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












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The formula of change of basis $[T]_{B'} = P_{B'leftarrow B}[T]_{B}P_{Bleftarrow B'}$.



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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  • $begingroup$
    @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 $[T]_{B'} = P_{B'leftarrow B}[T]_{B}P_{Bleftarrow B'}$.



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







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 $[T]_{B'} = P_{B'leftarrow B}[T]_{B}P_{Bleftarrow B'}$.



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







share|cite|improve this question











$endgroup$




The formula of change of basis $[T]_{B'} = P_{B'leftarrow B}[T]_{B}P_{Bleftarrow B'}$.



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




linear-algebra






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edited 10 mins ago









Carmeister

2,8792924




2,8792924










asked 4 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
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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.






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$endgroup$





















    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






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      2 Answers
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      2 Answers
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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.






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        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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