Three bases, one linear map, find vector $x$












0












$begingroup$


Given three bases $B, C$ and $D$ and linear map $f:Bbb R^2toBbb R^2$, and $x$ from $Bbb R^2$. We also know that $[x]_B=(x_1,x_2)^T$.



$$[f]_{Bto C}=begin{bmatrix}2 & 3 \3 & 1 end{bmatrix}$$



$$[f]_{Dto C}=begin{bmatrix}1 & 3 \-1 & 5 end{bmatrix}$$



Find $x$ in $D$ with respect to $x_1$ and $x_2$.






linear-algebra linear-transformations change-of-basis







share|cite|improve this question











$endgroup$












  • $begingroup$
    "from $B$ to $D$" is equivalent to "from $B$ to $C$ AND from $C$ to $D$". You know how to go from $B$ to $C$. You also know how to go from $D$ to $C$... so, ¿How to do the inverse walk?
    $endgroup$
    – Tito Eliatron
    Jan 9 at 11:28










  • $begingroup$
    the inverse is 1/8 begin{bmatrix}5 & -3 \1 & 1 end{bmatrix} . But is f necessarily identity?
    $endgroup$
    – Jan Lhoták
    Jan 9 at 11:32










  • $begingroup$
    Because if f is identity you can just generate x in D by multiplying [f]C→D [f]B→C x in B. Which would be begin{bmatrix}1/8 & 3/2 \5/8 & 1/2 end{bmatrix} -> (1/8x1 + 3/2x2; 5/8x1 + 1/2x2)T.
    $endgroup$
    – Jan Lhoták
    Jan 9 at 11:37












  • $begingroup$
    Ok is the answer: $[x]_D=(1/8 x_1 + 3/2 x_2 ,5/8 x_1 + 1/2 x_2)^T$ ?
    $endgroup$
    – Jan Lhoták
    Jan 9 at 12:04












  • $begingroup$
    Yes @JanLhoták, correct
    $endgroup$
    – Shubham Johri
    Jan 9 at 12:10
















0












$begingroup$


Given three bases $B, C$ and $D$ and linear map $f:Bbb R^2toBbb R^2$, and $x$ from $Bbb R^2$. We also know that $[x]_B=(x_1,x_2)^T$.



$$[f]_{Bto C}=begin{bmatrix}2 & 3 \3 & 1 end{bmatrix}$$



$$[f]_{Dto C}=begin{bmatrix}1 & 3 \-1 & 5 end{bmatrix}$$



Find $x$ in $D$ with respect to $x_1$ and $x_2$.






linear-algebra linear-transformations change-of-basis







share|cite|improve this question











$endgroup$












  • $begingroup$
    "from $B$ to $D$" is equivalent to "from $B$ to $C$ AND from $C$ to $D$". You know how to go from $B$ to $C$. You also know how to go from $D$ to $C$... so, ¿How to do the inverse walk?
    $endgroup$
    – Tito Eliatron
    Jan 9 at 11:28










  • $begingroup$
    the inverse is 1/8 begin{bmatrix}5 & -3 \1 & 1 end{bmatrix} . But is f necessarily identity?
    $endgroup$
    – Jan Lhoták
    Jan 9 at 11:32










  • $begingroup$
    Because if f is identity you can just generate x in D by multiplying [f]C→D [f]B→C x in B. Which would be begin{bmatrix}1/8 & 3/2 \5/8 & 1/2 end{bmatrix} -> (1/8x1 + 3/2x2; 5/8x1 + 1/2x2)T.
    $endgroup$
    – Jan Lhoták
    Jan 9 at 11:37












  • $begingroup$
    Ok is the answer: $[x]_D=(1/8 x_1 + 3/2 x_2 ,5/8 x_1 + 1/2 x_2)^T$ ?
    $endgroup$
    – Jan Lhoták
    Jan 9 at 12:04












  • $begingroup$
    Yes @JanLhoták, correct
    $endgroup$
    – Shubham Johri
    Jan 9 at 12:10














0












0








0





$begingroup$


Given three bases $B, C$ and $D$ and linear map $f:Bbb R^2toBbb R^2$, and $x$ from $Bbb R^2$. We also know that $[x]_B=(x_1,x_2)^T$.



$$[f]_{Bto C}=begin{bmatrix}2 & 3 \3 & 1 end{bmatrix}$$



$$[f]_{Dto C}=begin{bmatrix}1 & 3 \-1 & 5 end{bmatrix}$$



Find $x$ in $D$ with respect to $x_1$ and $x_2$.






linear-algebra linear-transformations change-of-basis







share|cite|improve this question











$endgroup$




Given three bases $B, C$ and $D$ and linear map $f:Bbb R^2toBbb R^2$, and $x$ from $Bbb R^2$. We also know that $[x]_B=(x_1,x_2)^T$.



$$[f]_{Bto C}=begin{bmatrix}2 & 3 \3 & 1 end{bmatrix}$$



$$[f]_{Dto C}=begin{bmatrix}1 & 3 \-1 & 5 end{bmatrix}$$



Find $x$ in $D$ with respect to $x_1$ and $x_2$.






linear-algebra linear-transformations change-of-basis




linear-algebra linear-transformations change-of-basis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 12:03







Jan Lhoták

















asked Jan 9 at 11:24









Jan LhotákJan Lhoták

275




275












  • $begingroup$
    "from $B$ to $D$" is equivalent to "from $B$ to $C$ AND from $C$ to $D$". You know how to go from $B$ to $C$. You also know how to go from $D$ to $C$... so, ¿How to do the inverse walk?
    $endgroup$
    – Tito Eliatron
    Jan 9 at 11:28










  • $begingroup$
    the inverse is 1/8 begin{bmatrix}5 & -3 \1 & 1 end{bmatrix} . But is f necessarily identity?
    $endgroup$
    – Jan Lhoták
    Jan 9 at 11:32










  • $begingroup$
    Because if f is identity you can just generate x in D by multiplying [f]C→D [f]B→C x in B. Which would be begin{bmatrix}1/8 & 3/2 \5/8 & 1/2 end{bmatrix} -> (1/8x1 + 3/2x2; 5/8x1 + 1/2x2)T.
    $endgroup$
    – Jan Lhoták
    Jan 9 at 11:37












  • $begingroup$
    Ok is the answer: $[x]_D=(1/8 x_1 + 3/2 x_2 ,5/8 x_1 + 1/2 x_2)^T$ ?
    $endgroup$
    – Jan Lhoták
    Jan 9 at 12:04












  • $begingroup$
    Yes @JanLhoták, correct
    $endgroup$
    – Shubham Johri
    Jan 9 at 12:10


















  • $begingroup$
    "from $B$ to $D$" is equivalent to "from $B$ to $C$ AND from $C$ to $D$". You know how to go from $B$ to $C$. You also know how to go from $D$ to $C$... so, ¿How to do the inverse walk?
    $endgroup$
    – Tito Eliatron
    Jan 9 at 11:28










  • $begingroup$
    the inverse is 1/8 begin{bmatrix}5 & -3 \1 & 1 end{bmatrix} . But is f necessarily identity?
    $endgroup$
    – Jan Lhoták
    Jan 9 at 11:32










  • $begingroup$
    Because if f is identity you can just generate x in D by multiplying [f]C→D [f]B→C x in B. Which would be begin{bmatrix}1/8 & 3/2 \5/8 & 1/2 end{bmatrix} -> (1/8x1 + 3/2x2; 5/8x1 + 1/2x2)T.
    $endgroup$
    – Jan Lhoták
    Jan 9 at 11:37












  • $begingroup$
    Ok is the answer: $[x]_D=(1/8 x_1 + 3/2 x_2 ,5/8 x_1 + 1/2 x_2)^T$ ?
    $endgroup$
    – Jan Lhoták
    Jan 9 at 12:04












  • $begingroup$
    Yes @JanLhoták, correct
    $endgroup$
    – Shubham Johri
    Jan 9 at 12:10
















$begingroup$
"from $B$ to $D$" is equivalent to "from $B$ to $C$ AND from $C$ to $D$". You know how to go from $B$ to $C$. You also know how to go from $D$ to $C$... so, ¿How to do the inverse walk?
$endgroup$
– Tito Eliatron
Jan 9 at 11:28




$begingroup$
"from $B$ to $D$" is equivalent to "from $B$ to $C$ AND from $C$ to $D$". You know how to go from $B$ to $C$. You also know how to go from $D$ to $C$... so, ¿How to do the inverse walk?
$endgroup$
– Tito Eliatron
Jan 9 at 11:28












$begingroup$
the inverse is 1/8 begin{bmatrix}5 & -3 \1 & 1 end{bmatrix} . But is f necessarily identity?
$endgroup$
– Jan Lhoták
Jan 9 at 11:32




$begingroup$
the inverse is 1/8 begin{bmatrix}5 & -3 \1 & 1 end{bmatrix} . But is f necessarily identity?
$endgroup$
– Jan Lhoták
Jan 9 at 11:32












$begingroup$
Because if f is identity you can just generate x in D by multiplying [f]C→D [f]B→C x in B. Which would be begin{bmatrix}1/8 & 3/2 \5/8 & 1/2 end{bmatrix} -> (1/8x1 + 3/2x2; 5/8x1 + 1/2x2)T.
$endgroup$
– Jan Lhoták
Jan 9 at 11:37






$begingroup$
Because if f is identity you can just generate x in D by multiplying [f]C→D [f]B→C x in B. Which would be begin{bmatrix}1/8 & 3/2 \5/8 & 1/2 end{bmatrix} -> (1/8x1 + 3/2x2; 5/8x1 + 1/2x2)T.
$endgroup$
– Jan Lhoták
Jan 9 at 11:37














$begingroup$
Ok is the answer: $[x]_D=(1/8 x_1 + 3/2 x_2 ,5/8 x_1 + 1/2 x_2)^T$ ?
$endgroup$
– Jan Lhoták
Jan 9 at 12:04






$begingroup$
Ok is the answer: $[x]_D=(1/8 x_1 + 3/2 x_2 ,5/8 x_1 + 1/2 x_2)^T$ ?
$endgroup$
– Jan Lhoták
Jan 9 at 12:04














$begingroup$
Yes @JanLhoták, correct
$endgroup$
– Shubham Johri
Jan 9 at 12:10




$begingroup$
Yes @JanLhoták, correct
$endgroup$
– Shubham Johri
Jan 9 at 12:10










1 Answer
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$begingroup$

$$[f]_{Dto C}=begin{bmatrix}1&3\-1&5end{bmatrix}implies [f^{-1}]_{Cto D}=begin{bmatrix}1&3\-1&5end{bmatrix}^{-1}$$



$v=[f]_{Bto C}x$ gives you the image of $x$ under $f$ with respect to basis $C$. $[f^{-1}]_{Cto D}v$ gives you the vector in $Bbb R^2$ with respect to basis $D$ whose image under $f$ is $v$ with respect to basis $C$; in other words, the representation of $x$ in $D$.



$$[x]_D=begin{bmatrix}1&3\-1&5end{bmatrix}^{-1}begin{bmatrix}2 & 3 \3 & 1 end{bmatrix}[x]_B=frac18begin{bmatrix}1&12\5&4end{bmatrix}begin{bmatrix}x_1\x_2end{bmatrix}$$






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    $begingroup$

    $$[f]_{Dto C}=begin{bmatrix}1&3\-1&5end{bmatrix}implies [f^{-1}]_{Cto D}=begin{bmatrix}1&3\-1&5end{bmatrix}^{-1}$$



    $v=[f]_{Bto C}x$ gives you the image of $x$ under $f$ with respect to basis $C$. $[f^{-1}]_{Cto D}v$ gives you the vector in $Bbb R^2$ with respect to basis $D$ whose image under $f$ is $v$ with respect to basis $C$; in other words, the representation of $x$ in $D$.



    $$[x]_D=begin{bmatrix}1&3\-1&5end{bmatrix}^{-1}begin{bmatrix}2 & 3 \3 & 1 end{bmatrix}[x]_B=frac18begin{bmatrix}1&12\5&4end{bmatrix}begin{bmatrix}x_1\x_2end{bmatrix}$$






    share|cite|improve this answer









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      0












      $begingroup$

      $$[f]_{Dto C}=begin{bmatrix}1&3\-1&5end{bmatrix}implies [f^{-1}]_{Cto D}=begin{bmatrix}1&3\-1&5end{bmatrix}^{-1}$$



      $v=[f]_{Bto C}x$ gives you the image of $x$ under $f$ with respect to basis $C$. $[f^{-1}]_{Cto D}v$ gives you the vector in $Bbb R^2$ with respect to basis $D$ whose image under $f$ is $v$ with respect to basis $C$; in other words, the representation of $x$ in $D$.



      $$[x]_D=begin{bmatrix}1&3\-1&5end{bmatrix}^{-1}begin{bmatrix}2 & 3 \3 & 1 end{bmatrix}[x]_B=frac18begin{bmatrix}1&12\5&4end{bmatrix}begin{bmatrix}x_1\x_2end{bmatrix}$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $$[f]_{Dto C}=begin{bmatrix}1&3\-1&5end{bmatrix}implies [f^{-1}]_{Cto D}=begin{bmatrix}1&3\-1&5end{bmatrix}^{-1}$$



        $v=[f]_{Bto C}x$ gives you the image of $x$ under $f$ with respect to basis $C$. $[f^{-1}]_{Cto D}v$ gives you the vector in $Bbb R^2$ with respect to basis $D$ whose image under $f$ is $v$ with respect to basis $C$; in other words, the representation of $x$ in $D$.



        $$[x]_D=begin{bmatrix}1&3\-1&5end{bmatrix}^{-1}begin{bmatrix}2 & 3 \3 & 1 end{bmatrix}[x]_B=frac18begin{bmatrix}1&12\5&4end{bmatrix}begin{bmatrix}x_1\x_2end{bmatrix}$$






        share|cite|improve this answer









        $endgroup$



        $$[f]_{Dto C}=begin{bmatrix}1&3\-1&5end{bmatrix}implies [f^{-1}]_{Cto D}=begin{bmatrix}1&3\-1&5end{bmatrix}^{-1}$$



        $v=[f]_{Bto C}x$ gives you the image of $x$ under $f$ with respect to basis $C$. $[f^{-1}]_{Cto D}v$ gives you the vector in $Bbb R^2$ with respect to basis $D$ whose image under $f$ is $v$ with respect to basis $C$; in other words, the representation of $x$ in $D$.



        $$[x]_D=begin{bmatrix}1&3\-1&5end{bmatrix}^{-1}begin{bmatrix}2 & 3 \3 & 1 end{bmatrix}[x]_B=frac18begin{bmatrix}1&12\5&4end{bmatrix}begin{bmatrix}x_1\x_2end{bmatrix}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 12:10









        Shubham JohriShubham Johri

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