Gradient of $||Ax - y||^2$ with respect to $A$
How do I proceed to find $nabla_A||Ax - y||^2$ where $A in mathbb{R}^{ntimes n}$ and $x,y in mathbb{R}^n$ and the norm is the Euclidean norm.
Attempt so far
$$||Ax - y||^2 = (Ax-y)^T(Ax-y) = x^TA^TAx - 2x^TAy + y^Ty $$
$$ nabla_A(x^TAy) = xy^T$$
Where I am stuck
I don't know how to tackle the $x^TA^TAx$ term since if I try to apply chain rule, I will have to differentiate a matrix with respect to a matrix.
vector-analysis matrix-calculus
edited Nov 22 at 9:31
user550103
6711315
asked Nov 22 at 6:38
Anant Joshi
451211
add a comment |
How do I proceed to find $nabla_A||Ax - y||^2$ where $A in mathbb{R}^{ntimes n}$ and $x,y in mathbb{R}^n$ and the norm is the Euclidean norm.
Attempt so far
$$||Ax - y||^2 = (Ax-y)^T(Ax-y) = x^TA^TAx - 2x^TAy + y^Ty $$
$$ nabla_A(x^TAy) = xy^T$$
Where I am stuck
I don't know how to tackle the $x^TA^TAx$ term since if I try to apply chain rule, I will have to differentiate a matrix with respect to a matrix.
vector-analysis matrix-calculus
edited Nov 22 at 9:31
user550103
6711315
asked Nov 22 at 6:38
Anant Joshi
451211
add a comment |
How do I proceed to find $nabla_A||Ax - y||^2$ where $A in mathbb{R}^{ntimes n}$ and $x,y in mathbb{R}^n$ and the norm is the Euclidean norm.
Attempt so far
$$||Ax - y||^2 = (Ax-y)^T(Ax-y) = x^TA^TAx - 2x^TAy + y^Ty $$
$$ nabla_A(x^TAy) = xy^T$$
Where I am stuck
I don't know how to tackle the $x^TA^TAx$ term since if I try to apply chain rule, I will have to differentiate a matrix with respect to a matrix.
vector-analysis matrix-calculus
edited Nov 22 at 9:31
user550103
6711315
asked Nov 22 at 6:38
Anant Joshi
451211
How do I proceed to find $nabla_A||Ax - y||^2$ where $A in mathbb{R}^{ntimes n}$ and $x,y in mathbb{R}^n$ and the norm is the Euclidean norm.
Attempt so far
$$||Ax - y||^2 = (Ax-y)^T(Ax-y) = x^TA^TAx - 2x^TAy + y^Ty $$
$$ nabla_A(x^TAy) = xy^T$$
Where I am stuck
I don't know how to tackle the $x^TA^TAx$ term since if I try to apply chain rule, I will have to differentiate a matrix with respect to a matrix.
vector-analysis matrix-calculus
vector-analysis matrix-calculus
edited Nov 22 at 9:31
user550103
6711315
asked Nov 22 at 6:38
Anant Joshi
451211
edited Nov 22 at 9:31
user550103
6711315
asked Nov 22 at 6:38
Anant Joshi
451211
edited Nov 22 at 9:31
user550103
6711315
edited Nov 22 at 9:31
user550103
6711315
edited Nov 22 at 9:31
user550103
6711315
6711315
asked Nov 22 at 6:38
Anant Joshi
451211
asked Nov 22 at 6:38
Anant Joshi
451211
asked Nov 22 at 6:38
Anant Joshi
451211
451211
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Before we start deriving the gradient, some facts and notations for brevity:
- Trace and Frobenius product relation $$leftlangle A, B Crightrangle={rm tr}(A^TBC) := A : B C$$
- Cyclic properties of Trace/Frobenius product
begin{align}
A : B C
&= BC : A \
&= A C^T : B \
&= {text{etc.}} cr
end{align}
Let $f := left|Ax-y right|^2 = Ax-y:Ax-y$.
Now, we can obtain the differential first, and then the gradient.
begin{align}
df
&= dleft( Ax-y:Ax-y right) \
&= left(dA x : Ax-yright) + left(Ax-y : dA xright) \
&= 2 left(Ax - yright) : dA x \
&= 2left( Ax-yright)x^T : dA\
end{align}
Thus, the gradient is
begin{align}
frac{partial}{partial A} left( left|Ax-y right|^2 right)= 2left( Ax-yright)x^T.
end{align}
answered Nov 22 at 9:08
user550103
6711315
What is the difference between differential and gradient?
– Anant Joshi
Nov 22 at 9:19
see this math.stackexchange.com/questions/289923/…
– user550103
Nov 22 at 9:20
add a comment |
For matrices, the most easy way is often to get back to definition of differentiability, i.e. :
$$||(A+H)x - y||^2-||Ax - y||^2=L_A(H) + o(|H|)$$
With $L_A$ a linear map.
We begin with :
$$||(A+H)x - y||^2=langle Ax+Hx-y , Ax+Hx-y rangle.$$
Then we have :
$$||(A+H)x - y||^2= langle Ax-y,Ax-yrangle + 2langle Hx,Ax -yrangle + langle Hx,Hxrangle.$$
To conclude : $$||(A+H)x - y||^2-||Ax - y||^2=2langle Hx,Ax -yrangle + o (|H|)$$
Thus :
$$left(nabla_A||Ax - y||^2right)_{i,j}=2langle E_{i,j}x,Ax -yrangle$$
With $E_{i,j}$ the matrix with a $1$ at row $i$ and column $j$ and $0$ otherwise.
answered Nov 22 at 9:11
nicomezi
4,0741820
Why is the final answer a number and not a matrix?
– Anant Joshi
Nov 22 at 9:13
Because I misunderstood the notation and went a bit fast at the end. Is it better now ?
– nicomezi
Nov 22 at 9:21
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Before we start deriving the gradient, some facts and notations for brevity:
- Trace and Frobenius product relation $$leftlangle A, B Crightrangle={rm tr}(A^TBC) := A : B C$$
- Cyclic properties of Trace/Frobenius product
begin{align}
A : B C
&= BC : A \
&= A C^T : B \
&= {text{etc.}} cr
end{align}
Let $f := left|Ax-y right|^2 = Ax-y:Ax-y$.
Now, we can obtain the differential first, and then the gradient.
begin{align}
df
&= dleft( Ax-y:Ax-y right) \
&= left(dA x : Ax-yright) + left(Ax-y : dA xright) \
&= 2 left(Ax - yright) : dA x \
&= 2left( Ax-yright)x^T : dA\
end{align}
Thus, the gradient is
begin{align}
frac{partial}{partial A} left( left|Ax-y right|^2 right)= 2left( Ax-yright)x^T.
end{align}
answered Nov 22 at 9:08
user550103
6711315
What is the difference between differential and gradient?
– Anant Joshi
Nov 22 at 9:19
see this math.stackexchange.com/questions/289923/…
– user550103
Nov 22 at 9:20
add a comment |
Before we start deriving the gradient, some facts and notations for brevity:
- Trace and Frobenius product relation $$leftlangle A, B Crightrangle={rm tr}(A^TBC) := A : B C$$
- Cyclic properties of Trace/Frobenius product
begin{align}
A : B C
&= BC : A \
&= A C^T : B \
&= {text{etc.}} cr
end{align}
Let $f := left|Ax-y right|^2 = Ax-y:Ax-y$.
Now, we can obtain the differential first, and then the gradient.
begin{align}
df
&= dleft( Ax-y:Ax-y right) \
&= left(dA x : Ax-yright) + left(Ax-y : dA xright) \
&= 2 left(Ax - yright) : dA x \
&= 2left( Ax-yright)x^T : dA\
end{align}
Thus, the gradient is
begin{align}
frac{partial}{partial A} left( left|Ax-y right|^2 right)= 2left( Ax-yright)x^T.
end{align}
answered Nov 22 at 9:08
user550103
6711315
What is the difference between differential and gradient?
– Anant Joshi
Nov 22 at 9:19
see this math.stackexchange.com/questions/289923/…
– user550103
Nov 22 at 9:20
add a comment |
Before we start deriving the gradient, some facts and notations for brevity:
- Trace and Frobenius product relation $$leftlangle A, B Crightrangle={rm tr}(A^TBC) := A : B C$$
- Cyclic properties of Trace/Frobenius product
begin{align}
A : B C
&= BC : A \
&= A C^T : B \
&= {text{etc.}} cr
end{align}
Let $f := left|Ax-y right|^2 = Ax-y:Ax-y$.
Now, we can obtain the differential first, and then the gradient.
begin{align}
df
&= dleft( Ax-y:Ax-y right) \
&= left(dA x : Ax-yright) + left(Ax-y : dA xright) \
&= 2 left(Ax - yright) : dA x \
&= 2left( Ax-yright)x^T : dA\
end{align}
Thus, the gradient is
begin{align}
frac{partial}{partial A} left( left|Ax-y right|^2 right)= 2left( Ax-yright)x^T.
end{align}
answered Nov 22 at 9:08
user550103
6711315
Before we start deriving the gradient, some facts and notations for brevity:
- Trace and Frobenius product relation $$leftlangle A, B Crightrangle={rm tr}(A^TBC) := A : B C$$
- Cyclic properties of Trace/Frobenius product
begin{align}
A : B C
&= BC : A \
&= A C^T : B \
&= {text{etc.}} cr
end{align}
Let $f := left|Ax-y right|^2 = Ax-y:Ax-y$.
Now, we can obtain the differential first, and then the gradient.
begin{align}
df
&= dleft( Ax-y:Ax-y right) \
&= left(dA x : Ax-yright) + left(Ax-y : dA xright) \
&= 2 left(Ax - yright) : dA x \
&= 2left( Ax-yright)x^T : dA\
end{align}
Thus, the gradient is
begin{align}
frac{partial}{partial A} left( left|Ax-y right|^2 right)= 2left( Ax-yright)x^T.
end{align}
answered Nov 22 at 9:08
user550103
6711315
edited Nov 22 at 9:17
answered Nov 22 at 9:08
user550103
6711315
answered Nov 22 at 9:08
user550103
6711315
answered Nov 22 at 9:08
user550103
6711315
6711315
What is the difference between differential and gradient?
– Anant Joshi
Nov 22 at 9:19
see this math.stackexchange.com/questions/289923/…
– user550103
Nov 22 at 9:20
add a comment |
What is the difference between differential and gradient?
– Anant Joshi
Nov 22 at 9:19
see this math.stackexchange.com/questions/289923/…
– user550103
Nov 22 at 9:20
What is the difference between differential and gradient?
– Anant Joshi
Nov 22 at 9:19
What is the difference between differential and gradient?
– Anant Joshi
Nov 22 at 9:19
see this math.stackexchange.com/questions/289923/…
– user550103
Nov 22 at 9:20
see this math.stackexchange.com/questions/289923/…
– user550103
Nov 22 at 9:20
add a comment |
For matrices, the most easy way is often to get back to definition of differentiability, i.e. :
$$||(A+H)x - y||^2-||Ax - y||^2=L_A(H) + o(|H|)$$
With $L_A$ a linear map.
We begin with :
$$||(A+H)x - y||^2=langle Ax+Hx-y , Ax+Hx-y rangle.$$
Then we have :
$$||(A+H)x - y||^2= langle Ax-y,Ax-yrangle + 2langle Hx,Ax -yrangle + langle Hx,Hxrangle.$$
To conclude : $$||(A+H)x - y||^2-||Ax - y||^2=2langle Hx,Ax -yrangle + o (|H|)$$
Thus :
$$left(nabla_A||Ax - y||^2right)_{i,j}=2langle E_{i,j}x,Ax -yrangle$$
With $E_{i,j}$ the matrix with a $1$ at row $i$ and column $j$ and $0$ otherwise.
answered Nov 22 at 9:11
nicomezi
4,0741820
Why is the final answer a number and not a matrix?
– Anant Joshi
Nov 22 at 9:13
Because I misunderstood the notation and went a bit fast at the end. Is it better now ?
– nicomezi
Nov 22 at 9:21
add a comment |
For matrices, the most easy way is often to get back to definition of differentiability, i.e. :
$$||(A+H)x - y||^2-||Ax - y||^2=L_A(H) + o(|H|)$$
With $L_A$ a linear map.
We begin with :
$$||(A+H)x - y||^2=langle Ax+Hx-y , Ax+Hx-y rangle.$$
Then we have :
$$||(A+H)x - y||^2= langle Ax-y,Ax-yrangle + 2langle Hx,Ax -yrangle + langle Hx,Hxrangle.$$
To conclude : $$||(A+H)x - y||^2-||Ax - y||^2=2langle Hx,Ax -yrangle + o (|H|)$$
Thus :
$$left(nabla_A||Ax - y||^2right)_{i,j}=2langle E_{i,j}x,Ax -yrangle$$
With $E_{i,j}$ the matrix with a $1$ at row $i$ and column $j$ and $0$ otherwise.
answered Nov 22 at 9:11
nicomezi
4,0741820
Why is the final answer a number and not a matrix?
– Anant Joshi
Nov 22 at 9:13
Because I misunderstood the notation and went a bit fast at the end. Is it better now ?
– nicomezi
Nov 22 at 9:21
add a comment |
For matrices, the most easy way is often to get back to definition of differentiability, i.e. :
$$||(A+H)x - y||^2-||Ax - y||^2=L_A(H) + o(|H|)$$
With $L_A$ a linear map.
We begin with :
$$||(A+H)x - y||^2=langle Ax+Hx-y , Ax+Hx-y rangle.$$
Then we have :
$$||(A+H)x - y||^2= langle Ax-y,Ax-yrangle + 2langle Hx,Ax -yrangle + langle Hx,Hxrangle.$$
To conclude : $$||(A+H)x - y||^2-||Ax - y||^2=2langle Hx,Ax -yrangle + o (|H|)$$
Thus :
$$left(nabla_A||Ax - y||^2right)_{i,j}=2langle E_{i,j}x,Ax -yrangle$$
With $E_{i,j}$ the matrix with a $1$ at row $i$ and column $j$ and $0$ otherwise.
answered Nov 22 at 9:11
nicomezi
4,0741820
For matrices, the most easy way is often to get back to definition of differentiability, i.e. :
$$||(A+H)x - y||^2-||Ax - y||^2=L_A(H) + o(|H|)$$
With $L_A$ a linear map.
We begin with :
$$||(A+H)x - y||^2=langle Ax+Hx-y , Ax+Hx-y rangle.$$
Then we have :
$$||(A+H)x - y||^2= langle Ax-y,Ax-yrangle + 2langle Hx,Ax -yrangle + langle Hx,Hxrangle.$$
To conclude : $$||(A+H)x - y||^2-||Ax - y||^2=2langle Hx,Ax -yrangle + o (|H|)$$
Thus :
$$left(nabla_A||Ax - y||^2right)_{i,j}=2langle E_{i,j}x,Ax -yrangle$$
With $E_{i,j}$ the matrix with a $1$ at row $i$ and column $j$ and $0$ otherwise.
answered Nov 22 at 9:11
nicomezi
4,0741820
edited Nov 22 at 9:20
answered Nov 22 at 9:11
nicomezi
4,0741820
answered Nov 22 at 9:11
nicomezi
4,0741820
answered Nov 22 at 9:11
nicomezi
4,0741820
4,0741820
Why is the final answer a number and not a matrix?
– Anant Joshi
Nov 22 at 9:13
Because I misunderstood the notation and went a bit fast at the end. Is it better now ?
– nicomezi
Nov 22 at 9:21
add a comment |
Why is the final answer a number and not a matrix?
– Anant Joshi
Nov 22 at 9:13
Because I misunderstood the notation and went a bit fast at the end. Is it better now ?
– nicomezi
Nov 22 at 9:21
Why is the final answer a number and not a matrix?
– Anant Joshi
Nov 22 at 9:13
Why is the final answer a number and not a matrix?
– Anant Joshi
Nov 22 at 9:13
Because I misunderstood the notation and went a bit fast at the end. Is it better now ?
– nicomezi
Nov 22 at 9:21
Because I misunderstood the notation and went a bit fast at the end. Is it better now ?
– nicomezi
Nov 22 at 9:21
add a comment |
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