If u is orthogonal to both v and w, and u not equal to 0, argue that u is not in the span of v and w. (












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QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.



Where I am at:
enter image description here



I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.



I also tried formulating the following steps to solve the problem.




  1. Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa


  2. Set u = av + bw = u (where a and b are constants)


  3. Disprove (2)


However, I could not get past step 1.



Any pointers would be greatly appreciated.






linear-algebra matrices







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    1












    $begingroup$


    QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.



    Where I am at:
    enter image description here



    I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.



    I also tried formulating the following steps to solve the problem.




    1. Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa


    2. Set u = av + bw = u (where a and b are constants)


    3. Disprove (2)


    However, I could not get past step 1.



    Any pointers would be greatly appreciated.






    linear-algebra matrices







    share|cite|improve this question









    New contributor




    Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      1












      1








      1





      $begingroup$


      QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.



      Where I am at:
      enter image description here



      I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.



      I also tried formulating the following steps to solve the problem.




      1. Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa


      2. Set u = av + bw = u (where a and b are constants)


      3. Disprove (2)


      However, I could not get past step 1.



      Any pointers would be greatly appreciated.






      linear-algebra matrices







      share|cite|improve this question









      New contributor




      Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.



      Where I am at:
      enter image description here



      I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.



      I also tried formulating the following steps to solve the problem.




      1. Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa


      2. Set u = av + bw = u (where a and b are constants)


      3. Disprove (2)


      However, I could not get past step 1.



      Any pointers would be greatly appreciated.






      linear-algebra matrices




      linear-algebra matrices






      share|cite|improve this question









      New contributor




      Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









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      share|cite|improve this question








      edited 46 mins ago









      YuiTo Cheng

      2,52341037




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      asked 58 mins ago









      Dimen3ionalDimen3ional

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          2 Answers
          2






          active

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          2












          $begingroup$

          $defmyvec#1{{bf#1}}$
          This is the same as saying




          if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.




          So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
          $$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
          Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
            $endgroup$
            – Dimen3ional
            43 mins ago










          • $begingroup$
            Have you studied proof by contradiction or by contrapositive? That's what this is.
            $endgroup$
            – David
            41 mins ago





















          2












          $begingroup$

          If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.






          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









            2












            $begingroup$

            $defmyvec#1{{bf#1}}$
            This is the same as saying




            if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.




            So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
            $$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
            Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
              $endgroup$
              – Dimen3ional
              43 mins ago










            • $begingroup$
              Have you studied proof by contradiction or by contrapositive? That's what this is.
              $endgroup$
              – David
              41 mins ago


















            2












            $begingroup$

            $defmyvec#1{{bf#1}}$
            This is the same as saying




            if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.




            So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
            $$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
            Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
              $endgroup$
              – Dimen3ional
              43 mins ago










            • $begingroup$
              Have you studied proof by contradiction or by contrapositive? That's what this is.
              $endgroup$
              – David
              41 mins ago
















            2












            2








            2





            $begingroup$

            $defmyvec#1{{bf#1}}$
            This is the same as saying




            if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.




            So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
            $$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
            Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.






            share|cite|improve this answer









            $endgroup$



            $defmyvec#1{{bf#1}}$
            This is the same as saying




            if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.




            So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
            $$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
            Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 49 mins ago









            DavidDavid

            70.1k668131




            70.1k668131












            • $begingroup$
              I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
              $endgroup$
              – Dimen3ional
              43 mins ago










            • $begingroup$
              Have you studied proof by contradiction or by contrapositive? That's what this is.
              $endgroup$
              – David
              41 mins ago




















            • $begingroup$
              I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
              $endgroup$
              – Dimen3ional
              43 mins ago










            • $begingroup$
              Have you studied proof by contradiction or by contrapositive? That's what this is.
              $endgroup$
              – David
              41 mins ago


















            $begingroup$
            I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
            $endgroup$
            – Dimen3ional
            43 mins ago




            $begingroup$
            I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
            $endgroup$
            – Dimen3ional
            43 mins ago












            $begingroup$
            Have you studied proof by contradiction or by contrapositive? That's what this is.
            $endgroup$
            – David
            41 mins ago






            $begingroup$
            Have you studied proof by contradiction or by contrapositive? That's what this is.
            $endgroup$
            – David
            41 mins ago













            2












            $begingroup$

            If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.






                share|cite|improve this answer









                $endgroup$



                If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 49 mins ago









                Kavi Rama MurthyKavi Rama Murthy

                75.6k53270




                75.6k53270






















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