Show that the associated matrix is $PAP^{-1}$












0












$begingroup$


Let $eta in operatorname{End}_{R}(R^n)$ and let $A$ be the associated matrix of $eta$ with respect to the basis $(e_1, dots e_n)$. Let $f_i = sum p_{i,j}e_j$ where the matrix $P=(p_{i,j}) in GL_n(R)$. Show that $PAP^{-1}$ is the associated matrix of the basis $(f_1 dots f_n)$.



$underline{My space attempt:}$



$$eta(f_i) = etaleft(sum_{j=1}^n p_{i,j}e_j right) = sum_{j=1}^n eta(p_{i,j} e_j) = sum_{j=1}^n p_{i,j} eta(e_j)$$



$$=sum_{j=1}^n p_{i,j} left( sum_{k=1}^n a_{i,k} e_k right) = sum_{j=1}^n sum_{k=1}^n(p_{i,j}a_{i,k})e_k$$



$$=sum_{j=1}^nsum_{k=1}^n (p_{i,j}a_{i,k}p_{i,j}^{-1})p_{i,j}e_k$$



From here I see that I am very close but I'm not sure how to turn that $p_{i,j}e_k$ into the $f_{k}$ because the sum includes all the other terms. Any help is appreciated, thanks!






linear-algebra abstract-algebra change-of-basis free-modules







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$endgroup$












  • $begingroup$
    The $i,j$ element of $P^{-1}$ is not the inverse of $p_{i,j}$.
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:14










  • $begingroup$
    Are you sure of the formulæ in this question? Normally the coordinates of the $f-i$s in basis $(e_1,dots,e_n)$ should be column vectors in $P$, not row vectors.
    $endgroup$
    – Bernard
    Dec 29 '18 at 21:15










  • $begingroup$
    Yeah I double checked what I have written is exactly what is in the book. Perhaps its a typo on the books part?
    $endgroup$
    – Justin Stevenson
    Dec 29 '18 at 21:28






  • 1




    $begingroup$
    I think $p^{-1}_{i,j}$ is supposed to mean the $i,j$ element of $P^{-1}$, rather than $(p_{ij})^{-1}$
    $endgroup$
    – Omnomnomnom
    Dec 29 '18 at 22:17
















0












$begingroup$


Let $eta in operatorname{End}_{R}(R^n)$ and let $A$ be the associated matrix of $eta$ with respect to the basis $(e_1, dots e_n)$. Let $f_i = sum p_{i,j}e_j$ where the matrix $P=(p_{i,j}) in GL_n(R)$. Show that $PAP^{-1}$ is the associated matrix of the basis $(f_1 dots f_n)$.



$underline{My space attempt:}$



$$eta(f_i) = etaleft(sum_{j=1}^n p_{i,j}e_j right) = sum_{j=1}^n eta(p_{i,j} e_j) = sum_{j=1}^n p_{i,j} eta(e_j)$$



$$=sum_{j=1}^n p_{i,j} left( sum_{k=1}^n a_{i,k} e_k right) = sum_{j=1}^n sum_{k=1}^n(p_{i,j}a_{i,k})e_k$$



$$=sum_{j=1}^nsum_{k=1}^n (p_{i,j}a_{i,k}p_{i,j}^{-1})p_{i,j}e_k$$



From here I see that I am very close but I'm not sure how to turn that $p_{i,j}e_k$ into the $f_{k}$ because the sum includes all the other terms. Any help is appreciated, thanks!






linear-algebra abstract-algebra change-of-basis free-modules







share|cite|improve this question











$endgroup$












  • $begingroup$
    The $i,j$ element of $P^{-1}$ is not the inverse of $p_{i,j}$.
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:14










  • $begingroup$
    Are you sure of the formulæ in this question? Normally the coordinates of the $f-i$s in basis $(e_1,dots,e_n)$ should be column vectors in $P$, not row vectors.
    $endgroup$
    – Bernard
    Dec 29 '18 at 21:15










  • $begingroup$
    Yeah I double checked what I have written is exactly what is in the book. Perhaps its a typo on the books part?
    $endgroup$
    – Justin Stevenson
    Dec 29 '18 at 21:28






  • 1




    $begingroup$
    I think $p^{-1}_{i,j}$ is supposed to mean the $i,j$ element of $P^{-1}$, rather than $(p_{ij})^{-1}$
    $endgroup$
    – Omnomnomnom
    Dec 29 '18 at 22:17














0












0








0





$begingroup$


Let $eta in operatorname{End}_{R}(R^n)$ and let $A$ be the associated matrix of $eta$ with respect to the basis $(e_1, dots e_n)$. Let $f_i = sum p_{i,j}e_j$ where the matrix $P=(p_{i,j}) in GL_n(R)$. Show that $PAP^{-1}$ is the associated matrix of the basis $(f_1 dots f_n)$.



$underline{My space attempt:}$



$$eta(f_i) = etaleft(sum_{j=1}^n p_{i,j}e_j right) = sum_{j=1}^n eta(p_{i,j} e_j) = sum_{j=1}^n p_{i,j} eta(e_j)$$



$$=sum_{j=1}^n p_{i,j} left( sum_{k=1}^n a_{i,k} e_k right) = sum_{j=1}^n sum_{k=1}^n(p_{i,j}a_{i,k})e_k$$



$$=sum_{j=1}^nsum_{k=1}^n (p_{i,j}a_{i,k}p_{i,j}^{-1})p_{i,j}e_k$$



From here I see that I am very close but I'm not sure how to turn that $p_{i,j}e_k$ into the $f_{k}$ because the sum includes all the other terms. Any help is appreciated, thanks!






linear-algebra abstract-algebra change-of-basis free-modules







share|cite|improve this question











$endgroup$




Let $eta in operatorname{End}_{R}(R^n)$ and let $A$ be the associated matrix of $eta$ with respect to the basis $(e_1, dots e_n)$. Let $f_i = sum p_{i,j}e_j$ where the matrix $P=(p_{i,j}) in GL_n(R)$. Show that $PAP^{-1}$ is the associated matrix of the basis $(f_1 dots f_n)$.



$underline{My space attempt:}$



$$eta(f_i) = etaleft(sum_{j=1}^n p_{i,j}e_j right) = sum_{j=1}^n eta(p_{i,j} e_j) = sum_{j=1}^n p_{i,j} eta(e_j)$$



$$=sum_{j=1}^n p_{i,j} left( sum_{k=1}^n a_{i,k} e_k right) = sum_{j=1}^n sum_{k=1}^n(p_{i,j}a_{i,k})e_k$$



$$=sum_{j=1}^nsum_{k=1}^n (p_{i,j}a_{i,k}p_{i,j}^{-1})p_{i,j}e_k$$



From here I see that I am very close but I'm not sure how to turn that $p_{i,j}e_k$ into the $f_{k}$ because the sum includes all the other terms. Any help is appreciated, thanks!






linear-algebra abstract-algebra change-of-basis free-modules




linear-algebra abstract-algebra change-of-basis free-modules






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 21:26







Justin Stevenson

















asked Dec 29 '18 at 21:02









Justin StevensonJustin Stevenson

957519




957519












  • $begingroup$
    The $i,j$ element of $P^{-1}$ is not the inverse of $p_{i,j}$.
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:14










  • $begingroup$
    Are you sure of the formulæ in this question? Normally the coordinates of the $f-i$s in basis $(e_1,dots,e_n)$ should be column vectors in $P$, not row vectors.
    $endgroup$
    – Bernard
    Dec 29 '18 at 21:15










  • $begingroup$
    Yeah I double checked what I have written is exactly what is in the book. Perhaps its a typo on the books part?
    $endgroup$
    – Justin Stevenson
    Dec 29 '18 at 21:28






  • 1




    $begingroup$
    I think $p^{-1}_{i,j}$ is supposed to mean the $i,j$ element of $P^{-1}$, rather than $(p_{ij})^{-1}$
    $endgroup$
    – Omnomnomnom
    Dec 29 '18 at 22:17


















  • $begingroup$
    The $i,j$ element of $P^{-1}$ is not the inverse of $p_{i,j}$.
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:14










  • $begingroup$
    Are you sure of the formulæ in this question? Normally the coordinates of the $f-i$s in basis $(e_1,dots,e_n)$ should be column vectors in $P$, not row vectors.
    $endgroup$
    – Bernard
    Dec 29 '18 at 21:15










  • $begingroup$
    Yeah I double checked what I have written is exactly what is in the book. Perhaps its a typo on the books part?
    $endgroup$
    – Justin Stevenson
    Dec 29 '18 at 21:28






  • 1




    $begingroup$
    I think $p^{-1}_{i,j}$ is supposed to mean the $i,j$ element of $P^{-1}$, rather than $(p_{ij})^{-1}$
    $endgroup$
    – Omnomnomnom
    Dec 29 '18 at 22:17
















$begingroup$
The $i,j$ element of $P^{-1}$ is not the inverse of $p_{i,j}$.
$endgroup$
– angryavian
Dec 29 '18 at 21:14




$begingroup$
The $i,j$ element of $P^{-1}$ is not the inverse of $p_{i,j}$.
$endgroup$
– angryavian
Dec 29 '18 at 21:14












$begingroup$
Are you sure of the formulæ in this question? Normally the coordinates of the $f-i$s in basis $(e_1,dots,e_n)$ should be column vectors in $P$, not row vectors.
$endgroup$
– Bernard
Dec 29 '18 at 21:15




$begingroup$
Are you sure of the formulæ in this question? Normally the coordinates of the $f-i$s in basis $(e_1,dots,e_n)$ should be column vectors in $P$, not row vectors.
$endgroup$
– Bernard
Dec 29 '18 at 21:15












$begingroup$
Yeah I double checked what I have written is exactly what is in the book. Perhaps its a typo on the books part?
$endgroup$
– Justin Stevenson
Dec 29 '18 at 21:28




$begingroup$
Yeah I double checked what I have written is exactly what is in the book. Perhaps its a typo on the books part?
$endgroup$
– Justin Stevenson
Dec 29 '18 at 21:28




1




1




$begingroup$
I think $p^{-1}_{i,j}$ is supposed to mean the $i,j$ element of $P^{-1}$, rather than $(p_{ij})^{-1}$
$endgroup$
– Omnomnomnom
Dec 29 '18 at 22:17




$begingroup$
I think $p^{-1}_{i,j}$ is supposed to mean the $i,j$ element of $P^{-1}$, rather than $(p_{ij})^{-1}$
$endgroup$
– Omnomnomnom
Dec 29 '18 at 22:17










1 Answer
1






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$begingroup$

I'll use the notation $p^{-1}_{ij}$ to refer to the $i,j$ entry of $P^{-1}$, which is what I suspect your book did.



We could finish your proof as follows:
$$
eta(f_i) = eta left( sum_{j=1}^n p_{ij} e_{j}right) = sum_{j=1}^n p_{ij} eta(e_{j})\
= sum_{j=1}^n p_{ij} left(sum_{k=1}^n a_{jk}e_{k}right) =
sum_{k=1}^n sum_{k=1}^n p_{ij}a_{jk} e_k
\= sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk} left(sum_{ell=1}^n p^{-1}_{kell}f_{ell}right)
= sum_{j=1}^n sum_{k=1}^n sum_{ell=1}^n p_{ij}a_{jk}p^{-1}_{kell} f_ell
$$
Now, it suffices to note that $sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk}p^{-1}_{kell}$ is indeed the $(ell,i)$ of the matrix $PAP^{-1}$.






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    $begingroup$

    I'll use the notation $p^{-1}_{ij}$ to refer to the $i,j$ entry of $P^{-1}$, which is what I suspect your book did.



    We could finish your proof as follows:
    $$
    eta(f_i) = eta left( sum_{j=1}^n p_{ij} e_{j}right) = sum_{j=1}^n p_{ij} eta(e_{j})\
    = sum_{j=1}^n p_{ij} left(sum_{k=1}^n a_{jk}e_{k}right) =
    sum_{k=1}^n sum_{k=1}^n p_{ij}a_{jk} e_k
    \= sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk} left(sum_{ell=1}^n p^{-1}_{kell}f_{ell}right)
    = sum_{j=1}^n sum_{k=1}^n sum_{ell=1}^n p_{ij}a_{jk}p^{-1}_{kell} f_ell
    $$
    Now, it suffices to note that $sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk}p^{-1}_{kell}$ is indeed the $(ell,i)$ of the matrix $PAP^{-1}$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I'll use the notation $p^{-1}_{ij}$ to refer to the $i,j$ entry of $P^{-1}$, which is what I suspect your book did.



      We could finish your proof as follows:
      $$
      eta(f_i) = eta left( sum_{j=1}^n p_{ij} e_{j}right) = sum_{j=1}^n p_{ij} eta(e_{j})\
      = sum_{j=1}^n p_{ij} left(sum_{k=1}^n a_{jk}e_{k}right) =
      sum_{k=1}^n sum_{k=1}^n p_{ij}a_{jk} e_k
      \= sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk} left(sum_{ell=1}^n p^{-1}_{kell}f_{ell}right)
      = sum_{j=1}^n sum_{k=1}^n sum_{ell=1}^n p_{ij}a_{jk}p^{-1}_{kell} f_ell
      $$
      Now, it suffices to note that $sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk}p^{-1}_{kell}$ is indeed the $(ell,i)$ of the matrix $PAP^{-1}$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I'll use the notation $p^{-1}_{ij}$ to refer to the $i,j$ entry of $P^{-1}$, which is what I suspect your book did.



        We could finish your proof as follows:
        $$
        eta(f_i) = eta left( sum_{j=1}^n p_{ij} e_{j}right) = sum_{j=1}^n p_{ij} eta(e_{j})\
        = sum_{j=1}^n p_{ij} left(sum_{k=1}^n a_{jk}e_{k}right) =
        sum_{k=1}^n sum_{k=1}^n p_{ij}a_{jk} e_k
        \= sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk} left(sum_{ell=1}^n p^{-1}_{kell}f_{ell}right)
        = sum_{j=1}^n sum_{k=1}^n sum_{ell=1}^n p_{ij}a_{jk}p^{-1}_{kell} f_ell
        $$
        Now, it suffices to note that $sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk}p^{-1}_{kell}$ is indeed the $(ell,i)$ of the matrix $PAP^{-1}$.






        share|cite|improve this answer









        $endgroup$



        I'll use the notation $p^{-1}_{ij}$ to refer to the $i,j$ entry of $P^{-1}$, which is what I suspect your book did.



        We could finish your proof as follows:
        $$
        eta(f_i) = eta left( sum_{j=1}^n p_{ij} e_{j}right) = sum_{j=1}^n p_{ij} eta(e_{j})\
        = sum_{j=1}^n p_{ij} left(sum_{k=1}^n a_{jk}e_{k}right) =
        sum_{k=1}^n sum_{k=1}^n p_{ij}a_{jk} e_k
        \= sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk} left(sum_{ell=1}^n p^{-1}_{kell}f_{ell}right)
        = sum_{j=1}^n sum_{k=1}^n sum_{ell=1}^n p_{ij}a_{jk}p^{-1}_{kell} f_ell
        $$
        Now, it suffices to note that $sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk}p^{-1}_{kell}$ is indeed the $(ell,i)$ of the matrix $PAP^{-1}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 30 '18 at 10:33









        OmnomnomnomOmnomnomnom

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