Gradient of matrix $b^{T}x$
up vote
0
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favorite
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 at 20:12
JIM BOY
356
add a comment |
up vote
0
down vote
favorite
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 at 20:12
JIM BOY
356
1
Yes, $nabla b^{T}x = b$.
– Math Lover
Nov 27 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 at 20:29
Give a look to this other answer: math.stackexchange.com/questions/3013384/…
– caverac
Nov 27 at 21:00
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 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 at 20:12
JIM BOY
356
asked Nov 27 at 20:12
JIM BOY
356
edited Nov 27 at 21:21
asked Nov 27 at 20:12
JIM BOY
356
asked Nov 27 at 20:12
JIM BOY
356
asked Nov 27 at 20:12
JIM BOY
356
356
1
Yes, $nabla b^{T}x = b$.
– Math Lover
Nov 27 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 at 20:29
Give a look to this other answer: math.stackexchange.com/questions/3013384/…
– caverac
Nov 27 at 21:00
add a comment |
1
Yes, $nabla b^{T}x = b$.
– Math Lover
Nov 27 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 at 20:29
Give a look to this other answer: math.stackexchange.com/questions/3013384/…
– caverac
Nov 27 at 21:00
1
1
Yes, $nabla b^{T}x = b$.
– Math Lover
Nov 27 at 20:15
Yes, $nabla b^{T}x = b$.
– Math Lover
Nov 27 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 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 at 20:29
Give a look to this other answer: math.stackexchange.com/questions/3013384/…
– caverac
Nov 27 at 21:00
Give a look to this other answer: math.stackexchange.com/questions/3013384/…
– caverac
Nov 27 at 21:00
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
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 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 at 21:14
1
Oh, sure, that is right.
– alrigazzi
Nov 28 at 19:15
add a comment |
up vote
1
down vote
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 at 21:50
TurlocTheRed
818311
Thanks. What I mean by typo is on the answer. My lecturer wrote $−b^T$. It should be $−b$ right?
– JIM BOY
Nov 27 at 23:04
1
Yep. the $delta^i_k$ makes that happen.
– TurlocTheRed
Nov 27 at 23:08
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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 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 at 21:14
1
Oh, sure, that is right.
– alrigazzi
Nov 28 at 19:15
add a comment |
up vote
2
down vote
accepted
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 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 at 21:14
1
Oh, sure, that is right.
– alrigazzi
Nov 28 at 19:15
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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 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 at 21:10
alrigazzi
363
answered Nov 27 at 21:10
alrigazzi
363
answered Nov 27 at 21:10
alrigazzi
363
answered Nov 27 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 at 21:14
1
Oh, sure, that is right.
– alrigazzi
Nov 28 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 at 21:14
1
Oh, sure, that is right.
– alrigazzi
Nov 28 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 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 at 21:14
1
1
Oh, sure, that is right.
– alrigazzi
Nov 28 at 19:15
Oh, sure, that is right.
– alrigazzi
Nov 28 at 19:15
add a comment |
up vote
1
down vote
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 at 21:50
TurlocTheRed
818311
Thanks. What I mean by typo is on the answer. My lecturer wrote $−b^T$. It should be $−b$ right?
– JIM BOY
Nov 27 at 23:04
1
Yep. the $delta^i_k$ makes that happen.
– TurlocTheRed
Nov 27 at 23:08
add a comment |
up vote
1
down vote
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 at 21:50
TurlocTheRed
818311
Thanks. What I mean by typo is on the answer. My lecturer wrote $−b^T$. It should be $−b$ right?
– JIM BOY
Nov 27 at 23:04
1
Yep. the $delta^i_k$ makes that happen.
– TurlocTheRed
Nov 27 at 23:08
add a comment |
up vote
1
down vote
up vote
1
down vote
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 at 21:50
TurlocTheRed
818311
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 at 21:50
TurlocTheRed
818311
edited Nov 28 at 16:56
answered Nov 27 at 21:50
TurlocTheRed
818311
answered Nov 27 at 21:50
TurlocTheRed
818311
answered Nov 27 at 21:50
TurlocTheRed
818311
818311
Thanks. What I mean by typo is on the answer. My lecturer wrote $−b^T$. It should be $−b$ right?
– JIM BOY
Nov 27 at 23:04
1
Yep. the $delta^i_k$ makes that happen.
– TurlocTheRed
Nov 27 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 at 23:04
1
Yep. the $delta^i_k$ makes that happen.
– TurlocTheRed
Nov 27 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 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 at 23:04
1
1
Yep. the $delta^i_k$ makes that happen.
– TurlocTheRed
Nov 27 at 23:08
Yep. the $delta^i_k$ makes that happen.
– TurlocTheRed
Nov 27 at 23:08
add a comment |
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Yes, $nabla b^{T}x = b$.
– Math Lover
Nov 27 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 at 20:29
Give a look to this other answer: math.stackexchange.com/questions/3013384/…
– caverac
Nov 27 at 21:00