Linear operator and inner product












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Theorem: Let $V$ be an inner product finite space with an orthonormal basis $mathcal B$. Let $L$ be an operator on $V$, and let $A = [L]_mathcal{B}$, the matrix associate to $L$. Then the matrix elements of $A$ are $$A_{ij} = langle b_i, Lb_jrangle.$$




If the basis $mathcal B$ is only orthogonal, is it true that$$A_{ij}=frac{langle b_i, Lb_jrangle}{langle b_i, b_irangle}?$$






linear-algebra vector-spaces linear-transformations norm







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  • $begingroup$
    @AlexVong Sorry, it should be finite, but I would also like to know if that is true or not when $V$ is of infinite dimension.
    $endgroup$
    – user398843
    Dec 13 '18 at 0:41
















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$begingroup$



Theorem: Let $V$ be an inner product finite space with an orthonormal basis $mathcal B$. Let $L$ be an operator on $V$, and let $A = [L]_mathcal{B}$, the matrix associate to $L$. Then the matrix elements of $A$ are $$A_{ij} = langle b_i, Lb_jrangle.$$




If the basis $mathcal B$ is only orthogonal, is it true that$$A_{ij}=frac{langle b_i, Lb_jrangle}{langle b_i, b_irangle}?$$






linear-algebra vector-spaces linear-transformations norm







share|cite|improve this question











$endgroup$












  • $begingroup$
    @AlexVong Sorry, it should be finite, but I would also like to know if that is true or not when $V$ is of infinite dimension.
    $endgroup$
    – user398843
    Dec 13 '18 at 0:41














1












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$begingroup$



Theorem: Let $V$ be an inner product finite space with an orthonormal basis $mathcal B$. Let $L$ be an operator on $V$, and let $A = [L]_mathcal{B}$, the matrix associate to $L$. Then the matrix elements of $A$ are $$A_{ij} = langle b_i, Lb_jrangle.$$




If the basis $mathcal B$ is only orthogonal, is it true that$$A_{ij}=frac{langle b_i, Lb_jrangle}{langle b_i, b_irangle}?$$






linear-algebra vector-spaces linear-transformations norm







share|cite|improve this question











$endgroup$





Theorem: Let $V$ be an inner product finite space with an orthonormal basis $mathcal B$. Let $L$ be an operator on $V$, and let $A = [L]_mathcal{B}$, the matrix associate to $L$. Then the matrix elements of $A$ are $$A_{ij} = langle b_i, Lb_jrangle.$$




If the basis $mathcal B$ is only orthogonal, is it true that$$A_{ij}=frac{langle b_i, Lb_jrangle}{langle b_i, b_irangle}?$$






linear-algebra vector-spaces linear-transformations norm




linear-algebra vector-spaces linear-transformations norm






share|cite|improve this question















share|cite|improve this question













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edited Dec 13 '18 at 0:45









Saucy O'Path

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asked Dec 13 '18 at 0:18









user398843user398843

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  • $begingroup$
    @AlexVong Sorry, it should be finite, but I would also like to know if that is true or not when $V$ is of infinite dimension.
    $endgroup$
    – user398843
    Dec 13 '18 at 0:41


















  • $begingroup$
    @AlexVong Sorry, it should be finite, but I would also like to know if that is true or not when $V$ is of infinite dimension.
    $endgroup$
    – user398843
    Dec 13 '18 at 0:41
















$begingroup$
@AlexVong Sorry, it should be finite, but I would also like to know if that is true or not when $V$ is of infinite dimension.
$endgroup$
– user398843
Dec 13 '18 at 0:41




$begingroup$
@AlexVong Sorry, it should be finite, but I would also like to know if that is true or not when $V$ is of infinite dimension.
$endgroup$
– user398843
Dec 13 '18 at 0:41










2 Answers
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Assume that $mathcal{B}={b_1,b_2,ldots,b_n}$, where $n:= dim(V)$. Since $$L(b_j)=sum_{k=1}^n,A_{k,j},b_ktext{ for each }jin{1,2,ldots,n}=:[n],,$$
we have
$$biglangle b_i,L(b_j)bigrangle=sum_{k=1}^n,A_{k,j},langle b_i,b_krangletext{ for all }i,jin[n],.$$
If $mathcal{B}$ is an orthogonal basis, then
$$biglangle b_i,L(b_j)bigrangle=A_{i,j},langle b_i,b_irangletext{ for all }i,jin[n],,$$ proving your claim.



In general, let $langle_,_rangle$ be a nondegenerate symmetric bilinear form on $V$ and ${beta_1,beta_2,ldots,beta_n}$ the dual basis of ${b_1,b_2,ldots,b_n}$. Then, $langle beta_i,b_jrangle =delta_{i,j}$ for all $i,jin[n]$, where $delta$ is the Kronecker delta. Then, the matrix $[A_{i,j}]_{i,jin[n]}$ of $L$ in the basis $mathcal{B}={b_1,b_2,ldots,b_n}$ is given by
$$A_{i,j}=biglangle beta_i,L(b_j)bigrangletext{ for all }i,jin [n],.$$
In your particular case,
$$beta_i=frac{b_i}{langle b_i,b_irangle}text{ for every }iin[n],.$$



Remark. In the case the base field is $mathbb{C}$, we can also take $langle _,_rangle$ to be a nondegenerate sesquilinear form on $V$ that is antilinear in the first entry, and linear in the second entry. The work is the same.






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    Yes, it is, because $[L]_{mathcal B}=[id]^{mathcal B'}_{mathcal B}[L]_{mathcal B'}[id]_{mathcal B'}^{mathcal B}$, where $mathcal B'=left{frac{b_i}{sqrt{langle b_i,b_irangle}}right}_{i=1}^n$.



    By the previous lemma, $left([L]_{mathcal B'}right)_{ij}=frac{langle b_i,Lb_jrangle}{sqrt{langle b_j,b_jranglelangle b_i,b_irangle}}$. Moreover, $left([id]^{mathcal B'}_{mathcal B}right)_{ij}=delta_{ij}frac1{sqrt{langle b_i,b_irangle}}$ and $left([id]_{mathcal B'}^{mathcal B}right)_{ij}=delta_{ij}sqrt{langle b_i,b_irangle}$, so $$left([L]_{mathcal B}right)_{ij}=sum_{k,h}left([id]^{mathcal B'}_{mathcal B}right)_{ik}left([L]_{mathcal B'}right)_{kh}left([id]_{mathcal B'}^{mathcal B}right)_{hj}=\=sum_{k,h}delta_{ik}frac1{sqrt{langle b_i,b_irangle}}frac{langle b_k,Lb_hrangle}{sqrt{langle b_k,b_kranglelangle b_h,b_hrangle}}delta_{hj}sqrt{langle b_h,b_hrangle}=frac{langle b_i,Lb_jrangle}{langle b_i,b_irangle}$$






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      2 Answers
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      2 Answers
      2






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      $begingroup$

      Assume that $mathcal{B}={b_1,b_2,ldots,b_n}$, where $n:= dim(V)$. Since $$L(b_j)=sum_{k=1}^n,A_{k,j},b_ktext{ for each }jin{1,2,ldots,n}=:[n],,$$
      we have
      $$biglangle b_i,L(b_j)bigrangle=sum_{k=1}^n,A_{k,j},langle b_i,b_krangletext{ for all }i,jin[n],.$$
      If $mathcal{B}$ is an orthogonal basis, then
      $$biglangle b_i,L(b_j)bigrangle=A_{i,j},langle b_i,b_irangletext{ for all }i,jin[n],,$$ proving your claim.



      In general, let $langle_,_rangle$ be a nondegenerate symmetric bilinear form on $V$ and ${beta_1,beta_2,ldots,beta_n}$ the dual basis of ${b_1,b_2,ldots,b_n}$. Then, $langle beta_i,b_jrangle =delta_{i,j}$ for all $i,jin[n]$, where $delta$ is the Kronecker delta. Then, the matrix $[A_{i,j}]_{i,jin[n]}$ of $L$ in the basis $mathcal{B}={b_1,b_2,ldots,b_n}$ is given by
      $$A_{i,j}=biglangle beta_i,L(b_j)bigrangletext{ for all }i,jin [n],.$$
      In your particular case,
      $$beta_i=frac{b_i}{langle b_i,b_irangle}text{ for every }iin[n],.$$



      Remark. In the case the base field is $mathbb{C}$, we can also take $langle _,_rangle$ to be a nondegenerate sesquilinear form on $V$ that is antilinear in the first entry, and linear in the second entry. The work is the same.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        Assume that $mathcal{B}={b_1,b_2,ldots,b_n}$, where $n:= dim(V)$. Since $$L(b_j)=sum_{k=1}^n,A_{k,j},b_ktext{ for each }jin{1,2,ldots,n}=:[n],,$$
        we have
        $$biglangle b_i,L(b_j)bigrangle=sum_{k=1}^n,A_{k,j},langle b_i,b_krangletext{ for all }i,jin[n],.$$
        If $mathcal{B}$ is an orthogonal basis, then
        $$biglangle b_i,L(b_j)bigrangle=A_{i,j},langle b_i,b_irangletext{ for all }i,jin[n],,$$ proving your claim.



        In general, let $langle_,_rangle$ be a nondegenerate symmetric bilinear form on $V$ and ${beta_1,beta_2,ldots,beta_n}$ the dual basis of ${b_1,b_2,ldots,b_n}$. Then, $langle beta_i,b_jrangle =delta_{i,j}$ for all $i,jin[n]$, where $delta$ is the Kronecker delta. Then, the matrix $[A_{i,j}]_{i,jin[n]}$ of $L$ in the basis $mathcal{B}={b_1,b_2,ldots,b_n}$ is given by
        $$A_{i,j}=biglangle beta_i,L(b_j)bigrangletext{ for all }i,jin [n],.$$
        In your particular case,
        $$beta_i=frac{b_i}{langle b_i,b_irangle}text{ for every }iin[n],.$$



        Remark. In the case the base field is $mathbb{C}$, we can also take $langle _,_rangle$ to be a nondegenerate sesquilinear form on $V$ that is antilinear in the first entry, and linear in the second entry. The work is the same.






        share|cite|improve this answer











        $endgroup$
















          1












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          $begingroup$

          Assume that $mathcal{B}={b_1,b_2,ldots,b_n}$, where $n:= dim(V)$. Since $$L(b_j)=sum_{k=1}^n,A_{k,j},b_ktext{ for each }jin{1,2,ldots,n}=:[n],,$$
          we have
          $$biglangle b_i,L(b_j)bigrangle=sum_{k=1}^n,A_{k,j},langle b_i,b_krangletext{ for all }i,jin[n],.$$
          If $mathcal{B}$ is an orthogonal basis, then
          $$biglangle b_i,L(b_j)bigrangle=A_{i,j},langle b_i,b_irangletext{ for all }i,jin[n],,$$ proving your claim.



          In general, let $langle_,_rangle$ be a nondegenerate symmetric bilinear form on $V$ and ${beta_1,beta_2,ldots,beta_n}$ the dual basis of ${b_1,b_2,ldots,b_n}$. Then, $langle beta_i,b_jrangle =delta_{i,j}$ for all $i,jin[n]$, where $delta$ is the Kronecker delta. Then, the matrix $[A_{i,j}]_{i,jin[n]}$ of $L$ in the basis $mathcal{B}={b_1,b_2,ldots,b_n}$ is given by
          $$A_{i,j}=biglangle beta_i,L(b_j)bigrangletext{ for all }i,jin [n],.$$
          In your particular case,
          $$beta_i=frac{b_i}{langle b_i,b_irangle}text{ for every }iin[n],.$$



          Remark. In the case the base field is $mathbb{C}$, we can also take $langle _,_rangle$ to be a nondegenerate sesquilinear form on $V$ that is antilinear in the first entry, and linear in the second entry. The work is the same.






          share|cite|improve this answer











          $endgroup$



          Assume that $mathcal{B}={b_1,b_2,ldots,b_n}$, where $n:= dim(V)$. Since $$L(b_j)=sum_{k=1}^n,A_{k,j},b_ktext{ for each }jin{1,2,ldots,n}=:[n],,$$
          we have
          $$biglangle b_i,L(b_j)bigrangle=sum_{k=1}^n,A_{k,j},langle b_i,b_krangletext{ for all }i,jin[n],.$$
          If $mathcal{B}$ is an orthogonal basis, then
          $$biglangle b_i,L(b_j)bigrangle=A_{i,j},langle b_i,b_irangletext{ for all }i,jin[n],,$$ proving your claim.



          In general, let $langle_,_rangle$ be a nondegenerate symmetric bilinear form on $V$ and ${beta_1,beta_2,ldots,beta_n}$ the dual basis of ${b_1,b_2,ldots,b_n}$. Then, $langle beta_i,b_jrangle =delta_{i,j}$ for all $i,jin[n]$, where $delta$ is the Kronecker delta. Then, the matrix $[A_{i,j}]_{i,jin[n]}$ of $L$ in the basis $mathcal{B}={b_1,b_2,ldots,b_n}$ is given by
          $$A_{i,j}=biglangle beta_i,L(b_j)bigrangletext{ for all }i,jin [n],.$$
          In your particular case,
          $$beta_i=frac{b_i}{langle b_i,b_irangle}text{ for every }iin[n],.$$



          Remark. In the case the base field is $mathbb{C}$, we can also take $langle _,_rangle$ to be a nondegenerate sesquilinear form on $V$ that is antilinear in the first entry, and linear in the second entry. The work is the same.







          share|cite|improve this answer














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          share|cite|improve this answer








          edited Dec 13 '18 at 0:50

























          answered Dec 13 '18 at 0:42









          BatominovskiBatominovski

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              $begingroup$

              Yes, it is, because $[L]_{mathcal B}=[id]^{mathcal B'}_{mathcal B}[L]_{mathcal B'}[id]_{mathcal B'}^{mathcal B}$, where $mathcal B'=left{frac{b_i}{sqrt{langle b_i,b_irangle}}right}_{i=1}^n$.



              By the previous lemma, $left([L]_{mathcal B'}right)_{ij}=frac{langle b_i,Lb_jrangle}{sqrt{langle b_j,b_jranglelangle b_i,b_irangle}}$. Moreover, $left([id]^{mathcal B'}_{mathcal B}right)_{ij}=delta_{ij}frac1{sqrt{langle b_i,b_irangle}}$ and $left([id]_{mathcal B'}^{mathcal B}right)_{ij}=delta_{ij}sqrt{langle b_i,b_irangle}$, so $$left([L]_{mathcal B}right)_{ij}=sum_{k,h}left([id]^{mathcal B'}_{mathcal B}right)_{ik}left([L]_{mathcal B'}right)_{kh}left([id]_{mathcal B'}^{mathcal B}right)_{hj}=\=sum_{k,h}delta_{ik}frac1{sqrt{langle b_i,b_irangle}}frac{langle b_k,Lb_hrangle}{sqrt{langle b_k,b_kranglelangle b_h,b_hrangle}}delta_{hj}sqrt{langle b_h,b_hrangle}=frac{langle b_i,Lb_jrangle}{langle b_i,b_irangle}$$






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                1












                $begingroup$

                Yes, it is, because $[L]_{mathcal B}=[id]^{mathcal B'}_{mathcal B}[L]_{mathcal B'}[id]_{mathcal B'}^{mathcal B}$, where $mathcal B'=left{frac{b_i}{sqrt{langle b_i,b_irangle}}right}_{i=1}^n$.



                By the previous lemma, $left([L]_{mathcal B'}right)_{ij}=frac{langle b_i,Lb_jrangle}{sqrt{langle b_j,b_jranglelangle b_i,b_irangle}}$. Moreover, $left([id]^{mathcal B'}_{mathcal B}right)_{ij}=delta_{ij}frac1{sqrt{langle b_i,b_irangle}}$ and $left([id]_{mathcal B'}^{mathcal B}right)_{ij}=delta_{ij}sqrt{langle b_i,b_irangle}$, so $$left([L]_{mathcal B}right)_{ij}=sum_{k,h}left([id]^{mathcal B'}_{mathcal B}right)_{ik}left([L]_{mathcal B'}right)_{kh}left([id]_{mathcal B'}^{mathcal B}right)_{hj}=\=sum_{k,h}delta_{ik}frac1{sqrt{langle b_i,b_irangle}}frac{langle b_k,Lb_hrangle}{sqrt{langle b_k,b_kranglelangle b_h,b_hrangle}}delta_{hj}sqrt{langle b_h,b_hrangle}=frac{langle b_i,Lb_jrangle}{langle b_i,b_irangle}$$






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                  $begingroup$

                  Yes, it is, because $[L]_{mathcal B}=[id]^{mathcal B'}_{mathcal B}[L]_{mathcal B'}[id]_{mathcal B'}^{mathcal B}$, where $mathcal B'=left{frac{b_i}{sqrt{langle b_i,b_irangle}}right}_{i=1}^n$.



                  By the previous lemma, $left([L]_{mathcal B'}right)_{ij}=frac{langle b_i,Lb_jrangle}{sqrt{langle b_j,b_jranglelangle b_i,b_irangle}}$. Moreover, $left([id]^{mathcal B'}_{mathcal B}right)_{ij}=delta_{ij}frac1{sqrt{langle b_i,b_irangle}}$ and $left([id]_{mathcal B'}^{mathcal B}right)_{ij}=delta_{ij}sqrt{langle b_i,b_irangle}$, so $$left([L]_{mathcal B}right)_{ij}=sum_{k,h}left([id]^{mathcal B'}_{mathcal B}right)_{ik}left([L]_{mathcal B'}right)_{kh}left([id]_{mathcal B'}^{mathcal B}right)_{hj}=\=sum_{k,h}delta_{ik}frac1{sqrt{langle b_i,b_irangle}}frac{langle b_k,Lb_hrangle}{sqrt{langle b_k,b_kranglelangle b_h,b_hrangle}}delta_{hj}sqrt{langle b_h,b_hrangle}=frac{langle b_i,Lb_jrangle}{langle b_i,b_irangle}$$






                  share|cite|improve this answer











                  $endgroup$



                  Yes, it is, because $[L]_{mathcal B}=[id]^{mathcal B'}_{mathcal B}[L]_{mathcal B'}[id]_{mathcal B'}^{mathcal B}$, where $mathcal B'=left{frac{b_i}{sqrt{langle b_i,b_irangle}}right}_{i=1}^n$.



                  By the previous lemma, $left([L]_{mathcal B'}right)_{ij}=frac{langle b_i,Lb_jrangle}{sqrt{langle b_j,b_jranglelangle b_i,b_irangle}}$. Moreover, $left([id]^{mathcal B'}_{mathcal B}right)_{ij}=delta_{ij}frac1{sqrt{langle b_i,b_irangle}}$ and $left([id]_{mathcal B'}^{mathcal B}right)_{ij}=delta_{ij}sqrt{langle b_i,b_irangle}$, so $$left([L]_{mathcal B}right)_{ij}=sum_{k,h}left([id]^{mathcal B'}_{mathcal B}right)_{ik}left([L]_{mathcal B'}right)_{kh}left([id]_{mathcal B'}^{mathcal B}right)_{hj}=\=sum_{k,h}delta_{ik}frac1{sqrt{langle b_i,b_irangle}}frac{langle b_k,Lb_hrangle}{sqrt{langle b_k,b_kranglelangle b_h,b_hrangle}}delta_{hj}sqrt{langle b_h,b_hrangle}=frac{langle b_i,Lb_jrangle}{langle b_i,b_irangle}$$







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                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 13 '18 at 0:40

























                  answered Dec 13 '18 at 0:24









                  Saucy O'PathSaucy O'Path

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