The matrix P is the transition matrix from what basis B to the basis B'












2












$begingroup$


The Matrix



begin{equation}
P = begin{bmatrix}1 & 1 & 0 \ 0 & 1 & 3 \ 3 & 0 & 1
end{bmatrix}
end{equation} is the transition matrix from what basis B to the basis B' = {(1,0,0),(1,1,0),(1,1,1)} for R3?
I'm having a hard time with this one, any help is appreciated, thanks :)






linear-algebra matrices







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  • $begingroup$
    Is that the matrix you want?
    $endgroup$
    – user99914
    Apr 6 '15 at 22:38
















2












$begingroup$


The Matrix



begin{equation}
P = begin{bmatrix}1 & 1 & 0 \ 0 & 1 & 3 \ 3 & 0 & 1
end{bmatrix}
end{equation} is the transition matrix from what basis B to the basis B' = {(1,0,0),(1,1,0),(1,1,1)} for R3?
I'm having a hard time with this one, any help is appreciated, thanks :)






linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is that the matrix you want?
    $endgroup$
    – user99914
    Apr 6 '15 at 22:38














2












2








2


1



$begingroup$


The Matrix



begin{equation}
P = begin{bmatrix}1 & 1 & 0 \ 0 & 1 & 3 \ 3 & 0 & 1
end{bmatrix}
end{equation} is the transition matrix from what basis B to the basis B' = {(1,0,0),(1,1,0),(1,1,1)} for R3?
I'm having a hard time with this one, any help is appreciated, thanks :)






linear-algebra matrices







share|cite|improve this question











$endgroup$




The Matrix



begin{equation}
P = begin{bmatrix}1 & 1 & 0 \ 0 & 1 & 3 \ 3 & 0 & 1
end{bmatrix}
end{equation} is the transition matrix from what basis B to the basis B' = {(1,0,0),(1,1,0),(1,1,1)} for R3?
I'm having a hard time with this one, any help is appreciated, thanks :)






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 6 '15 at 22:38







user99914

















asked Apr 6 '15 at 22:32









fatty cakezfatty cakez

2114




2114












  • $begingroup$
    Is that the matrix you want?
    $endgroup$
    – user99914
    Apr 6 '15 at 22:38


















  • $begingroup$
    Is that the matrix you want?
    $endgroup$
    – user99914
    Apr 6 '15 at 22:38
















$begingroup$
Is that the matrix you want?
$endgroup$
– user99914
Apr 6 '15 at 22:38




$begingroup$
Is that the matrix you want?
$endgroup$
– user99914
Apr 6 '15 at 22:38










3 Answers
3






active

oldest

votes


















0












$begingroup$

let the basis $B = {u, v, w}.$ then $$Pu = 1(1,0,0)^T+3(1,1,1)^T to u = P^{-1}(4,3,3)^T = pmatrix{0.1&-0.1&0.3\0.9&0.1&-0.3\-0.3&0.3&0.1}pmatrix{4\3\3}=pmatrix{1\3\0.}$$



can you find $v, w?$






share|cite|improve this answer









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    0












    $begingroup$

    $mathbf{Hint}:$Recall that the transition matrix has the coordinates of the basis you are looking for as its columns, that is suppose your basis is B=${[(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3)]}={(b_1,b_2,b_3)}$



    $$1(1,0,0)+0(1,1,0)+3(1,1,1)=(4,3,3)$$
    $$1(1,0,0)+1(1,1,0)+0(1,1,1)=(2,1,0)$$



    what next?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hey. If you don't mind me asking. Is (4,3,3) , (2,1,0) and ( 4,4,1) the basis B? I arrived at the answer
      $endgroup$
      – Zhi J Teoh
      Nov 11 '15 at 4:01





















    0












    $begingroup$

    Possibly, the best approach to the matter is in this way.

    Given a matrix $bf P$, you can see it as converting the standard base ($bf I$)
    into the vertical vectors of $bf P$.
    $$
    begin{array}{l}
    {bf P} = {bf P};{bf I}quad Rightarrow quad left( {{bf p}_1 |{bf p}_2 |{bf p}_3 } right) = {bf P}left( {{bf u}_1 |{bf u}_2 |{bf u}_3 } right)quad Rightarrow quad {bf p}_k = {bf P};{bf u}_k \
    quad Downarrow ;{rm if},{bf P},{rm invertible} \
    {bf I} = {bf P}^{, - ,{bf 1}} ;{bf P} = {bf P};{bf P}^{, - ,{bf 1}} \
    end{array}
    $$



    If the vectors of $bf P$ also constitute a basis, that implies that the matrix is invertible
    and that you can reverse the process, getting

    $ {bf P}^{, - ,{bf 1}}$ converts ${bf P}$ to ${bf I}$, and

    ${bf P}$ converts $ {bf P}^{, - ,{bf 1}}$ to ${bf I}$.



    So, for any matrix $bf X$ and any invertible $bf P$ we can write
    $$
    {bf X} = {bf I};{bf X} = {bf P};{bf P}^{, - ,{bf 1}} ;{bf X} = {bf P};left( {{bf P}^{, - ,{bf 1}} ;{bf X}} right)
    $$



    Coming to your case, the above simply translates into
    $$ bbox[lightyellow] {
    {bf X} = left( {begin{array}{*{20}c}
    1 & 1 & 1 \
    0 & 1 & 1 \
    0 & 0 & 1 \
    end{array}} right) = {bf P};left( {{bf P}^{, - ,{bf 1}} ;{bf X}} right) = left( {begin{array}{*{20}c}
    1 & 1 & 0 \
    0 & 1 & 3 \
    3 & 0 & 1 \
    end{array}} right)left( {frac{1}{{10}}left( {begin{array}{*{20}c}
    1 & 0 & 3 \
    9 & {10} & 7 \
    { - 3} & 0 & 1 \
    end{array}} right)} right)
    } $$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      0












      $begingroup$

      let the basis $B = {u, v, w}.$ then $$Pu = 1(1,0,0)^T+3(1,1,1)^T to u = P^{-1}(4,3,3)^T = pmatrix{0.1&-0.1&0.3\0.9&0.1&-0.3\-0.3&0.3&0.1}pmatrix{4\3\3}=pmatrix{1\3\0.}$$



      can you find $v, w?$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        let the basis $B = {u, v, w}.$ then $$Pu = 1(1,0,0)^T+3(1,1,1)^T to u = P^{-1}(4,3,3)^T = pmatrix{0.1&-0.1&0.3\0.9&0.1&-0.3\-0.3&0.3&0.1}pmatrix{4\3\3}=pmatrix{1\3\0.}$$



        can you find $v, w?$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          let the basis $B = {u, v, w}.$ then $$Pu = 1(1,0,0)^T+3(1,1,1)^T to u = P^{-1}(4,3,3)^T = pmatrix{0.1&-0.1&0.3\0.9&0.1&-0.3\-0.3&0.3&0.1}pmatrix{4\3\3}=pmatrix{1\3\0.}$$



          can you find $v, w?$






          share|cite|improve this answer









          $endgroup$



          let the basis $B = {u, v, w}.$ then $$Pu = 1(1,0,0)^T+3(1,1,1)^T to u = P^{-1}(4,3,3)^T = pmatrix{0.1&-0.1&0.3\0.9&0.1&-0.3\-0.3&0.3&0.1}pmatrix{4\3\3}=pmatrix{1\3\0.}$$



          can you find $v, w?$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 6 '15 at 22:57









          abelabel

          26.6k12048




          26.6k12048























              0












              $begingroup$

              $mathbf{Hint}:$Recall that the transition matrix has the coordinates of the basis you are looking for as its columns, that is suppose your basis is B=${[(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3)]}={(b_1,b_2,b_3)}$



              $$1(1,0,0)+0(1,1,0)+3(1,1,1)=(4,3,3)$$
              $$1(1,0,0)+1(1,1,0)+0(1,1,1)=(2,1,0)$$



              what next?






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Hey. If you don't mind me asking. Is (4,3,3) , (2,1,0) and ( 4,4,1) the basis B? I arrived at the answer
                $endgroup$
                – Zhi J Teoh
                Nov 11 '15 at 4:01


















              0












              $begingroup$

              $mathbf{Hint}:$Recall that the transition matrix has the coordinates of the basis you are looking for as its columns, that is suppose your basis is B=${[(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3)]}={(b_1,b_2,b_3)}$



              $$1(1,0,0)+0(1,1,0)+3(1,1,1)=(4,3,3)$$
              $$1(1,0,0)+1(1,1,0)+0(1,1,1)=(2,1,0)$$



              what next?






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Hey. If you don't mind me asking. Is (4,3,3) , (2,1,0) and ( 4,4,1) the basis B? I arrived at the answer
                $endgroup$
                – Zhi J Teoh
                Nov 11 '15 at 4:01
















              0












              0








              0





              $begingroup$

              $mathbf{Hint}:$Recall that the transition matrix has the coordinates of the basis you are looking for as its columns, that is suppose your basis is B=${[(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3)]}={(b_1,b_2,b_3)}$



              $$1(1,0,0)+0(1,1,0)+3(1,1,1)=(4,3,3)$$
              $$1(1,0,0)+1(1,1,0)+0(1,1,1)=(2,1,0)$$



              what next?






              share|cite|improve this answer









              $endgroup$



              $mathbf{Hint}:$Recall that the transition matrix has the coordinates of the basis you are looking for as its columns, that is suppose your basis is B=${[(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3)]}={(b_1,b_2,b_3)}$



              $$1(1,0,0)+0(1,1,0)+3(1,1,1)=(4,3,3)$$
              $$1(1,0,0)+1(1,1,0)+0(1,1,1)=(2,1,0)$$



              what next?







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Apr 6 '15 at 22:57









              QualityQuality

              2,99141543




              2,99141543












              • $begingroup$
                Hey. If you don't mind me asking. Is (4,3,3) , (2,1,0) and ( 4,4,1) the basis B? I arrived at the answer
                $endgroup$
                – Zhi J Teoh
                Nov 11 '15 at 4:01




















              • $begingroup$
                Hey. If you don't mind me asking. Is (4,3,3) , (2,1,0) and ( 4,4,1) the basis B? I arrived at the answer
                $endgroup$
                – Zhi J Teoh
                Nov 11 '15 at 4:01


















              $begingroup$
              Hey. If you don't mind me asking. Is (4,3,3) , (2,1,0) and ( 4,4,1) the basis B? I arrived at the answer
              $endgroup$
              – Zhi J Teoh
              Nov 11 '15 at 4:01






              $begingroup$
              Hey. If you don't mind me asking. Is (4,3,3) , (2,1,0) and ( 4,4,1) the basis B? I arrived at the answer
              $endgroup$
              – Zhi J Teoh
              Nov 11 '15 at 4:01













              0












              $begingroup$

              Possibly, the best approach to the matter is in this way.

              Given a matrix $bf P$, you can see it as converting the standard base ($bf I$)
              into the vertical vectors of $bf P$.
              $$
              begin{array}{l}
              {bf P} = {bf P};{bf I}quad Rightarrow quad left( {{bf p}_1 |{bf p}_2 |{bf p}_3 } right) = {bf P}left( {{bf u}_1 |{bf u}_2 |{bf u}_3 } right)quad Rightarrow quad {bf p}_k = {bf P};{bf u}_k \
              quad Downarrow ;{rm if},{bf P},{rm invertible} \
              {bf I} = {bf P}^{, - ,{bf 1}} ;{bf P} = {bf P};{bf P}^{, - ,{bf 1}} \
              end{array}
              $$



              If the vectors of $bf P$ also constitute a basis, that implies that the matrix is invertible
              and that you can reverse the process, getting

              $ {bf P}^{, - ,{bf 1}}$ converts ${bf P}$ to ${bf I}$, and

              ${bf P}$ converts $ {bf P}^{, - ,{bf 1}}$ to ${bf I}$.



              So, for any matrix $bf X$ and any invertible $bf P$ we can write
              $$
              {bf X} = {bf I};{bf X} = {bf P};{bf P}^{, - ,{bf 1}} ;{bf X} = {bf P};left( {{bf P}^{, - ,{bf 1}} ;{bf X}} right)
              $$



              Coming to your case, the above simply translates into
              $$ bbox[lightyellow] {
              {bf X} = left( {begin{array}{*{20}c}
              1 & 1 & 1 \
              0 & 1 & 1 \
              0 & 0 & 1 \
              end{array}} right) = {bf P};left( {{bf P}^{, - ,{bf 1}} ;{bf X}} right) = left( {begin{array}{*{20}c}
              1 & 1 & 0 \
              0 & 1 & 3 \
              3 & 0 & 1 \
              end{array}} right)left( {frac{1}{{10}}left( {begin{array}{*{20}c}
              1 & 0 & 3 \
              9 & {10} & 7 \
              { - 3} & 0 & 1 \
              end{array}} right)} right)
              } $$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Possibly, the best approach to the matter is in this way.

                Given a matrix $bf P$, you can see it as converting the standard base ($bf I$)
                into the vertical vectors of $bf P$.
                $$
                begin{array}{l}
                {bf P} = {bf P};{bf I}quad Rightarrow quad left( {{bf p}_1 |{bf p}_2 |{bf p}_3 } right) = {bf P}left( {{bf u}_1 |{bf u}_2 |{bf u}_3 } right)quad Rightarrow quad {bf p}_k = {bf P};{bf u}_k \
                quad Downarrow ;{rm if},{bf P},{rm invertible} \
                {bf I} = {bf P}^{, - ,{bf 1}} ;{bf P} = {bf P};{bf P}^{, - ,{bf 1}} \
                end{array}
                $$



                If the vectors of $bf P$ also constitute a basis, that implies that the matrix is invertible
                and that you can reverse the process, getting

                $ {bf P}^{, - ,{bf 1}}$ converts ${bf P}$ to ${bf I}$, and

                ${bf P}$ converts $ {bf P}^{, - ,{bf 1}}$ to ${bf I}$.



                So, for any matrix $bf X$ and any invertible $bf P$ we can write
                $$
                {bf X} = {bf I};{bf X} = {bf P};{bf P}^{, - ,{bf 1}} ;{bf X} = {bf P};left( {{bf P}^{, - ,{bf 1}} ;{bf X}} right)
                $$



                Coming to your case, the above simply translates into
                $$ bbox[lightyellow] {
                {bf X} = left( {begin{array}{*{20}c}
                1 & 1 & 1 \
                0 & 1 & 1 \
                0 & 0 & 1 \
                end{array}} right) = {bf P};left( {{bf P}^{, - ,{bf 1}} ;{bf X}} right) = left( {begin{array}{*{20}c}
                1 & 1 & 0 \
                0 & 1 & 3 \
                3 & 0 & 1 \
                end{array}} right)left( {frac{1}{{10}}left( {begin{array}{*{20}c}
                1 & 0 & 3 \
                9 & {10} & 7 \
                { - 3} & 0 & 1 \
                end{array}} right)} right)
                } $$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Possibly, the best approach to the matter is in this way.

                  Given a matrix $bf P$, you can see it as converting the standard base ($bf I$)
                  into the vertical vectors of $bf P$.
                  $$
                  begin{array}{l}
                  {bf P} = {bf P};{bf I}quad Rightarrow quad left( {{bf p}_1 |{bf p}_2 |{bf p}_3 } right) = {bf P}left( {{bf u}_1 |{bf u}_2 |{bf u}_3 } right)quad Rightarrow quad {bf p}_k = {bf P};{bf u}_k \
                  quad Downarrow ;{rm if},{bf P},{rm invertible} \
                  {bf I} = {bf P}^{, - ,{bf 1}} ;{bf P} = {bf P};{bf P}^{, - ,{bf 1}} \
                  end{array}
                  $$



                  If the vectors of $bf P$ also constitute a basis, that implies that the matrix is invertible
                  and that you can reverse the process, getting

                  $ {bf P}^{, - ,{bf 1}}$ converts ${bf P}$ to ${bf I}$, and

                  ${bf P}$ converts $ {bf P}^{, - ,{bf 1}}$ to ${bf I}$.



                  So, for any matrix $bf X$ and any invertible $bf P$ we can write
                  $$
                  {bf X} = {bf I};{bf X} = {bf P};{bf P}^{, - ,{bf 1}} ;{bf X} = {bf P};left( {{bf P}^{, - ,{bf 1}} ;{bf X}} right)
                  $$



                  Coming to your case, the above simply translates into
                  $$ bbox[lightyellow] {
                  {bf X} = left( {begin{array}{*{20}c}
                  1 & 1 & 1 \
                  0 & 1 & 1 \
                  0 & 0 & 1 \
                  end{array}} right) = {bf P};left( {{bf P}^{, - ,{bf 1}} ;{bf X}} right) = left( {begin{array}{*{20}c}
                  1 & 1 & 0 \
                  0 & 1 & 3 \
                  3 & 0 & 1 \
                  end{array}} right)left( {frac{1}{{10}}left( {begin{array}{*{20}c}
                  1 & 0 & 3 \
                  9 & {10} & 7 \
                  { - 3} & 0 & 1 \
                  end{array}} right)} right)
                  } $$






                  share|cite|improve this answer









                  $endgroup$



                  Possibly, the best approach to the matter is in this way.

                  Given a matrix $bf P$, you can see it as converting the standard base ($bf I$)
                  into the vertical vectors of $bf P$.
                  $$
                  begin{array}{l}
                  {bf P} = {bf P};{bf I}quad Rightarrow quad left( {{bf p}_1 |{bf p}_2 |{bf p}_3 } right) = {bf P}left( {{bf u}_1 |{bf u}_2 |{bf u}_3 } right)quad Rightarrow quad {bf p}_k = {bf P};{bf u}_k \
                  quad Downarrow ;{rm if},{bf P},{rm invertible} \
                  {bf I} = {bf P}^{, - ,{bf 1}} ;{bf P} = {bf P};{bf P}^{, - ,{bf 1}} \
                  end{array}
                  $$



                  If the vectors of $bf P$ also constitute a basis, that implies that the matrix is invertible
                  and that you can reverse the process, getting

                  $ {bf P}^{, - ,{bf 1}}$ converts ${bf P}$ to ${bf I}$, and

                  ${bf P}$ converts $ {bf P}^{, - ,{bf 1}}$ to ${bf I}$.



                  So, for any matrix $bf X$ and any invertible $bf P$ we can write
                  $$
                  {bf X} = {bf I};{bf X} = {bf P};{bf P}^{, - ,{bf 1}} ;{bf X} = {bf P};left( {{bf P}^{, - ,{bf 1}} ;{bf X}} right)
                  $$



                  Coming to your case, the above simply translates into
                  $$ bbox[lightyellow] {
                  {bf X} = left( {begin{array}{*{20}c}
                  1 & 1 & 1 \
                  0 & 1 & 1 \
                  0 & 0 & 1 \
                  end{array}} right) = {bf P};left( {{bf P}^{, - ,{bf 1}} ;{bf X}} right) = left( {begin{array}{*{20}c}
                  1 & 1 & 0 \
                  0 & 1 & 3 \
                  3 & 0 & 1 \
                  end{array}} right)left( {frac{1}{{10}}left( {begin{array}{*{20}c}
                  1 & 0 & 3 \
                  9 & {10} & 7 \
                  { - 3} & 0 & 1 \
                  end{array}} right)} right)
                  } $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered May 4 '17 at 13:15









                  G CabG Cab

                  19.6k31239




                  19.6k31239






























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                      Listed building (literalmente: prédio listado ) é toda estrutura protegida por lei pelo complexo sistema de tombamento do Reino Unido. Um listed building é qualquer estrutura oficialmente designada como tendo especial interesse arquitetônico, histórico ou cultural. É um estatuto amplamente usado, aplicado a centenas de milhares de estruturas. Estruturas tombadas não podem ser alteradas sem permissão especial, e desfrutam de isenções fiscais e outros benefícios. Cada um dos países do Reino Unido trata tombamento de forma diferente: Índice 1 Inglaterra 2 País de Gales 3 Irlanda do Norte 4 Escócia 5 Referências Inglaterra | O Palácio de Buckingham, Londres. As estruturas tombadas são listadas no Statutory List of Buildings of Special Architectural or Historic Interest , e incluem campos, sinais de trânsito, piers e áreas geográficas. A principal legislação pertinente é o Ancient Monuments and Archaeological Areas Act 1979 , [ 1 ] embora legisla...
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