-algebra







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      These questions refer to subspace.



      One requirement can be met while the other fails. Show this by finding

      (a) A set of vectors in $R^2$ for which $x + y$ stays in the set but $frac12x$ may be outside.

      (b) A set of vectors in $R^2$ (other than two quarter-planes) for which every $cx$ stays in the set but $x + y$ may be outside.



      these are the answers my textbook gives,

      (a) The vectors with integer components allow addition, but not multiplication by $frac12$

      This is saying that you can't multiply a vector by a non-integer? why not?



      (b) Remove the x axis from the xy plane (but leave the origin). Multiplication by any c is allowed but not all vector additions.

      If you remove the x-axis then you are only left with vectors on the y-axis, so shouldn't it still be closed under addition?






      linear-algebra







      share|cite|improve this question









      $endgroup$




      These questions refer to subspace.



      One requirement can be met while the other fails. Show this by finding

      (a) A set of vectors in $R^2$ for which $x + y$ stays in the set but $frac12x$ may be outside.

      (b) A set of vectors in $R^2$ (other than two quarter-planes) for which every $cx$ stays in the set but $x + y$ may be outside.



      these are the answers my textbook gives,

      (a) The vectors with integer components allow addition, but not multiplication by $frac12$

      This is saying that you can't multiply a vector by a non-integer? why not?



      (b) Remove the x axis from the xy plane (but leave the origin). Multiplication by any c is allowed but not all vector additions.

      If you remove the x-axis then you are only left with vectors on the y-axis, so shouldn't it still be closed under addition?






      linear-algebra




      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Feb 26 '16 at 22:21









      idknuttinidknuttin

      1,2071631




      1,2071631






















          4 Answers
          4






          active

          oldest

          votes


















          1












          $begingroup$


          1. $frac12(1, 1)$ is not an integer vector.

          2. The plane with the $x$-axis removed contains all points with the first coordinates non-zero.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            a set of integer vectors can be considered a subset?
            $endgroup$
            – idknuttin
            Feb 26 '16 at 22:26










          • $begingroup$
            @idknuttin ??? If it is a set, why would it not be a subset?
            $endgroup$
            – Igor Rivin
            Feb 26 '16 at 22:27










          • $begingroup$
            I meant a set of integer vectors can be considered a subspace?
            $endgroup$
            – idknuttin
            Feb 26 '16 at 22:30










          • $begingroup$
            @idknuttin No, but it does not say it should be a subspace.
            $endgroup$
            – Igor Rivin
            Feb 26 '16 at 22:30



















          2












          $begingroup$

          (a) If you multiply a nonzero vector with integer components by a noninteger, the result is not in "the set of vectors with integer components" anymore.



          (b) if you remove the $x$-axis, you still have points like $(1,1)$ and $(1,-1)$ even though you don't have their sum.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Technically scaling by 0 makes b) not work.
            $endgroup$
            – Improve
            Feb 26 '16 at 22:44












          • $begingroup$
            @improve since the book said "but leave the origin" and the question was about closure under addition, I didn't think that was worth mentioning here. But you're certainly correct.
            $endgroup$
            – Mark S.
            Feb 26 '16 at 22:46










          • $begingroup$
            I did not read the entire question.
            $endgroup$
            – Improve
            Feb 26 '16 at 22:47





















          1












          $begingroup$

          They are asking for sets that satisfy some property of vector spaces, but that are not vector spaces.



          (a) the only operation they are allowing is addition, thus clearly vectors with entire componends only add to vectors with entire components, it's not a vector space since $mathbb{Z}$ is not a field (you don't have $n^{-1}$.



          (b) You are assuming it's a vector space, it is not, thats why the set doesn't collapses to only the y-axis.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            b)



            imagine a line that is exactly on the y-axis.
            you have a vector v = (0,1) , so cv always on the y-axis but for example
            (0,1) + (1,1) = (1,2) wich is not on the line (y-axix)






            share|cite|improve this answer









            $endgroup$














              Your Answer








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              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$


              1. $frac12(1, 1)$ is not an integer vector.

              2. The plane with the $x$-axis removed contains all points with the first coordinates non-zero.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                a set of integer vectors can be considered a subset?
                $endgroup$
                – idknuttin
                Feb 26 '16 at 22:26










              • $begingroup$
                @idknuttin ??? If it is a set, why would it not be a subset?
                $endgroup$
                – Igor Rivin
                Feb 26 '16 at 22:27










              • $begingroup$
                I meant a set of integer vectors can be considered a subspace?
                $endgroup$
                – idknuttin
                Feb 26 '16 at 22:30










              • $begingroup$
                @idknuttin No, but it does not say it should be a subspace.
                $endgroup$
                – Igor Rivin
                Feb 26 '16 at 22:30
















              1












              $begingroup$


              1. $frac12(1, 1)$ is not an integer vector.

              2. The plane with the $x$-axis removed contains all points with the first coordinates non-zero.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                a set of integer vectors can be considered a subset?
                $endgroup$
                – idknuttin
                Feb 26 '16 at 22:26










              • $begingroup$
                @idknuttin ??? If it is a set, why would it not be a subset?
                $endgroup$
                – Igor Rivin
                Feb 26 '16 at 22:27










              • $begingroup$
                I meant a set of integer vectors can be considered a subspace?
                $endgroup$
                – idknuttin
                Feb 26 '16 at 22:30










              • $begingroup$
                @idknuttin No, but it does not say it should be a subspace.
                $endgroup$
                – Igor Rivin
                Feb 26 '16 at 22:30














              1












              1








              1





              $begingroup$


              1. $frac12(1, 1)$ is not an integer vector.

              2. The plane with the $x$-axis removed contains all points with the first coordinates non-zero.






              share|cite|improve this answer









              $endgroup$




              1. $frac12(1, 1)$ is not an integer vector.

              2. The plane with the $x$-axis removed contains all points with the first coordinates non-zero.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Feb 26 '16 at 22:24









              Igor RivinIgor Rivin

              16.1k11234




              16.1k11234












              • $begingroup$
                a set of integer vectors can be considered a subset?
                $endgroup$
                – idknuttin
                Feb 26 '16 at 22:26










              • $begingroup$
                @idknuttin ??? If it is a set, why would it not be a subset?
                $endgroup$
                – Igor Rivin
                Feb 26 '16 at 22:27










              • $begingroup$
                I meant a set of integer vectors can be considered a subspace?
                $endgroup$
                – idknuttin
                Feb 26 '16 at 22:30










              • $begingroup$
                @idknuttin No, but it does not say it should be a subspace.
                $endgroup$
                – Igor Rivin
                Feb 26 '16 at 22:30


















              • $begingroup$
                a set of integer vectors can be considered a subset?
                $endgroup$
                – idknuttin
                Feb 26 '16 at 22:26










              • $begingroup$
                @idknuttin ??? If it is a set, why would it not be a subset?
                $endgroup$
                – Igor Rivin
                Feb 26 '16 at 22:27










              • $begingroup$
                I meant a set of integer vectors can be considered a subspace?
                $endgroup$
                – idknuttin
                Feb 26 '16 at 22:30










              • $begingroup$
                @idknuttin No, but it does not say it should be a subspace.
                $endgroup$
                – Igor Rivin
                Feb 26 '16 at 22:30
















              $begingroup$
              a set of integer vectors can be considered a subset?
              $endgroup$
              – idknuttin
              Feb 26 '16 at 22:26




              $begingroup$
              a set of integer vectors can be considered a subset?
              $endgroup$
              – idknuttin
              Feb 26 '16 at 22:26












              $begingroup$
              @idknuttin ??? If it is a set, why would it not be a subset?
              $endgroup$
              – Igor Rivin
              Feb 26 '16 at 22:27




              $begingroup$
              @idknuttin ??? If it is a set, why would it not be a subset?
              $endgroup$
              – Igor Rivin
              Feb 26 '16 at 22:27












              $begingroup$
              I meant a set of integer vectors can be considered a subspace?
              $endgroup$
              – idknuttin
              Feb 26 '16 at 22:30




              $begingroup$
              I meant a set of integer vectors can be considered a subspace?
              $endgroup$
              – idknuttin
              Feb 26 '16 at 22:30












              $begingroup$
              @idknuttin No, but it does not say it should be a subspace.
              $endgroup$
              – Igor Rivin
              Feb 26 '16 at 22:30




              $begingroup$
              @idknuttin No, but it does not say it should be a subspace.
              $endgroup$
              – Igor Rivin
              Feb 26 '16 at 22:30











              2












              $begingroup$

              (a) If you multiply a nonzero vector with integer components by a noninteger, the result is not in "the set of vectors with integer components" anymore.



              (b) if you remove the $x$-axis, you still have points like $(1,1)$ and $(1,-1)$ even though you don't have their sum.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Technically scaling by 0 makes b) not work.
                $endgroup$
                – Improve
                Feb 26 '16 at 22:44












              • $begingroup$
                @improve since the book said "but leave the origin" and the question was about closure under addition, I didn't think that was worth mentioning here. But you're certainly correct.
                $endgroup$
                – Mark S.
                Feb 26 '16 at 22:46










              • $begingroup$
                I did not read the entire question.
                $endgroup$
                – Improve
                Feb 26 '16 at 22:47


















              2












              $begingroup$

              (a) If you multiply a nonzero vector with integer components by a noninteger, the result is not in "the set of vectors with integer components" anymore.



              (b) if you remove the $x$-axis, you still have points like $(1,1)$ and $(1,-1)$ even though you don't have their sum.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Technically scaling by 0 makes b) not work.
                $endgroup$
                – Improve
                Feb 26 '16 at 22:44












              • $begingroup$
                @improve since the book said "but leave the origin" and the question was about closure under addition, I didn't think that was worth mentioning here. But you're certainly correct.
                $endgroup$
                – Mark S.
                Feb 26 '16 at 22:46










              • $begingroup$
                I did not read the entire question.
                $endgroup$
                – Improve
                Feb 26 '16 at 22:47
















              2












              2








              2





              $begingroup$

              (a) If you multiply a nonzero vector with integer components by a noninteger, the result is not in "the set of vectors with integer components" anymore.



              (b) if you remove the $x$-axis, you still have points like $(1,1)$ and $(1,-1)$ even though you don't have their sum.






              share|cite|improve this answer









              $endgroup$



              (a) If you multiply a nonzero vector with integer components by a noninteger, the result is not in "the set of vectors with integer components" anymore.



              (b) if you remove the $x$-axis, you still have points like $(1,1)$ and $(1,-1)$ even though you don't have their sum.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Feb 26 '16 at 22:26









              Mark S.Mark S.

              12.5k22772




              12.5k22772












              • $begingroup$
                Technically scaling by 0 makes b) not work.
                $endgroup$
                – Improve
                Feb 26 '16 at 22:44












              • $begingroup$
                @improve since the book said "but leave the origin" and the question was about closure under addition, I didn't think that was worth mentioning here. But you're certainly correct.
                $endgroup$
                – Mark S.
                Feb 26 '16 at 22:46










              • $begingroup$
                I did not read the entire question.
                $endgroup$
                – Improve
                Feb 26 '16 at 22:47




















              • $begingroup$
                Technically scaling by 0 makes b) not work.
                $endgroup$
                – Improve
                Feb 26 '16 at 22:44












              • $begingroup$
                @improve since the book said "but leave the origin" and the question was about closure under addition, I didn't think that was worth mentioning here. But you're certainly correct.
                $endgroup$
                – Mark S.
                Feb 26 '16 at 22:46










              • $begingroup$
                I did not read the entire question.
                $endgroup$
                – Improve
                Feb 26 '16 at 22:47


















              $begingroup$
              Technically scaling by 0 makes b) not work.
              $endgroup$
              – Improve
              Feb 26 '16 at 22:44






              $begingroup$
              Technically scaling by 0 makes b) not work.
              $endgroup$
              – Improve
              Feb 26 '16 at 22:44














              $begingroup$
              @improve since the book said "but leave the origin" and the question was about closure under addition, I didn't think that was worth mentioning here. But you're certainly correct.
              $endgroup$
              – Mark S.
              Feb 26 '16 at 22:46




              $begingroup$
              @improve since the book said "but leave the origin" and the question was about closure under addition, I didn't think that was worth mentioning here. But you're certainly correct.
              $endgroup$
              – Mark S.
              Feb 26 '16 at 22:46












              $begingroup$
              I did not read the entire question.
              $endgroup$
              – Improve
              Feb 26 '16 at 22:47






              $begingroup$
              I did not read the entire question.
              $endgroup$
              – Improve
              Feb 26 '16 at 22:47













              1












              $begingroup$

              They are asking for sets that satisfy some property of vector spaces, but that are not vector spaces.



              (a) the only operation they are allowing is addition, thus clearly vectors with entire componends only add to vectors with entire components, it's not a vector space since $mathbb{Z}$ is not a field (you don't have $n^{-1}$.



              (b) You are assuming it's a vector space, it is not, thats why the set doesn't collapses to only the y-axis.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                They are asking for sets that satisfy some property of vector spaces, but that are not vector spaces.



                (a) the only operation they are allowing is addition, thus clearly vectors with entire componends only add to vectors with entire components, it's not a vector space since $mathbb{Z}$ is not a field (you don't have $n^{-1}$.



                (b) You are assuming it's a vector space, it is not, thats why the set doesn't collapses to only the y-axis.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  They are asking for sets that satisfy some property of vector spaces, but that are not vector spaces.



                  (a) the only operation they are allowing is addition, thus clearly vectors with entire componends only add to vectors with entire components, it's not a vector space since $mathbb{Z}$ is not a field (you don't have $n^{-1}$.



                  (b) You are assuming it's a vector space, it is not, thats why the set doesn't collapses to only the y-axis.






                  share|cite|improve this answer









                  $endgroup$



                  They are asking for sets that satisfy some property of vector spaces, but that are not vector spaces.



                  (a) the only operation they are allowing is addition, thus clearly vectors with entire componends only add to vectors with entire components, it's not a vector space since $mathbb{Z}$ is not a field (you don't have $n^{-1}$.



                  (b) You are assuming it's a vector space, it is not, thats why the set doesn't collapses to only the y-axis.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 26 '16 at 22:29









                  panchopancho

                  573312




                  573312























                      0












                      $begingroup$

                      b)



                      imagine a line that is exactly on the y-axis.
                      you have a vector v = (0,1) , so cv always on the y-axis but for example
                      (0,1) + (1,1) = (1,2) wich is not on the line (y-axix)






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        b)



                        imagine a line that is exactly on the y-axis.
                        you have a vector v = (0,1) , so cv always on the y-axis but for example
                        (0,1) + (1,1) = (1,2) wich is not on the line (y-axix)






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          b)



                          imagine a line that is exactly on the y-axis.
                          you have a vector v = (0,1) , so cv always on the y-axis but for example
                          (0,1) + (1,1) = (1,2) wich is not on the line (y-axix)






                          share|cite|improve this answer









                          $endgroup$



                          b)



                          imagine a line that is exactly on the y-axis.
                          you have a vector v = (0,1) , so cv always on the y-axis but for example
                          (0,1) + (1,1) = (1,2) wich is not on the line (y-axix)







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 24 '18 at 18:59









                          Bishoy AbdBishoy Abd

                          1012




                          1012






























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                              IC on Digikey is 5x more expensive than board containing same IC on Alibaba: How? [on hold]

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                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 12 2 $begingroup$ I built a prototype using an NFC board I bought from Alibaba for 1$ (https://www.alibaba.com/showroom/rfid-rc522.html). It claims to use the NXP MFRC522 IC, and if I look on the board the chip has that part number and even the little NXP logo. It works like a charm on the prototype so now I'd like to build the NFC antenna into the product and use the same chip (I'll go through the antenna tuning/matching process). The problem is this: If I want to buy this IC on Digikey, it costs $5.36 for QTY 1 and $3.61 for QTY 1,500 (https://www.digikey.com/products/en/rf-if-and-rfid/rfid-rf-access-monitoring-ics/880?k=rc522). The whole board fro
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