Element not orthogonal to any other non zero element












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Consider an inner product space. Does there(always) exist a non zero element which is not orthogonal to any other non zero element? In other way, can I always find a non zero element which is orthogonal to a given non zero element?



For example, in the Cartesian plane to any non zero vector one can consider the vector perpendicular to it, which is orthogonal. But is it true for any general inner product space?






inner-product-space







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    0












    $begingroup$


    Consider an inner product space. Does there(always) exist a non zero element which is not orthogonal to any other non zero element? In other way, can I always find a non zero element which is orthogonal to a given non zero element?



    For example, in the Cartesian plane to any non zero vector one can consider the vector perpendicular to it, which is orthogonal. But is it true for any general inner product space?






    inner-product-space







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Consider an inner product space. Does there(always) exist a non zero element which is not orthogonal to any other non zero element? In other way, can I always find a non zero element which is orthogonal to a given non zero element?



      For example, in the Cartesian plane to any non zero vector one can consider the vector perpendicular to it, which is orthogonal. But is it true for any general inner product space?






      inner-product-space







      share|cite|improve this question









      $endgroup$




      Consider an inner product space. Does there(always) exist a non zero element which is not orthogonal to any other non zero element? In other way, can I always find a non zero element which is orthogonal to a given non zero element?



      For example, in the Cartesian plane to any non zero vector one can consider the vector perpendicular to it, which is orthogonal. But is it true for any general inner product space?






      inner-product-space




      inner-product-space






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 16 '18 at 12:08









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          2 Answers
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          $begingroup$

          If the inner product is of dimension $1$ or $0$, then yes. The zero vector in a $0$-dimensional space will be orthogonal only to the $0$ vector (trivially). In a $1$-dimensional space, we can form a basis out of any non-zero vector $v$, and we have, for any scalar $a$,
          $$langle av, v rangle = a langle v, v rangle neq 0 text{ if }a neq 0.$$
          Otherwise, no, there is no such vector. In an inner product space of dimension greater than $1$ (possibly infinite), we may find two linearly independent vectors $v$ and $w$. Consider,
          $$u = w - frac{langle w, v rangle}{langle v, v rangle} v.$$
          Then
          $$langle u, v rangle = leftlangle w - frac{langle w, v rangle}{langle v, v rangle} v, v rightrangle = langle w, v rangle - frac{langle w, v rangle}{langle v, v rangle} langle v, vrangle = 0.$$
          If it were the case that $v$ were only orthogonal with the $0$ vector, then we would have to have
          $$w - frac{langle w, v rangle}{langle v, v rangle} v = 0,$$
          which would contradict the linear independence of $v$ and $w$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is Gram-Schmidt orthogonalisation right?
            $endgroup$
            – 1.414212
            Dec 16 '18 at 12:24










          • $begingroup$
            Yes, that's right.
            $endgroup$
            – Theo Bendit
            Dec 16 '18 at 12:28



















          0












          $begingroup$

          No, not in general. Take $mathbb R$ with the inner product $langle x,yrangle=xy$. But in a space with dimension greater than $1$, it is true.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I should have asked clearly, space in my mind is having dimension more than 2 actually. Can you explain how is it true? (Can you post it in answer?)
            $endgroup$
            – 1.414212
            Dec 16 '18 at 12:16










          • $begingroup$
            Why? Another poster has already provided an answer to that question.
            $endgroup$
            – José Carlos Santos
            Dec 16 '18 at 12:21











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          If the inner product is of dimension $1$ or $0$, then yes. The zero vector in a $0$-dimensional space will be orthogonal only to the $0$ vector (trivially). In a $1$-dimensional space, we can form a basis out of any non-zero vector $v$, and we have, for any scalar $a$,
          $$langle av, v rangle = a langle v, v rangle neq 0 text{ if }a neq 0.$$
          Otherwise, no, there is no such vector. In an inner product space of dimension greater than $1$ (possibly infinite), we may find two linearly independent vectors $v$ and $w$. Consider,
          $$u = w - frac{langle w, v rangle}{langle v, v rangle} v.$$
          Then
          $$langle u, v rangle = leftlangle w - frac{langle w, v rangle}{langle v, v rangle} v, v rightrangle = langle w, v rangle - frac{langle w, v rangle}{langle v, v rangle} langle v, vrangle = 0.$$
          If it were the case that $v$ were only orthogonal with the $0$ vector, then we would have to have
          $$w - frac{langle w, v rangle}{langle v, v rangle} v = 0,$$
          which would contradict the linear independence of $v$ and $w$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is Gram-Schmidt orthogonalisation right?
            $endgroup$
            – 1.414212
            Dec 16 '18 at 12:24










          • $begingroup$
            Yes, that's right.
            $endgroup$
            – Theo Bendit
            Dec 16 '18 at 12:28
















          1












          $begingroup$

          If the inner product is of dimension $1$ or $0$, then yes. The zero vector in a $0$-dimensional space will be orthogonal only to the $0$ vector (trivially). In a $1$-dimensional space, we can form a basis out of any non-zero vector $v$, and we have, for any scalar $a$,
          $$langle av, v rangle = a langle v, v rangle neq 0 text{ if }a neq 0.$$
          Otherwise, no, there is no such vector. In an inner product space of dimension greater than $1$ (possibly infinite), we may find two linearly independent vectors $v$ and $w$. Consider,
          $$u = w - frac{langle w, v rangle}{langle v, v rangle} v.$$
          Then
          $$langle u, v rangle = leftlangle w - frac{langle w, v rangle}{langle v, v rangle} v, v rightrangle = langle w, v rangle - frac{langle w, v rangle}{langle v, v rangle} langle v, vrangle = 0.$$
          If it were the case that $v$ were only orthogonal with the $0$ vector, then we would have to have
          $$w - frac{langle w, v rangle}{langle v, v rangle} v = 0,$$
          which would contradict the linear independence of $v$ and $w$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is Gram-Schmidt orthogonalisation right?
            $endgroup$
            – 1.414212
            Dec 16 '18 at 12:24










          • $begingroup$
            Yes, that's right.
            $endgroup$
            – Theo Bendit
            Dec 16 '18 at 12:28














          1












          1








          1





          $begingroup$

          If the inner product is of dimension $1$ or $0$, then yes. The zero vector in a $0$-dimensional space will be orthogonal only to the $0$ vector (trivially). In a $1$-dimensional space, we can form a basis out of any non-zero vector $v$, and we have, for any scalar $a$,
          $$langle av, v rangle = a langle v, v rangle neq 0 text{ if }a neq 0.$$
          Otherwise, no, there is no such vector. In an inner product space of dimension greater than $1$ (possibly infinite), we may find two linearly independent vectors $v$ and $w$. Consider,
          $$u = w - frac{langle w, v rangle}{langle v, v rangle} v.$$
          Then
          $$langle u, v rangle = leftlangle w - frac{langle w, v rangle}{langle v, v rangle} v, v rightrangle = langle w, v rangle - frac{langle w, v rangle}{langle v, v rangle} langle v, vrangle = 0.$$
          If it were the case that $v$ were only orthogonal with the $0$ vector, then we would have to have
          $$w - frac{langle w, v rangle}{langle v, v rangle} v = 0,$$
          which would contradict the linear independence of $v$ and $w$.






          share|cite|improve this answer









          $endgroup$



          If the inner product is of dimension $1$ or $0$, then yes. The zero vector in a $0$-dimensional space will be orthogonal only to the $0$ vector (trivially). In a $1$-dimensional space, we can form a basis out of any non-zero vector $v$, and we have, for any scalar $a$,
          $$langle av, v rangle = a langle v, v rangle neq 0 text{ if }a neq 0.$$
          Otherwise, no, there is no such vector. In an inner product space of dimension greater than $1$ (possibly infinite), we may find two linearly independent vectors $v$ and $w$. Consider,
          $$u = w - frac{langle w, v rangle}{langle v, v rangle} v.$$
          Then
          $$langle u, v rangle = leftlangle w - frac{langle w, v rangle}{langle v, v rangle} v, v rightrangle = langle w, v rangle - frac{langle w, v rangle}{langle v, v rangle} langle v, vrangle = 0.$$
          If it were the case that $v$ were only orthogonal with the $0$ vector, then we would have to have
          $$w - frac{langle w, v rangle}{langle v, v rangle} v = 0,$$
          which would contradict the linear independence of $v$ and $w$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 12:19









          Theo BenditTheo Bendit

          17.2k12149




          17.2k12149












          • $begingroup$
            This is Gram-Schmidt orthogonalisation right?
            $endgroup$
            – 1.414212
            Dec 16 '18 at 12:24










          • $begingroup$
            Yes, that's right.
            $endgroup$
            – Theo Bendit
            Dec 16 '18 at 12:28


















          • $begingroup$
            This is Gram-Schmidt orthogonalisation right?
            $endgroup$
            – 1.414212
            Dec 16 '18 at 12:24










          • $begingroup$
            Yes, that's right.
            $endgroup$
            – Theo Bendit
            Dec 16 '18 at 12:28
















          $begingroup$
          This is Gram-Schmidt orthogonalisation right?
          $endgroup$
          – 1.414212
          Dec 16 '18 at 12:24




          $begingroup$
          This is Gram-Schmidt orthogonalisation right?
          $endgroup$
          – 1.414212
          Dec 16 '18 at 12:24












          $begingroup$
          Yes, that's right.
          $endgroup$
          – Theo Bendit
          Dec 16 '18 at 12:28




          $begingroup$
          Yes, that's right.
          $endgroup$
          – Theo Bendit
          Dec 16 '18 at 12:28











          0












          $begingroup$

          No, not in general. Take $mathbb R$ with the inner product $langle x,yrangle=xy$. But in a space with dimension greater than $1$, it is true.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I should have asked clearly, space in my mind is having dimension more than 2 actually. Can you explain how is it true? (Can you post it in answer?)
            $endgroup$
            – 1.414212
            Dec 16 '18 at 12:16










          • $begingroup$
            Why? Another poster has already provided an answer to that question.
            $endgroup$
            – José Carlos Santos
            Dec 16 '18 at 12:21
















          0












          $begingroup$

          No, not in general. Take $mathbb R$ with the inner product $langle x,yrangle=xy$. But in a space with dimension greater than $1$, it is true.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I should have asked clearly, space in my mind is having dimension more than 2 actually. Can you explain how is it true? (Can you post it in answer?)
            $endgroup$
            – 1.414212
            Dec 16 '18 at 12:16










          • $begingroup$
            Why? Another poster has already provided an answer to that question.
            $endgroup$
            – José Carlos Santos
            Dec 16 '18 at 12:21














          0












          0








          0





          $begingroup$

          No, not in general. Take $mathbb R$ with the inner product $langle x,yrangle=xy$. But in a space with dimension greater than $1$, it is true.






          share|cite|improve this answer









          $endgroup$



          No, not in general. Take $mathbb R$ with the inner product $langle x,yrangle=xy$. But in a space with dimension greater than $1$, it is true.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 12:13









          José Carlos SantosJosé Carlos Santos

          155k22124227




          155k22124227












          • $begingroup$
            I should have asked clearly, space in my mind is having dimension more than 2 actually. Can you explain how is it true? (Can you post it in answer?)
            $endgroup$
            – 1.414212
            Dec 16 '18 at 12:16










          • $begingroup$
            Why? Another poster has already provided an answer to that question.
            $endgroup$
            – José Carlos Santos
            Dec 16 '18 at 12:21


















          • $begingroup$
            I should have asked clearly, space in my mind is having dimension more than 2 actually. Can you explain how is it true? (Can you post it in answer?)
            $endgroup$
            – 1.414212
            Dec 16 '18 at 12:16










          • $begingroup$
            Why? Another poster has already provided an answer to that question.
            $endgroup$
            – José Carlos Santos
            Dec 16 '18 at 12:21
















          $begingroup$
          I should have asked clearly, space in my mind is having dimension more than 2 actually. Can you explain how is it true? (Can you post it in answer?)
          $endgroup$
          – 1.414212
          Dec 16 '18 at 12:16




          $begingroup$
          I should have asked clearly, space in my mind is having dimension more than 2 actually. Can you explain how is it true? (Can you post it in answer?)
          $endgroup$
          – 1.414212
          Dec 16 '18 at 12:16












          $begingroup$
          Why? Another poster has already provided an answer to that question.
          $endgroup$
          – José Carlos Santos
          Dec 16 '18 at 12:21




          $begingroup$
          Why? Another poster has already provided an answer to that question.
          $endgroup$
          – José Carlos Santos
          Dec 16 '18 at 12:21


















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