Gradient of matrix $b^{T}x$
I'm trying to understand my exam solution from the lecturer, but I got confused over one small thing in the solution.
Problem: Consider $f(x)=frac{1}{2}x^{T}Ax - b^{T}x$
Where $A=begin{bmatrix}2&1\1&4end{bmatrix}$ , $b=begin{bmatrix}0\7end{bmatrix}$ and $x=begin{bmatrix}x\yend{bmatrix}$
Caclulate $nabla f(x)$.
Solution:
From the theory we know that if $A$ is symmetric, we get
$nabla f(x)= Ax - b^{T}$
$nabla f(x)=begin{bmatrix}2x+y\x+4y-7end{bmatrix}$
Is it a typo on the answer, where $Ax-b^{T}$ should've been written as $Ax-b$?
If it should've been written as $b$, then I understand why we have $-7$ in the 2nd row.
Thank you
matrices vector-analysis differential
asked Nov 27 '18 at 20:12
JIM BOY
356
add a comment |
I'm trying to understand my exam solution from the lecturer, but I got confused over one small thing in the solution.
Problem: Consider $f(x)=frac{1}{2}x^{T}Ax - b^{T}x$
Where $A=begin{bmatrix}2&1\1&4end{bmatrix}$ , $b=begin{bmatrix}0\7end{bmatrix}$ and $x=begin{bmatrix}x\yend{bmatrix}$
Caclulate $nabla f(x)$.
Solution:
From the theory we know that if $A$ is symmetric, we get
$nabla f(x)= Ax - b^{T}$
$nabla f(x)=begin{bmatrix}2x+y\x+4y-7end{bmatrix}$
Is it a typo on the answer, where $Ax-b^{T}$ should've been written as $Ax-b$?
If it should've been written as $b$, then I understand why we have $-7$ in the 2nd row.
Thank you
matrices vector-analysis differential
asked Nov 27 '18 at 20:12
JIM BOY
356
1
Yes, $nabla b^{T}x = b$.
– Math Lover
Nov 27 '18 at 20:15
Could you please elaborate how the transpose of $b$ becomes $b$? And do you think it is a typo where $b^{T}$ in the answer should've been written as $b$?
– JIM BOY
Nov 27 '18 at 20:29
Give a look to this other answer: math.stackexchange.com/questions/3013384/…
– caverac
Nov 27 '18 at 21:00
add a comment |
I'm trying to understand my exam solution from the lecturer, but I got confused over one small thing in the solution.
Problem: Consider $f(x)=frac{1}{2}x^{T}Ax - b^{T}x$
Where $A=begin{bmatrix}2&1\1&4end{bmatrix}$ , $b=begin{bmatrix}0\7end{bmatrix}$ and $x=begin{bmatrix}x\yend{bmatrix}$
Caclulate $nabla f(x)$.
Solution:
From the theory we know that if $A$ is symmetric, we get
$nabla f(x)= Ax - b^{T}$
$nabla f(x)=begin{bmatrix}2x+y\x+4y-7end{bmatrix}$
Is it a typo on the answer, where $Ax-b^{T}$ should've been written as $Ax-b$?
If it should've been written as $b$, then I understand why we have $-7$ in the 2nd row.
Thank you
matrices vector-analysis differential
asked Nov 27 '18 at 20:12
JIM BOY
356
I'm trying to understand my exam solution from the lecturer, but I got confused over one small thing in the solution.
Problem: Consider $f(x)=frac{1}{2}x^{T}Ax - b^{T}x$
Where $A=begin{bmatrix}2&1\1&4end{bmatrix}$ , $b=begin{bmatrix}0\7end{bmatrix}$ and $x=begin{bmatrix}x\yend{bmatrix}$
Caclulate $nabla f(x)$.
Solution:
From the theory we know that if $A$ is symmetric, we get
$nabla f(x)= Ax - b^{T}$
$nabla f(x)=begin{bmatrix}2x+y\x+4y-7end{bmatrix}$
Is it a typo on the answer, where $Ax-b^{T}$ should've been written as $Ax-b$?
If it should've been written as $b$, then I understand why we have $-7$ in the 2nd row.
Thank you
matrices vector-analysis differential
matrices vector-analysis differential
asked Nov 27 '18 at 20:12
JIM BOY
356
asked Nov 27 '18 at 20:12
JIM BOY
356
edited Nov 27 '18 at 21:21
asked Nov 27 '18 at 20:12
JIM BOY
356
asked Nov 27 '18 at 20:12
JIM BOY
356
asked Nov 27 '18 at 20:12
JIM BOY
356
356
1
Yes, $nabla b^{T}x = b$.
– Math Lover
Nov 27 '18 at 20:15
Could you please elaborate how the transpose of $b$ becomes $b$? And do you think it is a typo where $b^{T}$ in the answer should've been written as $b$?
– JIM BOY
Nov 27 '18 at 20:29
Give a look to this other answer: math.stackexchange.com/questions/3013384/…
– caverac
Nov 27 '18 at 21:00
add a comment |
1
Yes, $nabla b^{T}x = b$.
– Math Lover
Nov 27 '18 at 20:15
Could you please elaborate how the transpose of $b$ becomes $b$? And do you think it is a typo where $b^{T}$ in the answer should've been written as $b$?
– JIM BOY
Nov 27 '18 at 20:29
Give a look to this other answer: math.stackexchange.com/questions/3013384/…
– caverac
Nov 27 '18 at 21:00
1
1
Yes, $nabla b^{T}x = b$.
– Math Lover
Nov 27 '18 at 20:15
Yes, $nabla b^{T}x = b$.
– Math Lover
Nov 27 '18 at 20:15
Could you please elaborate how the transpose of $b$ becomes $b$? And do you think it is a typo where $b^{T}$ in the answer should've been written as $b$?
– JIM BOY
Nov 27 '18 at 20:29
Could you please elaborate how the transpose of $b$ becomes $b$? And do you think it is a typo where $b^{T}$ in the answer should've been written as $b$?
– JIM BOY
Nov 27 '18 at 20:29
Give a look to this other answer: math.stackexchange.com/questions/3013384/…
– caverac
Nov 27 '18 at 21:00
Give a look to this other answer: math.stackexchange.com/questions/3013384/…
– caverac
Nov 27 '18 at 21:00
add a comment |
2 Answers
2
active
oldest
votes
There is no typo, you need to transpose $b$ if you want to compute its scalar product with $x$ (quick check, treat this as a normal matrix-matrix multiplication: the dimensions would not agree without transposition!).
And yes, $nabla (b^Tx) = b$.
In this case, since there are only two components, we can compute everything by hand. You can then understand how this generalizes, once you figure out what happens when applying the gradient.
We have:
$$b^Tx = begin{bmatrix} 0 & 7end{bmatrix} cdot begin{bmatrix} x \ y end{bmatrix} = 0x + 7y$$
and since
$$nabla = begin{bmatrix} frac{partial}{partial x} \ frac{partial}{partial y} end{bmatrix},$$
you can compute
$$nabla (b^Tx) = begin{bmatrix} frac{partial}{partial x} (0x + 7y) \ frac{partial}{partial y} (0x + 7y) end{bmatrix} = begin{bmatrix} 0 \ 7 end{bmatrix}.$$
answered Nov 27 '18 at 21:10
alrigazzi
363
Thanks. Now I get it how $nabla b^{T}x = b$. What I mean by typo is on the answer. My lecturer wrote $-b^{T}$. It should be $-b$ right?
– JIM BOY
Nov 27 '18 at 21:14
1
Oh, sure, that is right.
– alrigazzi
Nov 28 '18 at 19:15
add a comment |
Call a column vector $vec{x}$. Indicate $x^1$ for the 1st component, $x^2$ for the second, and so on. Square of a component will be $(x^1)^2$ Will be the first component (usually x coponent) squared.
$x^i$ will be used to represent the i_th component or the whole vector.
The transpose of vector will be represented with a lowered index $x_i$ is $vec{x}^T$
Inner products can only be performed between a column vector and a row vector, so one vector is addressed by lower indices, the other by vectors. Matching indicies imply a sum.
So $b^Tx=b_ix^i=b_xx_x+b_yx_y+b_zx_z$
$nabla(b^Tx)=frac{partial}{partial x^j}(b_ix^i)$
By the product rule:
$frac{partial}{partial x^j}(b_ix^i)=frac{partial b_i}{partial x_j}x^i+b_ifrac{partial x^i}{partial x^j}$
But $vec{b}$ isa constant, so the derivatives of any component is 0. $frac{partial x_j}{partial x_i}$ is 0 unless the indices match, in which case the value is 1.
So the other term gives us $b_j$, establishing the expected result.
answered Nov 27 '18 at 21:50
TurlocTheRed
838311
Thanks. What I mean by typo is on the answer. My lecturer wrote $−b^T$. It should be $−b$ right?
– JIM BOY
Nov 27 '18 at 23:04
1
Yep. the $delta^i_k$ makes that happen.
– TurlocTheRed
Nov 27 '18 at 23:08
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3016249%2fgradient-of-matrix-btx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is no typo, you need to transpose $b$ if you want to compute its scalar product with $x$ (quick check, treat this as a normal matrix-matrix multiplication: the dimensions would not agree without transposition!).
And yes, $nabla (b^Tx) = b$.
In this case, since there are only two components, we can compute everything by hand. You can then understand how this generalizes, once you figure out what happens when applying the gradient.
We have:
$$b^Tx = begin{bmatrix} 0 & 7end{bmatrix} cdot begin{bmatrix} x \ y end{bmatrix} = 0x + 7y$$
and since
$$nabla = begin{bmatrix} frac{partial}{partial x} \ frac{partial}{partial y} end{bmatrix},$$
you can compute
$$nabla (b^Tx) = begin{bmatrix} frac{partial}{partial x} (0x + 7y) \ frac{partial}{partial y} (0x + 7y) end{bmatrix} = begin{bmatrix} 0 \ 7 end{bmatrix}.$$
answered Nov 27 '18 at 21:10
alrigazzi
363
Thanks. Now I get it how $nabla b^{T}x = b$. What I mean by typo is on the answer. My lecturer wrote $-b^{T}$. It should be $-b$ right?
– JIM BOY
Nov 27 '18 at 21:14
1
Oh, sure, that is right.
– alrigazzi
Nov 28 '18 at 19:15
add a comment |
There is no typo, you need to transpose $b$ if you want to compute its scalar product with $x$ (quick check, treat this as a normal matrix-matrix multiplication: the dimensions would not agree without transposition!).
And yes, $nabla (b^Tx) = b$.
In this case, since there are only two components, we can compute everything by hand. You can then understand how this generalizes, once you figure out what happens when applying the gradient.
We have:
$$b^Tx = begin{bmatrix} 0 & 7end{bmatrix} cdot begin{bmatrix} x \ y end{bmatrix} = 0x + 7y$$
and since
$$nabla = begin{bmatrix} frac{partial}{partial x} \ frac{partial}{partial y} end{bmatrix},$$
you can compute
$$nabla (b^Tx) = begin{bmatrix} frac{partial}{partial x} (0x + 7y) \ frac{partial}{partial y} (0x + 7y) end{bmatrix} = begin{bmatrix} 0 \ 7 end{bmatrix}.$$
answered Nov 27 '18 at 21:10
alrigazzi
363
Thanks. Now I get it how $nabla b^{T}x = b$. What I mean by typo is on the answer. My lecturer wrote $-b^{T}$. It should be $-b$ right?
– JIM BOY
Nov 27 '18 at 21:14
1
Oh, sure, that is right.
– alrigazzi
Nov 28 '18 at 19:15
add a comment |
There is no typo, you need to transpose $b$ if you want to compute its scalar product with $x$ (quick check, treat this as a normal matrix-matrix multiplication: the dimensions would not agree without transposition!).
And yes, $nabla (b^Tx) = b$.
In this case, since there are only two components, we can compute everything by hand. You can then understand how this generalizes, once you figure out what happens when applying the gradient.
We have:
$$b^Tx = begin{bmatrix} 0 & 7end{bmatrix} cdot begin{bmatrix} x \ y end{bmatrix} = 0x + 7y$$
and since
$$nabla = begin{bmatrix} frac{partial}{partial x} \ frac{partial}{partial y} end{bmatrix},$$
you can compute
$$nabla (b^Tx) = begin{bmatrix} frac{partial}{partial x} (0x + 7y) \ frac{partial}{partial y} (0x + 7y) end{bmatrix} = begin{bmatrix} 0 \ 7 end{bmatrix}.$$
answered Nov 27 '18 at 21:10
alrigazzi
363
There is no typo, you need to transpose $b$ if you want to compute its scalar product with $x$ (quick check, treat this as a normal matrix-matrix multiplication: the dimensions would not agree without transposition!).
And yes, $nabla (b^Tx) = b$.
In this case, since there are only two components, we can compute everything by hand. You can then understand how this generalizes, once you figure out what happens when applying the gradient.
We have:
$$b^Tx = begin{bmatrix} 0 & 7end{bmatrix} cdot begin{bmatrix} x \ y end{bmatrix} = 0x + 7y$$
and since
$$nabla = begin{bmatrix} frac{partial}{partial x} \ frac{partial}{partial y} end{bmatrix},$$
you can compute
$$nabla (b^Tx) = begin{bmatrix} frac{partial}{partial x} (0x + 7y) \ frac{partial}{partial y} (0x + 7y) end{bmatrix} = begin{bmatrix} 0 \ 7 end{bmatrix}.$$
answered Nov 27 '18 at 21:10
alrigazzi
363
answered Nov 27 '18 at 21:10
alrigazzi
363
answered Nov 27 '18 at 21:10
alrigazzi
363
answered Nov 27 '18 at 21:10
alrigazzi
363
363
Thanks. Now I get it how $nabla b^{T}x = b$. What I mean by typo is on the answer. My lecturer wrote $-b^{T}$. It should be $-b$ right?
– JIM BOY
Nov 27 '18 at 21:14
1
Oh, sure, that is right.
– alrigazzi
Nov 28 '18 at 19:15
add a comment |
Thanks. Now I get it how $nabla b^{T}x = b$. What I mean by typo is on the answer. My lecturer wrote $-b^{T}$. It should be $-b$ right?
– JIM BOY
Nov 27 '18 at 21:14
1
Oh, sure, that is right.
– alrigazzi
Nov 28 '18 at 19:15
Thanks. Now I get it how $nabla b^{T}x = b$. What I mean by typo is on the answer. My lecturer wrote $-b^{T}$. It should be $-b$ right?
– JIM BOY
Nov 27 '18 at 21:14
Thanks. Now I get it how $nabla b^{T}x = b$. What I mean by typo is on the answer. My lecturer wrote $-b^{T}$. It should be $-b$ right?
– JIM BOY
Nov 27 '18 at 21:14
1
1
Oh, sure, that is right.
– alrigazzi
Nov 28 '18 at 19:15
Oh, sure, that is right.
– alrigazzi
Nov 28 '18 at 19:15
add a comment |
Call a column vector $vec{x}$. Indicate $x^1$ for the 1st component, $x^2$ for the second, and so on. Square of a component will be $(x^1)^2$ Will be the first component (usually x coponent) squared.
$x^i$ will be used to represent the i_th component or the whole vector.
The transpose of vector will be represented with a lowered index $x_i$ is $vec{x}^T$
Inner products can only be performed between a column vector and a row vector, so one vector is addressed by lower indices, the other by vectors. Matching indicies imply a sum.
So $b^Tx=b_ix^i=b_xx_x+b_yx_y+b_zx_z$
$nabla(b^Tx)=frac{partial}{partial x^j}(b_ix^i)$
By the product rule:
$frac{partial}{partial x^j}(b_ix^i)=frac{partial b_i}{partial x_j}x^i+b_ifrac{partial x^i}{partial x^j}$
But $vec{b}$ isa constant, so the derivatives of any component is 0. $frac{partial x_j}{partial x_i}$ is 0 unless the indices match, in which case the value is 1.
So the other term gives us $b_j$, establishing the expected result.
answered Nov 27 '18 at 21:50
TurlocTheRed
838311
Thanks. What I mean by typo is on the answer. My lecturer wrote $−b^T$. It should be $−b$ right?
– JIM BOY
Nov 27 '18 at 23:04
1
Yep. the $delta^i_k$ makes that happen.
– TurlocTheRed
Nov 27 '18 at 23:08
add a comment |
Call a column vector $vec{x}$. Indicate $x^1$ for the 1st component, $x^2$ for the second, and so on. Square of a component will be $(x^1)^2$ Will be the first component (usually x coponent) squared.
$x^i$ will be used to represent the i_th component or the whole vector.
The transpose of vector will be represented with a lowered index $x_i$ is $vec{x}^T$
Inner products can only be performed between a column vector and a row vector, so one vector is addressed by lower indices, the other by vectors. Matching indicies imply a sum.
So $b^Tx=b_ix^i=b_xx_x+b_yx_y+b_zx_z$
$nabla(b^Tx)=frac{partial}{partial x^j}(b_ix^i)$
By the product rule:
$frac{partial}{partial x^j}(b_ix^i)=frac{partial b_i}{partial x_j}x^i+b_ifrac{partial x^i}{partial x^j}$
But $vec{b}$ isa constant, so the derivatives of any component is 0. $frac{partial x_j}{partial x_i}$ is 0 unless the indices match, in which case the value is 1.
So the other term gives us $b_j$, establishing the expected result.
answered Nov 27 '18 at 21:50
TurlocTheRed
838311
Thanks. What I mean by typo is on the answer. My lecturer wrote $−b^T$. It should be $−b$ right?
– JIM BOY
Nov 27 '18 at 23:04
1
Yep. the $delta^i_k$ makes that happen.
– TurlocTheRed
Nov 27 '18 at 23:08
add a comment |
Call a column vector $vec{x}$. Indicate $x^1$ for the 1st component, $x^2$ for the second, and so on. Square of a component will be $(x^1)^2$ Will be the first component (usually x coponent) squared.
$x^i$ will be used to represent the i_th component or the whole vector.
The transpose of vector will be represented with a lowered index $x_i$ is $vec{x}^T$
Inner products can only be performed between a column vector and a row vector, so one vector is addressed by lower indices, the other by vectors. Matching indicies imply a sum.
So $b^Tx=b_ix^i=b_xx_x+b_yx_y+b_zx_z$
$nabla(b^Tx)=frac{partial}{partial x^j}(b_ix^i)$
By the product rule:
$frac{partial}{partial x^j}(b_ix^i)=frac{partial b_i}{partial x_j}x^i+b_ifrac{partial x^i}{partial x^j}$
But $vec{b}$ isa constant, so the derivatives of any component is 0. $frac{partial x_j}{partial x_i}$ is 0 unless the indices match, in which case the value is 1.
So the other term gives us $b_j$, establishing the expected result.
answered Nov 27 '18 at 21:50
TurlocTheRed
838311
Call a column vector $vec{x}$. Indicate $x^1$ for the 1st component, $x^2$ for the second, and so on. Square of a component will be $(x^1)^2$ Will be the first component (usually x coponent) squared.
$x^i$ will be used to represent the i_th component or the whole vector.
The transpose of vector will be represented with a lowered index $x_i$ is $vec{x}^T$
Inner products can only be performed between a column vector and a row vector, so one vector is addressed by lower indices, the other by vectors. Matching indicies imply a sum.
So $b^Tx=b_ix^i=b_xx_x+b_yx_y+b_zx_z$
$nabla(b^Tx)=frac{partial}{partial x^j}(b_ix^i)$
By the product rule:
$frac{partial}{partial x^j}(b_ix^i)=frac{partial b_i}{partial x_j}x^i+b_ifrac{partial x^i}{partial x^j}$
But $vec{b}$ isa constant, so the derivatives of any component is 0. $frac{partial x_j}{partial x_i}$ is 0 unless the indices match, in which case the value is 1.
So the other term gives us $b_j$, establishing the expected result.
answered Nov 27 '18 at 21:50
TurlocTheRed
838311
edited Nov 28 '18 at 16:56
answered Nov 27 '18 at 21:50
TurlocTheRed
838311
answered Nov 27 '18 at 21:50
TurlocTheRed
838311
answered Nov 27 '18 at 21:50
TurlocTheRed
838311
838311
Thanks. What I mean by typo is on the answer. My lecturer wrote $−b^T$. It should be $−b$ right?
– JIM BOY
Nov 27 '18 at 23:04
1
Yep. the $delta^i_k$ makes that happen.
– TurlocTheRed
Nov 27 '18 at 23:08
add a comment |
Thanks. What I mean by typo is on the answer. My lecturer wrote $−b^T$. It should be $−b$ right?
– JIM BOY
Nov 27 '18 at 23:04
1
Yep. the $delta^i_k$ makes that happen.
– TurlocTheRed
Nov 27 '18 at 23:08
Thanks. What I mean by typo is on the answer. My lecturer wrote $−b^T$. It should be $−b$ right?
– JIM BOY
Nov 27 '18 at 23:04
Thanks. What I mean by typo is on the answer. My lecturer wrote $−b^T$. It should be $−b$ right?
– JIM BOY
Nov 27 '18 at 23:04
1
1
Yep. the $delta^i_k$ makes that happen.
– TurlocTheRed
Nov 27 '18 at 23:08
Yep. the $delta^i_k$ makes that happen.
– TurlocTheRed
Nov 27 '18 at 23:08
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3016249%2fgradient-of-matrix-btx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Yes, $nabla b^{T}x = b$.
– Math Lover
Nov 27 '18 at 20:15
Could you please elaborate how the transpose of $b$ becomes $b$? And do you think it is a typo where $b^{T}$ in the answer should've been written as $b$?
– JIM BOY
Nov 27 '18 at 20:29
Give a look to this other answer: math.stackexchange.com/questions/3013384/…
– caverac
Nov 27 '18 at 21:00