Creating an orthonormal basis with Gram schmidt procedure error.












0












$begingroup$


I have a question which says the following:




Let $V$ be the span of $v_{1}=(0,1,2)$, $v_{2}=(-1,0,1)$ and $v_{3}=(-1,1,3)$.



Construct an orthonormal basis $B'$ for $V$ (usual dot product).




I know how to do the question. That's easy. I need to apply the gram schmidt procedure thrice and I have my answer. My problem is the 3rd iteration.



$U_{1}=V_{1}=(0,1,2)$



Orthonormalizing, we obtain $V_{1}'=(0,frac{1}{sqrt{5}},frac{2}{sqrt{5}})$



Then $U_2=V_2-frac{V_2 cdot U_1}{||U_1||^2}U_1$



Doing so gives: $U_2=(-1,frac{-2}{5},frac{1}{5})$



Orthonormalizing, we get $V_{2}'=(0,frac{-2}{sqrt{30}},frac{1}{sqrt{30}})$



Finally, $U_3=V_3-frac{V_3 cdot U_1}{||U_1||^2}U_1 - frac{V_3 cdot U_2}{||U_2||^2}U_2$



However this time, if I plug in everything, I get $U_3=(0,0,0)$. While that is indeed orthogonal to both ${v_1}$ and ${v_2}$ there is no way I can orthonormalize $U_3$ since I get division by $0$.



Is the question just badly asked? Because there is simply no way to obtain an orthonormal basis.






linear-algebra orthonormal gram-schmidt







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I have a question which says the following:




    Let $V$ be the span of $v_{1}=(0,1,2)$, $v_{2}=(-1,0,1)$ and $v_{3}=(-1,1,3)$.



    Construct an orthonormal basis $B'$ for $V$ (usual dot product).




    I know how to do the question. That's easy. I need to apply the gram schmidt procedure thrice and I have my answer. My problem is the 3rd iteration.



    $U_{1}=V_{1}=(0,1,2)$



    Orthonormalizing, we obtain $V_{1}'=(0,frac{1}{sqrt{5}},frac{2}{sqrt{5}})$



    Then $U_2=V_2-frac{V_2 cdot U_1}{||U_1||^2}U_1$



    Doing so gives: $U_2=(-1,frac{-2}{5},frac{1}{5})$



    Orthonormalizing, we get $V_{2}'=(0,frac{-2}{sqrt{30}},frac{1}{sqrt{30}})$



    Finally, $U_3=V_3-frac{V_3 cdot U_1}{||U_1||^2}U_1 - frac{V_3 cdot U_2}{||U_2||^2}U_2$



    However this time, if I plug in everything, I get $U_3=(0,0,0)$. While that is indeed orthogonal to both ${v_1}$ and ${v_2}$ there is no way I can orthonormalize $U_3$ since I get division by $0$.



    Is the question just badly asked? Because there is simply no way to obtain an orthonormal basis.






    linear-algebra orthonormal gram-schmidt







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have a question which says the following:




      Let $V$ be the span of $v_{1}=(0,1,2)$, $v_{2}=(-1,0,1)$ and $v_{3}=(-1,1,3)$.



      Construct an orthonormal basis $B'$ for $V$ (usual dot product).




      I know how to do the question. That's easy. I need to apply the gram schmidt procedure thrice and I have my answer. My problem is the 3rd iteration.



      $U_{1}=V_{1}=(0,1,2)$



      Orthonormalizing, we obtain $V_{1}'=(0,frac{1}{sqrt{5}},frac{2}{sqrt{5}})$



      Then $U_2=V_2-frac{V_2 cdot U_1}{||U_1||^2}U_1$



      Doing so gives: $U_2=(-1,frac{-2}{5},frac{1}{5})$



      Orthonormalizing, we get $V_{2}'=(0,frac{-2}{sqrt{30}},frac{1}{sqrt{30}})$



      Finally, $U_3=V_3-frac{V_3 cdot U_1}{||U_1||^2}U_1 - frac{V_3 cdot U_2}{||U_2||^2}U_2$



      However this time, if I plug in everything, I get $U_3=(0,0,0)$. While that is indeed orthogonal to both ${v_1}$ and ${v_2}$ there is no way I can orthonormalize $U_3$ since I get division by $0$.



      Is the question just badly asked? Because there is simply no way to obtain an orthonormal basis.






      linear-algebra orthonormal gram-schmidt







      share|cite|improve this question









      $endgroup$




      I have a question which says the following:




      Let $V$ be the span of $v_{1}=(0,1,2)$, $v_{2}=(-1,0,1)$ and $v_{3}=(-1,1,3)$.



      Construct an orthonormal basis $B'$ for $V$ (usual dot product).




      I know how to do the question. That's easy. I need to apply the gram schmidt procedure thrice and I have my answer. My problem is the 3rd iteration.



      $U_{1}=V_{1}=(0,1,2)$



      Orthonormalizing, we obtain $V_{1}'=(0,frac{1}{sqrt{5}},frac{2}{sqrt{5}})$



      Then $U_2=V_2-frac{V_2 cdot U_1}{||U_1||^2}U_1$



      Doing so gives: $U_2=(-1,frac{-2}{5},frac{1}{5})$



      Orthonormalizing, we get $V_{2}'=(0,frac{-2}{sqrt{30}},frac{1}{sqrt{30}})$



      Finally, $U_3=V_3-frac{V_3 cdot U_1}{||U_1||^2}U_1 - frac{V_3 cdot U_2}{||U_2||^2}U_2$



      However this time, if I plug in everything, I get $U_3=(0,0,0)$. While that is indeed orthogonal to both ${v_1}$ and ${v_2}$ there is no way I can orthonormalize $U_3$ since I get division by $0$.



      Is the question just badly asked? Because there is simply no way to obtain an orthonormal basis.






      linear-algebra orthonormal gram-schmidt




      linear-algebra orthonormal gram-schmidt






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 12 '18 at 6:09









      Future Math personFuture Math person

      972817




      972817






















          1 Answer
          1






          active

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          1












          $begingroup$

          Hint: what is the dimension of $V$? This should tell you how big a basis should be.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not?
            $endgroup$
            – Future Math person
            Dec 12 '18 at 6:12












          • $begingroup$
            This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span{(1,0),(2,0)}$ does not have dimension $2$.
            $endgroup$
            – platty
            Dec 12 '18 at 6:14










          • $begingroup$
            Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less.
            $endgroup$
            – PSG
            Dec 12 '18 at 6:15












          • $begingroup$
            Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks!
            $endgroup$
            – Future Math person
            Dec 12 '18 at 6:16











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Hint: what is the dimension of $V$? This should tell you how big a basis should be.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not?
            $endgroup$
            – Future Math person
            Dec 12 '18 at 6:12












          • $begingroup$
            This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span{(1,0),(2,0)}$ does not have dimension $2$.
            $endgroup$
            – platty
            Dec 12 '18 at 6:14










          • $begingroup$
            Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less.
            $endgroup$
            – PSG
            Dec 12 '18 at 6:15












          • $begingroup$
            Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks!
            $endgroup$
            – Future Math person
            Dec 12 '18 at 6:16
















          1












          $begingroup$

          Hint: what is the dimension of $V$? This should tell you how big a basis should be.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not?
            $endgroup$
            – Future Math person
            Dec 12 '18 at 6:12












          • $begingroup$
            This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span{(1,0),(2,0)}$ does not have dimension $2$.
            $endgroup$
            – platty
            Dec 12 '18 at 6:14










          • $begingroup$
            Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less.
            $endgroup$
            – PSG
            Dec 12 '18 at 6:15












          • $begingroup$
            Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks!
            $endgroup$
            – Future Math person
            Dec 12 '18 at 6:16














          1












          1








          1





          $begingroup$

          Hint: what is the dimension of $V$? This should tell you how big a basis should be.






          share|cite|improve this answer









          $endgroup$



          Hint: what is the dimension of $V$? This should tell you how big a basis should be.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 6:12









          plattyplatty

          3,370320




          3,370320












          • $begingroup$
            Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not?
            $endgroup$
            – Future Math person
            Dec 12 '18 at 6:12












          • $begingroup$
            This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span{(1,0),(2,0)}$ does not have dimension $2$.
            $endgroup$
            – platty
            Dec 12 '18 at 6:14










          • $begingroup$
            Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less.
            $endgroup$
            – PSG
            Dec 12 '18 at 6:15












          • $begingroup$
            Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks!
            $endgroup$
            – Future Math person
            Dec 12 '18 at 6:16


















          • $begingroup$
            Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not?
            $endgroup$
            – Future Math person
            Dec 12 '18 at 6:12












          • $begingroup$
            This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span{(1,0),(2,0)}$ does not have dimension $2$.
            $endgroup$
            – platty
            Dec 12 '18 at 6:14










          • $begingroup$
            Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less.
            $endgroup$
            – PSG
            Dec 12 '18 at 6:15












          • $begingroup$
            Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks!
            $endgroup$
            – Future Math person
            Dec 12 '18 at 6:16
















          $begingroup$
          Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not?
          $endgroup$
          – Future Math person
          Dec 12 '18 at 6:12






          $begingroup$
          Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not?
          $endgroup$
          – Future Math person
          Dec 12 '18 at 6:12














          $begingroup$
          This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span{(1,0),(2,0)}$ does not have dimension $2$.
          $endgroup$
          – platty
          Dec 12 '18 at 6:14




          $begingroup$
          This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span{(1,0),(2,0)}$ does not have dimension $2$.
          $endgroup$
          – platty
          Dec 12 '18 at 6:14












          $begingroup$
          Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less.
          $endgroup$
          – PSG
          Dec 12 '18 at 6:15






          $begingroup$
          Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less.
          $endgroup$
          – PSG
          Dec 12 '18 at 6:15














          $begingroup$
          Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks!
          $endgroup$
          – Future Math person
          Dec 12 '18 at 6:16




          $begingroup$
          Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks!
          $endgroup$
          – Future Math person
          Dec 12 '18 at 6:16


















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