For the following linear operators $ L$ on $mathbb{R}^3$, find a matrix $A$ such that $L(x)=Ax$ for every $x$...
For the following linear operators $L$ on $mathbb{R}^3$, find a matrix $A$ such that $L(x)=Ax$ for every $x$ in $mathbb{R}^3$. $L((x_1,x_2,x_3)^T)=(x_3,x_2,x_1)^T$
linear-algebra linear-transformations
asked Nov 27 at 5:02
Shadow Z
113
closed as off-topic by angryavian, Brahadeesh, max_zorn, Cyclohexanol., José Carlos Santos Nov 27 at 7:10
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For the following linear operators $L$ on $mathbb{R}^3$, find a matrix $A$ such that $L(x)=Ax$ for every $x$ in $mathbb{R}^3$. $L((x_1,x_2,x_3)^T)=(x_3,x_2,x_1)^T$
linear-algebra linear-transformations
asked Nov 27 at 5:02
Shadow Z
113
closed as off-topic by angryavian, Brahadeesh, max_zorn, Cyclohexanol., José Carlos Santos Nov 27 at 7:10
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – angryavian, Brahadeesh, max_zorn, Cyclohexanol., José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
Here $Ax=x$. What should $A$ be?
– Yadati Kiran
Nov 27 at 5:23
add a comment |
For the following linear operators $L$ on $mathbb{R}^3$, find a matrix $A$ such that $L(x)=Ax$ for every $x$ in $mathbb{R}^3$. $L((x_1,x_2,x_3)^T)=(x_3,x_2,x_1)^T$
linear-algebra linear-transformations
asked Nov 27 at 5:02
Shadow Z
113
For the following linear operators $L$ on $mathbb{R}^3$, find a matrix $A$ such that $L(x)=Ax$ for every $x$ in $mathbb{R}^3$. $L((x_1,x_2,x_3)^T)=(x_3,x_2,x_1)^T$
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Nov 27 at 5:02
Shadow Z
113
asked Nov 27 at 5:02
Shadow Z
113
edited Nov 27 at 7:25
asked Nov 27 at 5:02
Shadow Z
113
asked Nov 27 at 5:02
Shadow Z
113
asked Nov 27 at 5:02
Shadow Z
113
113
closed as off-topic by angryavian, Brahadeesh, max_zorn, Cyclohexanol., José Carlos Santos Nov 27 at 7:10
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – angryavian, Brahadeesh, max_zorn, Cyclohexanol., José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by angryavian, Brahadeesh, max_zorn, Cyclohexanol., José Carlos Santos Nov 27 at 7:10
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – angryavian, Brahadeesh, max_zorn, Cyclohexanol., José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
Here $Ax=x$. What should $A$ be?
– Yadati Kiran
Nov 27 at 5:23
add a comment |
Here $Ax=x$. What should $A$ be?
– Yadati Kiran
Nov 27 at 5:23
Here $Ax=x$. What should $A$ be?
– Yadati Kiran
Nov 27 at 5:23
Here $Ax=x$. What should $A$ be?
– Yadati Kiran
Nov 27 at 5:23
add a comment |
1 Answer
1
active
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votes
Since the equation $L(x)=Ax$ is true for all $x$, one way to find $A$ is to simply take $3$ linearly independent $x$ in $mathbb{R}^3$ and put them in the above equation and get the values of each term in $A$. You can take any linearly independent $x$, but I prefer to use the unit vectors $(0,0,1)^T, (0,1,0)^T, (1,0,0)^T$. After putting each of them in $L(x)=Ax$, you will quickly find that
$$A =
begin{matrix}
0 & 0 & 1 \
0 & 1 & 0 \
1 & 0 & 0 \
end{matrix}
$$
You can also verify this is the matrix by putting a general $y=(y_1,y_2,y_3)^T$ and see that the result is $(y_3,y_2,y_1)^T$
answered Nov 27 at 5:59
Sauhard Sharma
63213
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Since the equation $L(x)=Ax$ is true for all $x$, one way to find $A$ is to simply take $3$ linearly independent $x$ in $mathbb{R}^3$ and put them in the above equation and get the values of each term in $A$. You can take any linearly independent $x$, but I prefer to use the unit vectors $(0,0,1)^T, (0,1,0)^T, (1,0,0)^T$. After putting each of them in $L(x)=Ax$, you will quickly find that
$$A =
begin{matrix}
0 & 0 & 1 \
0 & 1 & 0 \
1 & 0 & 0 \
end{matrix}
$$
You can also verify this is the matrix by putting a general $y=(y_1,y_2,y_3)^T$ and see that the result is $(y_3,y_2,y_1)^T$
answered Nov 27 at 5:59
Sauhard Sharma
63213
add a comment |
Since the equation $L(x)=Ax$ is true for all $x$, one way to find $A$ is to simply take $3$ linearly independent $x$ in $mathbb{R}^3$ and put them in the above equation and get the values of each term in $A$. You can take any linearly independent $x$, but I prefer to use the unit vectors $(0,0,1)^T, (0,1,0)^T, (1,0,0)^T$. After putting each of them in $L(x)=Ax$, you will quickly find that
$$A =
begin{matrix}
0 & 0 & 1 \
0 & 1 & 0 \
1 & 0 & 0 \
end{matrix}
$$
You can also verify this is the matrix by putting a general $y=(y_1,y_2,y_3)^T$ and see that the result is $(y_3,y_2,y_1)^T$
answered Nov 27 at 5:59
Sauhard Sharma
63213
add a comment |
Since the equation $L(x)=Ax$ is true for all $x$, one way to find $A$ is to simply take $3$ linearly independent $x$ in $mathbb{R}^3$ and put them in the above equation and get the values of each term in $A$. You can take any linearly independent $x$, but I prefer to use the unit vectors $(0,0,1)^T, (0,1,0)^T, (1,0,0)^T$. After putting each of them in $L(x)=Ax$, you will quickly find that
$$A =
begin{matrix}
0 & 0 & 1 \
0 & 1 & 0 \
1 & 0 & 0 \
end{matrix}
$$
You can also verify this is the matrix by putting a general $y=(y_1,y_2,y_3)^T$ and see that the result is $(y_3,y_2,y_1)^T$
answered Nov 27 at 5:59
Sauhard Sharma
63213
Since the equation $L(x)=Ax$ is true for all $x$, one way to find $A$ is to simply take $3$ linearly independent $x$ in $mathbb{R}^3$ and put them in the above equation and get the values of each term in $A$. You can take any linearly independent $x$, but I prefer to use the unit vectors $(0,0,1)^T, (0,1,0)^T, (1,0,0)^T$. After putting each of them in $L(x)=Ax$, you will quickly find that
$$A =
begin{matrix}
0 & 0 & 1 \
0 & 1 & 0 \
1 & 0 & 0 \
end{matrix}
$$
You can also verify this is the matrix by putting a general $y=(y_1,y_2,y_3)^T$ and see that the result is $(y_3,y_2,y_1)^T$
answered Nov 27 at 5:59
Sauhard Sharma
63213
answered Nov 27 at 5:59
Sauhard Sharma
63213
answered Nov 27 at 5:59
Sauhard Sharma
63213
answered Nov 27 at 5:59
Sauhard Sharma
63213
63213
add a comment |
add a comment |
Here $Ax=x$. What should $A$ be?
– Yadati Kiran
Nov 27 at 5:23