Why is the change of basis formula counter-intuitive? [See details] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Change of basisChange of basis = similarity?Change of Basis vs. Linear TransformationMatrices for change of basis linear transformationsConfusion about change of basis matrixIntuitive understanding of the $BAB^-1$ formula for changing basis in linear transformations.Standard Basis and Change of Basis MatrixStandard matrix linear transformation - change of basisHard change of basis/ linear transformation problemChange of basis difference between linear and bilinear transformation

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Why is the change of basis formula counter-intuitive? [See details]



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Change of basisChange of basis = similarity?Change of Basis vs. Linear TransformationMatrices for change of basis linear transformationsConfusion about change of basis matrixIntuitive understanding of the $BAB^-1$ formula for changing basis in linear transformations.Standard Basis and Change of Basis MatrixStandard matrix linear transformation - change of basisHard change of basis/ linear transformation problemChange of basis difference between linear and bilinear transformation










1












$begingroup$


The formula of change of basis $[T]_B' = P_B'leftarrow B[T]_BP_Bleftarrow B'$.



I don't understand why you need $P_Bleftarrow B'$? It seems to me that if you have the transformation expressed in $B$ already with $[T]_B$ you just need to translate to $B'$ by using $P_B'leftarrow B$ to get $[T]_B'$ rendering $P_Bleftarrow B'$ as useless. Can someone explain what I am missing here?










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$endgroup$











  • $begingroup$
    @littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it.
    $endgroup$
    – Dr.Stone
    4 hours ago
















1












$begingroup$


The formula of change of basis $[T]_B' = P_B'leftarrow B[T]_BP_Bleftarrow B'$.



I don't understand why you need $P_Bleftarrow B'$? It seems to me that if you have the transformation expressed in $B$ already with $[T]_B$ you just need to translate to $B'$ by using $P_B'leftarrow B$ to get $[T]_B'$ rendering $P_Bleftarrow B'$ as useless. Can someone explain what I am missing here?










share|cite|improve this question











$endgroup$











  • $begingroup$
    @littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it.
    $endgroup$
    – Dr.Stone
    4 hours ago














1












1








1





$begingroup$


The formula of change of basis $[T]_B' = P_B'leftarrow B[T]_BP_Bleftarrow B'$.



I don't understand why you need $P_Bleftarrow B'$? It seems to me that if you have the transformation expressed in $B$ already with $[T]_B$ you just need to translate to $B'$ by using $P_B'leftarrow B$ to get $[T]_B'$ rendering $P_Bleftarrow B'$ as useless. Can someone explain what I am missing here?










share|cite|improve this question











$endgroup$




The formula of change of basis $[T]_B' = P_B'leftarrow B[T]_BP_Bleftarrow B'$.



I don't understand why you need $P_Bleftarrow B'$? It seems to me that if you have the transformation expressed in $B$ already with $[T]_B$ you just need to translate to $B'$ by using $P_B'leftarrow B$ to get $[T]_B'$ rendering $P_Bleftarrow B'$ as useless. Can someone explain what I am missing here?







linear-algebra






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edited 11 mins ago









Carmeister

2,8792924




2,8792924










asked 5 hours ago









Dr.StoneDr.Stone

626




626











  • $begingroup$
    @littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it.
    $endgroup$
    – Dr.Stone
    4 hours ago

















  • $begingroup$
    @littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it.
    $endgroup$
    – Dr.Stone
    4 hours ago
















$begingroup$
@littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it.
$endgroup$
– Dr.Stone
4 hours ago





$begingroup$
@littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it.
$endgroup$
– Dr.Stone
4 hours ago











2 Answers
2






active

oldest

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$begingroup$

Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$

    Write $B=e_1,...,e_n, B' =e_1',...,e_n'$



    If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.



    So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_B'to B(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_Bto B'$ comes from on the left. This gives the formula






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.






      share|cite|improve this answer









      $endgroup$

















        2












        $begingroup$

        Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.






        share|cite|improve this answer









        $endgroup$















          2












          2








          2





          $begingroup$

          Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.






          share|cite|improve this answer









          $endgroup$



          Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 4 hours ago









          littleOlittleO

          30.6k649111




          30.6k649111





















              1












              $begingroup$

              Write $B=e_1,...,e_n, B' =e_1',...,e_n'$



              If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.



              So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_B'to B(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_Bto B'$ comes from on the left. This gives the formula






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                Write $B=e_1,...,e_n, B' =e_1',...,e_n'$



                If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.



                So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_B'to B(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_Bto B'$ comes from on the left. This gives the formula






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  Write $B=e_1,...,e_n, B' =e_1',...,e_n'$



                  If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.



                  So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_B'to B(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_Bto B'$ comes from on the left. This gives the formula






                  share|cite|improve this answer









                  $endgroup$



                  Write $B=e_1,...,e_n, B' =e_1',...,e_n'$



                  If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.



                  So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_B'to B(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_Bto B'$ comes from on the left. This gives the formula







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 4 hours ago









                  MaxMax

                  16.6k11144




                  16.6k11144



























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