Standard matrix of a transformation, matrix representation [closed]












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I know that the answer is $left[begin{matrix} 2 & -1 \ 1 & 1 end{matrix}right]$, but how to get the answer?




Let $mathcal{B} = { mathbf{b}_1 , mathbf{b}_2 }$ be the basis for $mathbb{R}^2$ with $mathbf{b}_1 = left [ begin{matrix} 1 \ 1 end{matrix} right ]$, $mathbf{b}_2 = left [ begin{matrix} 0 \ 1 end{matrix} right ]$. Furthermore, let $T: mathbb{R}^2 to mathbb{R}^2$ be a linear transformation. The matrix representation of $T$ with respect to $mathcal{B}$ is $[T]_mathcal{B} = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]$.
What is the standard matrix of $T$?





Original problem: The Problem






linear-algebra matrices transformation







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closed as off-topic by Dietrich Burde, Namaste, max_zorn, metamorphy, Ali Caglayan Jan 5 at 19:31


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  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, Namaste, max_zorn, metamorphy, Ali Caglayan

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    You'll get a better response if you write the question here rather than linking to an image, and also explain what you have tried and which step you got stuck on.
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    – littleO
    Jan 5 at 13:50










  • $begingroup$
    @JohnDoe the hint helped, thanks :)
    $endgroup$
    – Antoni Malecki
    Jan 5 at 14:49












  • $begingroup$
    @AntoniMalecki great! :)
    $endgroup$
    – John Doe
    Jan 5 at 14:51
















-1












$begingroup$


I know that the answer is $left[begin{matrix} 2 & -1 \ 1 & 1 end{matrix}right]$, but how to get the answer?




Let $mathcal{B} = { mathbf{b}_1 , mathbf{b}_2 }$ be the basis for $mathbb{R}^2$ with $mathbf{b}_1 = left [ begin{matrix} 1 \ 1 end{matrix} right ]$, $mathbf{b}_2 = left [ begin{matrix} 0 \ 1 end{matrix} right ]$. Furthermore, let $T: mathbb{R}^2 to mathbb{R}^2$ be a linear transformation. The matrix representation of $T$ with respect to $mathcal{B}$ is $[T]_mathcal{B} = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]$.
What is the standard matrix of $T$?





Original problem: The Problem






linear-algebra matrices transformation







share|cite|improve this question











$endgroup$



closed as off-topic by Dietrich Burde, Namaste, max_zorn, metamorphy, Ali Caglayan Jan 5 at 19:31


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, Namaste, max_zorn, metamorphy, Ali Caglayan

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Hint
    $endgroup$
    – John Doe
    Jan 5 at 13:50








  • 1




    $begingroup$
    You'll get a better response if you write the question here rather than linking to an image, and also explain what you have tried and which step you got stuck on.
    $endgroup$
    – littleO
    Jan 5 at 13:50










  • $begingroup$
    @JohnDoe the hint helped, thanks :)
    $endgroup$
    – Antoni Malecki
    Jan 5 at 14:49












  • $begingroup$
    @AntoniMalecki great! :)
    $endgroup$
    – John Doe
    Jan 5 at 14:51














-1












-1








-1





$begingroup$


I know that the answer is $left[begin{matrix} 2 & -1 \ 1 & 1 end{matrix}right]$, but how to get the answer?




Let $mathcal{B} = { mathbf{b}_1 , mathbf{b}_2 }$ be the basis for $mathbb{R}^2$ with $mathbf{b}_1 = left [ begin{matrix} 1 \ 1 end{matrix} right ]$, $mathbf{b}_2 = left [ begin{matrix} 0 \ 1 end{matrix} right ]$. Furthermore, let $T: mathbb{R}^2 to mathbb{R}^2$ be a linear transformation. The matrix representation of $T$ with respect to $mathcal{B}$ is $[T]_mathcal{B} = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]$.
What is the standard matrix of $T$?





Original problem: The Problem






linear-algebra matrices transformation







share|cite|improve this question











$endgroup$




I know that the answer is $left[begin{matrix} 2 & -1 \ 1 & 1 end{matrix}right]$, but how to get the answer?




Let $mathcal{B} = { mathbf{b}_1 , mathbf{b}_2 }$ be the basis for $mathbb{R}^2$ with $mathbf{b}_1 = left [ begin{matrix} 1 \ 1 end{matrix} right ]$, $mathbf{b}_2 = left [ begin{matrix} 0 \ 1 end{matrix} right ]$. Furthermore, let $T: mathbb{R}^2 to mathbb{R}^2$ be a linear transformation. The matrix representation of $T$ with respect to $mathcal{B}$ is $[T]_mathcal{B} = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]$.
What is the standard matrix of $T$?





Original problem: The Problem






linear-algebra matrices transformation




linear-algebra matrices transformation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 14:24









Nominal Animal

7,1232617




7,1232617










asked Jan 5 at 13:47









Antoni MaleckiAntoni Malecki

52




52




closed as off-topic by Dietrich Burde, Namaste, max_zorn, metamorphy, Ali Caglayan Jan 5 at 19:31


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, Namaste, max_zorn, metamorphy, Ali Caglayan

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Dietrich Burde, Namaste, max_zorn, metamorphy, Ali Caglayan Jan 5 at 19:31


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, Namaste, max_zorn, metamorphy, Ali Caglayan

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Hint
    $endgroup$
    – John Doe
    Jan 5 at 13:50








  • 1




    $begingroup$
    You'll get a better response if you write the question here rather than linking to an image, and also explain what you have tried and which step you got stuck on.
    $endgroup$
    – littleO
    Jan 5 at 13:50










  • $begingroup$
    @JohnDoe the hint helped, thanks :)
    $endgroup$
    – Antoni Malecki
    Jan 5 at 14:49












  • $begingroup$
    @AntoniMalecki great! :)
    $endgroup$
    – John Doe
    Jan 5 at 14:51


















  • $begingroup$
    Hint
    $endgroup$
    – John Doe
    Jan 5 at 13:50








  • 1




    $begingroup$
    You'll get a better response if you write the question here rather than linking to an image, and also explain what you have tried and which step you got stuck on.
    $endgroup$
    – littleO
    Jan 5 at 13:50










  • $begingroup$
    @JohnDoe the hint helped, thanks :)
    $endgroup$
    – Antoni Malecki
    Jan 5 at 14:49












  • $begingroup$
    @AntoniMalecki great! :)
    $endgroup$
    – John Doe
    Jan 5 at 14:51
















$begingroup$
Hint
$endgroup$
– John Doe
Jan 5 at 13:50






$begingroup$
Hint
$endgroup$
– John Doe
Jan 5 at 13:50






1




1




$begingroup$
You'll get a better response if you write the question here rather than linking to an image, and also explain what you have tried and which step you got stuck on.
$endgroup$
– littleO
Jan 5 at 13:50




$begingroup$
You'll get a better response if you write the question here rather than linking to an image, and also explain what you have tried and which step you got stuck on.
$endgroup$
– littleO
Jan 5 at 13:50












$begingroup$
@JohnDoe the hint helped, thanks :)
$endgroup$
– Antoni Malecki
Jan 5 at 14:49






$begingroup$
@JohnDoe the hint helped, thanks :)
$endgroup$
– Antoni Malecki
Jan 5 at 14:49














$begingroup$
@AntoniMalecki great! :)
$endgroup$
– John Doe
Jan 5 at 14:51




$begingroup$
@AntoniMalecki great! :)
$endgroup$
– John Doe
Jan 5 at 14:51










1 Answer
1






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0












$begingroup$

You have
$$bbox{T = left [ begin{matrix} t_{11} & t_{12} \ t_{21} & t_{22} end{matrix} right ]}, quad bbox{mathcal{B} = left [ begin{matrix} 1 & 0 \ 1 & 1 end{matrix} right ]}$$
The change of basis is
$$bbox{[T]_mathcal{B} = mathcal{B}^{-1} T mathcal{B} = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]}$$
First step is to calculate $mathcal{B}^{-1}$. A 2×2 matrix is easiest to invert (if possible) via its adjugate matrix. Simply put,
$$bbox{mathbf{M} = left [ begin{matrix} m_{11} & m_{12} \ m_{21} & m_{22} end{matrix} right ]} quad iff quad bbox{mathbf{M}^{-1} = frac{1}{m_{11} m_{22} - m_{12} m_{21}} left [ begin{matrix} m_{22} & -m_{12} \ -m_{21} & m_{11} end{matrix} right ]}$$
Applying this to $mathcal{B}$, we get
$$bbox{mathcal{B}^{-1} = left [ begin{matrix} 1 & 0 \ -1 & 1 end{matrix} right ]}$$
Thus, you need to solve
$$bbox{ left [ begin{matrix} 1 & 0 \ -1 & 1 end{matrix} right ] left [ begin{matrix} t_{11} & t_{12} \ t_{21} & t_{22} end{matrix} right ] left [ begin{matrix} 1 & 0 \ 1 & 1 end{matrix} right ] = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]}$$
for $t_{11}$, $t_{12}$, $t_{21}$, and $t_{22}$.






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    You have
    $$bbox{T = left [ begin{matrix} t_{11} & t_{12} \ t_{21} & t_{22} end{matrix} right ]}, quad bbox{mathcal{B} = left [ begin{matrix} 1 & 0 \ 1 & 1 end{matrix} right ]}$$
    The change of basis is
    $$bbox{[T]_mathcal{B} = mathcal{B}^{-1} T mathcal{B} = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]}$$
    First step is to calculate $mathcal{B}^{-1}$. A 2×2 matrix is easiest to invert (if possible) via its adjugate matrix. Simply put,
    $$bbox{mathbf{M} = left [ begin{matrix} m_{11} & m_{12} \ m_{21} & m_{22} end{matrix} right ]} quad iff quad bbox{mathbf{M}^{-1} = frac{1}{m_{11} m_{22} - m_{12} m_{21}} left [ begin{matrix} m_{22} & -m_{12} \ -m_{21} & m_{11} end{matrix} right ]}$$
    Applying this to $mathcal{B}$, we get
    $$bbox{mathcal{B}^{-1} = left [ begin{matrix} 1 & 0 \ -1 & 1 end{matrix} right ]}$$
    Thus, you need to solve
    $$bbox{ left [ begin{matrix} 1 & 0 \ -1 & 1 end{matrix} right ] left [ begin{matrix} t_{11} & t_{12} \ t_{21} & t_{22} end{matrix} right ] left [ begin{matrix} 1 & 0 \ 1 & 1 end{matrix} right ] = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]}$$
    for $t_{11}$, $t_{12}$, $t_{21}$, and $t_{22}$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You have
      $$bbox{T = left [ begin{matrix} t_{11} & t_{12} \ t_{21} & t_{22} end{matrix} right ]}, quad bbox{mathcal{B} = left [ begin{matrix} 1 & 0 \ 1 & 1 end{matrix} right ]}$$
      The change of basis is
      $$bbox{[T]_mathcal{B} = mathcal{B}^{-1} T mathcal{B} = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]}$$
      First step is to calculate $mathcal{B}^{-1}$. A 2×2 matrix is easiest to invert (if possible) via its adjugate matrix. Simply put,
      $$bbox{mathbf{M} = left [ begin{matrix} m_{11} & m_{12} \ m_{21} & m_{22} end{matrix} right ]} quad iff quad bbox{mathbf{M}^{-1} = frac{1}{m_{11} m_{22} - m_{12} m_{21}} left [ begin{matrix} m_{22} & -m_{12} \ -m_{21} & m_{11} end{matrix} right ]}$$
      Applying this to $mathcal{B}$, we get
      $$bbox{mathcal{B}^{-1} = left [ begin{matrix} 1 & 0 \ -1 & 1 end{matrix} right ]}$$
      Thus, you need to solve
      $$bbox{ left [ begin{matrix} 1 & 0 \ -1 & 1 end{matrix} right ] left [ begin{matrix} t_{11} & t_{12} \ t_{21} & t_{22} end{matrix} right ] left [ begin{matrix} 1 & 0 \ 1 & 1 end{matrix} right ] = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]}$$
      for $t_{11}$, $t_{12}$, $t_{21}$, and $t_{22}$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You have
        $$bbox{T = left [ begin{matrix} t_{11} & t_{12} \ t_{21} & t_{22} end{matrix} right ]}, quad bbox{mathcal{B} = left [ begin{matrix} 1 & 0 \ 1 & 1 end{matrix} right ]}$$
        The change of basis is
        $$bbox{[T]_mathcal{B} = mathcal{B}^{-1} T mathcal{B} = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]}$$
        First step is to calculate $mathcal{B}^{-1}$. A 2×2 matrix is easiest to invert (if possible) via its adjugate matrix. Simply put,
        $$bbox{mathbf{M} = left [ begin{matrix} m_{11} & m_{12} \ m_{21} & m_{22} end{matrix} right ]} quad iff quad bbox{mathbf{M}^{-1} = frac{1}{m_{11} m_{22} - m_{12} m_{21}} left [ begin{matrix} m_{22} & -m_{12} \ -m_{21} & m_{11} end{matrix} right ]}$$
        Applying this to $mathcal{B}$, we get
        $$bbox{mathcal{B}^{-1} = left [ begin{matrix} 1 & 0 \ -1 & 1 end{matrix} right ]}$$
        Thus, you need to solve
        $$bbox{ left [ begin{matrix} 1 & 0 \ -1 & 1 end{matrix} right ] left [ begin{matrix} t_{11} & t_{12} \ t_{21} & t_{22} end{matrix} right ] left [ begin{matrix} 1 & 0 \ 1 & 1 end{matrix} right ] = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]}$$
        for $t_{11}$, $t_{12}$, $t_{21}$, and $t_{22}$.






        share|cite|improve this answer









        $endgroup$



        You have
        $$bbox{T = left [ begin{matrix} t_{11} & t_{12} \ t_{21} & t_{22} end{matrix} right ]}, quad bbox{mathcal{B} = left [ begin{matrix} 1 & 0 \ 1 & 1 end{matrix} right ]}$$
        The change of basis is
        $$bbox{[T]_mathcal{B} = mathcal{B}^{-1} T mathcal{B} = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]}$$
        First step is to calculate $mathcal{B}^{-1}$. A 2×2 matrix is easiest to invert (if possible) via its adjugate matrix. Simply put,
        $$bbox{mathbf{M} = left [ begin{matrix} m_{11} & m_{12} \ m_{21} & m_{22} end{matrix} right ]} quad iff quad bbox{mathbf{M}^{-1} = frac{1}{m_{11} m_{22} - m_{12} m_{21}} left [ begin{matrix} m_{22} & -m_{12} \ -m_{21} & m_{11} end{matrix} right ]}$$
        Applying this to $mathcal{B}$, we get
        $$bbox{mathcal{B}^{-1} = left [ begin{matrix} 1 & 0 \ -1 & 1 end{matrix} right ]}$$
        Thus, you need to solve
        $$bbox{ left [ begin{matrix} 1 & 0 \ -1 & 1 end{matrix} right ] left [ begin{matrix} t_{11} & t_{12} \ t_{21} & t_{22} end{matrix} right ] left [ begin{matrix} 1 & 0 \ 1 & 1 end{matrix} right ] = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]}$$
        for $t_{11}$, $t_{12}$, $t_{21}$, and $t_{22}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 14:55









        Nominal AnimalNominal Animal

        7,1232617




        7,1232617















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