Transformation Matrices: $hat{mathbf{e}}_j = sum_{i} H_j^i mathbf{e}_i$ and $hat{mathbf{mathrm{x}}} =...
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My textbook says the following in an appendix on tensor notation:
Consider a change of coordinate axes in which the basis vectors $mathbf{e}_i$ are replaced by a new basis set $hat{mathbf{e}}_j$, where $hat{mathbf{e}}_j = sum_{i} H_j^i mathbf{e}_i$, and $H$ is the basis transformation matrix with entries $H_j^i$. If $hat{mathbf{mathrm{x}}} = (hat{x}^1, hat{x}^2, hat{x}^3)^T$ are the coordinates of the vector with respect to the new basis, then we may verify that $hat{mathbf{mathrm{x}}} = H^{-1}mathbf{mathrm{x}}$. Thus, if the basis vectors transform according to $H$ the coordinates of points transform according to the inverse transformation $H^{-1}$.
I don't understand what the last part is saying (and yes, I've made sure to copy it exactly):
Thus, if the basis vectors transform according to $H$ the coordinates of points transform according to the inverse transformation $H^{-1}$.
I have 2 problems with this part:
- It seems to me that the syntax is poor, which makes it difficult to try and understand what it is trying to say. Even so, I think I have managed to decipher it (see 2).
- Beyond the syntax and with regards to mathematics, I'm wondering why is it that we use $H$ as the transformation matrix for the basis vectors in $hat{mathbf{e}}_j = sum_{i} H_j^i mathbf{e}_i$, whereas we use $H^{-1}$ as the transformation matrix for the coordinates of points $hat{mathbf{mathrm{x}}} = H^{-1}mathbf{mathrm{x}}$? In other words, I don't understand why we use $H$ for one and $H^{-1}$ for the other? I only have introductory-level linear algebra knowledge, so please explain gently.
I would greatly appreciate it if people could please take the time to help me understand this.
linear-algebra linear-transformations tensors change-of-basis
asked Nov 18 at 10:35
The Pointer
2,67921233
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show 3 more comments
up vote
1
down vote
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My textbook says the following in an appendix on tensor notation:
Consider a change of coordinate axes in which the basis vectors $mathbf{e}_i$ are replaced by a new basis set $hat{mathbf{e}}_j$, where $hat{mathbf{e}}_j = sum_{i} H_j^i mathbf{e}_i$, and $H$ is the basis transformation matrix with entries $H_j^i$. If $hat{mathbf{mathrm{x}}} = (hat{x}^1, hat{x}^2, hat{x}^3)^T$ are the coordinates of the vector with respect to the new basis, then we may verify that $hat{mathbf{mathrm{x}}} = H^{-1}mathbf{mathrm{x}}$. Thus, if the basis vectors transform according to $H$ the coordinates of points transform according to the inverse transformation $H^{-1}$.
I don't understand what the last part is saying (and yes, I've made sure to copy it exactly):
Thus, if the basis vectors transform according to $H$ the coordinates of points transform according to the inverse transformation $H^{-1}$.
I have 2 problems with this part:
- It seems to me that the syntax is poor, which makes it difficult to try and understand what it is trying to say. Even so, I think I have managed to decipher it (see 2).
- Beyond the syntax and with regards to mathematics, I'm wondering why is it that we use $H$ as the transformation matrix for the basis vectors in $hat{mathbf{e}}_j = sum_{i} H_j^i mathbf{e}_i$, whereas we use $H^{-1}$ as the transformation matrix for the coordinates of points $hat{mathbf{mathrm{x}}} = H^{-1}mathbf{mathrm{x}}$? In other words, I don't understand why we use $H$ for one and $H^{-1}$ for the other? I only have introductory-level linear algebra knowledge, so please explain gently.
I would greatly appreciate it if people could please take the time to help me understand this.
linear-algebra linear-transformations tensors change-of-basis
asked Nov 18 at 10:35
The Pointer
2,67921233
Note that if $hat{e}_i = H^j_i e_j$, then for any vector $x$ we have $x=hat{x}^i hat{e}_i =hat{x}^i H^j_i e_j = x^j e_j$. So $x^j= hat{x}^i H^j_i$. Which means $hat{x}^i =[H^{-1}]^i_j x^i$.
– Kelvin Lois
Nov 18 at 10:54
@KelvinLois Thanks for the comment. How did you get $hat{x}^i =[H^{-1}]^i_j x^i$ from $x^j= hat{x}^i H^j_i$?
– The Pointer
Nov 18 at 12:00
Consider it as a matrix multiplication. So just take the inverse.
– Kelvin Lois
Nov 18 at 12:53
@KelvinLois please show what matrix multiplication you did to get.
– The Pointer
Nov 18 at 12:59
$[x^i]= H [hat{x}^i] $ And then multiply both sides by H inverse.
– Kelvin Lois
Nov 18 at 13:02
|
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My textbook says the following in an appendix on tensor notation:
Consider a change of coordinate axes in which the basis vectors $mathbf{e}_i$ are replaced by a new basis set $hat{mathbf{e}}_j$, where $hat{mathbf{e}}_j = sum_{i} H_j^i mathbf{e}_i$, and $H$ is the basis transformation matrix with entries $H_j^i$. If $hat{mathbf{mathrm{x}}} = (hat{x}^1, hat{x}^2, hat{x}^3)^T$ are the coordinates of the vector with respect to the new basis, then we may verify that $hat{mathbf{mathrm{x}}} = H^{-1}mathbf{mathrm{x}}$. Thus, if the basis vectors transform according to $H$ the coordinates of points transform according to the inverse transformation $H^{-1}$.
I don't understand what the last part is saying (and yes, I've made sure to copy it exactly):
Thus, if the basis vectors transform according to $H$ the coordinates of points transform according to the inverse transformation $H^{-1}$.
I have 2 problems with this part:
- It seems to me that the syntax is poor, which makes it difficult to try and understand what it is trying to say. Even so, I think I have managed to decipher it (see 2).
- Beyond the syntax and with regards to mathematics, I'm wondering why is it that we use $H$ as the transformation matrix for the basis vectors in $hat{mathbf{e}}_j = sum_{i} H_j^i mathbf{e}_i$, whereas we use $H^{-1}$ as the transformation matrix for the coordinates of points $hat{mathbf{mathrm{x}}} = H^{-1}mathbf{mathrm{x}}$? In other words, I don't understand why we use $H$ for one and $H^{-1}$ for the other? I only have introductory-level linear algebra knowledge, so please explain gently.
I would greatly appreciate it if people could please take the time to help me understand this.
linear-algebra linear-transformations tensors change-of-basis
asked Nov 18 at 10:35
The Pointer
2,67921233
My textbook says the following in an appendix on tensor notation:
Consider a change of coordinate axes in which the basis vectors $mathbf{e}_i$ are replaced by a new basis set $hat{mathbf{e}}_j$, where $hat{mathbf{e}}_j = sum_{i} H_j^i mathbf{e}_i$, and $H$ is the basis transformation matrix with entries $H_j^i$. If $hat{mathbf{mathrm{x}}} = (hat{x}^1, hat{x}^2, hat{x}^3)^T$ are the coordinates of the vector with respect to the new basis, then we may verify that $hat{mathbf{mathrm{x}}} = H^{-1}mathbf{mathrm{x}}$. Thus, if the basis vectors transform according to $H$ the coordinates of points transform according to the inverse transformation $H^{-1}$.
I don't understand what the last part is saying (and yes, I've made sure to copy it exactly):
Thus, if the basis vectors transform according to $H$ the coordinates of points transform according to the inverse transformation $H^{-1}$.
I have 2 problems with this part:
- It seems to me that the syntax is poor, which makes it difficult to try and understand what it is trying to say. Even so, I think I have managed to decipher it (see 2).
- Beyond the syntax and with regards to mathematics, I'm wondering why is it that we use $H$ as the transformation matrix for the basis vectors in $hat{mathbf{e}}_j = sum_{i} H_j^i mathbf{e}_i$, whereas we use $H^{-1}$ as the transformation matrix for the coordinates of points $hat{mathbf{mathrm{x}}} = H^{-1}mathbf{mathrm{x}}$? In other words, I don't understand why we use $H$ for one and $H^{-1}$ for the other? I only have introductory-level linear algebra knowledge, so please explain gently.
I would greatly appreciate it if people could please take the time to help me understand this.
linear-algebra linear-transformations tensors change-of-basis
linear-algebra linear-transformations tensors change-of-basis
asked Nov 18 at 10:35
The Pointer
2,67921233
asked Nov 18 at 10:35
The Pointer
2,67921233
asked Nov 18 at 10:35
The Pointer
2,67921233
asked Nov 18 at 10:35
The Pointer
2,67921233
asked Nov 18 at 10:35
The Pointer
2,67921233
2,67921233
Note that if $hat{e}_i = H^j_i e_j$, then for any vector $x$ we have $x=hat{x}^i hat{e}_i =hat{x}^i H^j_i e_j = x^j e_j$. So $x^j= hat{x}^i H^j_i$. Which means $hat{x}^i =[H^{-1}]^i_j x^i$.
– Kelvin Lois
Nov 18 at 10:54
@KelvinLois Thanks for the comment. How did you get $hat{x}^i =[H^{-1}]^i_j x^i$ from $x^j= hat{x}^i H^j_i$?
– The Pointer
Nov 18 at 12:00
Consider it as a matrix multiplication. So just take the inverse.
– Kelvin Lois
Nov 18 at 12:53
@KelvinLois please show what matrix multiplication you did to get.
– The Pointer
Nov 18 at 12:59
$[x^i]= H [hat{x}^i] $ And then multiply both sides by H inverse.
– Kelvin Lois
Nov 18 at 13:02
|
show 3 more comments
Note that if $hat{e}_i = H^j_i e_j$, then for any vector $x$ we have $x=hat{x}^i hat{e}_i =hat{x}^i H^j_i e_j = x^j e_j$. So $x^j= hat{x}^i H^j_i$. Which means $hat{x}^i =[H^{-1}]^i_j x^i$.
– Kelvin Lois
Nov 18 at 10:54
@KelvinLois Thanks for the comment. How did you get $hat{x}^i =[H^{-1}]^i_j x^i$ from $x^j= hat{x}^i H^j_i$?
– The Pointer
Nov 18 at 12:00
Consider it as a matrix multiplication. So just take the inverse.
– Kelvin Lois
Nov 18 at 12:53
@KelvinLois please show what matrix multiplication you did to get.
– The Pointer
Nov 18 at 12:59
$[x^i]= H [hat{x}^i] $ And then multiply both sides by H inverse.
– Kelvin Lois
Nov 18 at 13:02
Note that if $hat{e}_i = H^j_i e_j$, then for any vector $x$ we have $x=hat{x}^i hat{e}_i =hat{x}^i H^j_i e_j = x^j e_j$. So $x^j= hat{x}^i H^j_i$. Which means $hat{x}^i =[H^{-1}]^i_j x^i$.
– Kelvin Lois
Nov 18 at 10:54
Note that if $hat{e}_i = H^j_i e_j$, then for any vector $x$ we have $x=hat{x}^i hat{e}_i =hat{x}^i H^j_i e_j = x^j e_j$. So $x^j= hat{x}^i H^j_i$. Which means $hat{x}^i =[H^{-1}]^i_j x^i$.
– Kelvin Lois
Nov 18 at 10:54
@KelvinLois Thanks for the comment. How did you get $hat{x}^i =[H^{-1}]^i_j x^i$ from $x^j= hat{x}^i H^j_i$?
– The Pointer
Nov 18 at 12:00
@KelvinLois Thanks for the comment. How did you get $hat{x}^i =[H^{-1}]^i_j x^i$ from $x^j= hat{x}^i H^j_i$?
– The Pointer
Nov 18 at 12:00
Consider it as a matrix multiplication. So just take the inverse.
– Kelvin Lois
Nov 18 at 12:53
Consider it as a matrix multiplication. So just take the inverse.
– Kelvin Lois
Nov 18 at 12:53
@KelvinLois please show what matrix multiplication you did to get.
– The Pointer
Nov 18 at 12:59
@KelvinLois please show what matrix multiplication you did to get.
– The Pointer
Nov 18 at 12:59
$[x^i]= H [hat{x}^i] $ And then multiply both sides by H inverse.
– Kelvin Lois
Nov 18 at 13:02
$[x^i]= H [hat{x}^i] $ And then multiply both sides by H inverse.
– Kelvin Lois
Nov 18 at 13:02
|
show 3 more comments
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Note that if $hat{e}_i = H^j_i e_j$, then for any vector $x$ we have $x=hat{x}^i hat{e}_i =hat{x}^i H^j_i e_j = x^j e_j$. So $x^j= hat{x}^i H^j_i$. Which means $hat{x}^i =[H^{-1}]^i_j x^i$.
– Kelvin Lois
Nov 18 at 10:54
@KelvinLois Thanks for the comment. How did you get $hat{x}^i =[H^{-1}]^i_j x^i$ from $x^j= hat{x}^i H^j_i$?
– The Pointer
Nov 18 at 12:00
Consider it as a matrix multiplication. So just take the inverse.
– Kelvin Lois
Nov 18 at 12:53
@KelvinLois please show what matrix multiplication you did to get.
– The Pointer
Nov 18 at 12:59
$[x^i]= H [hat{x}^i] $ And then multiply both sides by H inverse.
– Kelvin Lois
Nov 18 at 13:02