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: 
linear-algebra matrices transformation
edited Jan 5 at 14:24
Nominal Animal
7,1232617
asked Jan 5 at 13:47
Antoni MaleckiAntoni Malecki
52
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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.
add a comment |
$begingroup$
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:
linear-algebra matrices transformation
edited Jan 5 at 14:24
Nominal Animal
7,1232617
asked Jan 5 at 13:47
Antoni MaleckiAntoni Malecki
52
$endgroup$
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.
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Hint
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– John Doe
Jan 5 at 13:50
1
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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
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@JohnDoe the hint helped, thanks :)
$endgroup$
– Antoni Malecki
Jan 5 at 14:49
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@AntoniMalecki great! :)
$endgroup$
– John Doe
Jan 5 at 14:51
add a comment |
$begingroup$
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:
linear-algebra matrices transformation
edited Jan 5 at 14:24
Nominal Animal
7,1232617
asked Jan 5 at 13:47
Antoni MaleckiAntoni Malecki
52
$endgroup$
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:
linear-algebra matrices transformation
linear-algebra matrices transformation
edited Jan 5 at 14:24
Nominal Animal
7,1232617
asked Jan 5 at 13:47
Antoni MaleckiAntoni Malecki
52
edited Jan 5 at 14:24
Nominal Animal
7,1232617
asked Jan 5 at 13:47
Antoni MaleckiAntoni Malecki
52
edited Jan 5 at 14:24
Nominal Animal
7,1232617
edited Jan 5 at 14:24
Nominal Animal
7,1232617
edited Jan 5 at 14:24
Nominal Animal
7,1232617
7,1232617
asked Jan 5 at 13:47
Antoni MaleckiAntoni Malecki
52
asked Jan 5 at 13:47
Antoni MaleckiAntoni Malecki
52
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
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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 $t_{22}$.
answered Jan 5 at 14:55
Nominal AnimalNominal Animal
7,1232617
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $t_{22}$.
answered Jan 5 at 14:55
Nominal AnimalNominal Animal
7,1232617
$endgroup$
add a comment |
$begingroup$
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 $t_{22}$.
answered Jan 5 at 14:55
Nominal AnimalNominal Animal
7,1232617
$endgroup$
add a comment |
$begingroup$
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 $t_{22}$.
answered Jan 5 at 14:55
Nominal AnimalNominal Animal
7,1232617
$endgroup$
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 $t_{22}$.
answered Jan 5 at 14:55
Nominal AnimalNominal Animal
7,1232617
answered Jan 5 at 14:55
Nominal AnimalNominal Animal
7,1232617
answered Jan 5 at 14:55
Nominal AnimalNominal Animal
7,1232617
answered Jan 5 at 14:55
Nominal AnimalNominal Animal
7,1232617
7,1232617
add a comment |
add a comment |
$begingroup$
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