Mixing
Matrix derivatives, problem with dimensions
$begingroup$
I'm trying to find a derivative of function:
$$L = f cdot y; f = X cdot W + b$$
Matrices shapes: $X.shape=(1, m), W.shape=(m,10), b.shape=(1, 10), y.shape=(10, 1)$
I'm looking for $frac{partial L}{partial W}$
According to chain-rule:
$$frac{partial L}{partial W} = frac{partial L}{partial f} frac{partial f}{partial W} $$
Separately we can find:
$$ frac{partial L}{partial f} = y$$
$$ frac{partial f}{partial W} = X$$
And the problem is that the derivative's dimension of $frac{partial L}{partial W} $ according to my formula is $(10, m)$. However, the dimension should coincide with dimension of $W$.
Also I was advised to find differential of $L$:
$$ d(L) = d(f cdot y) = d(f) cdot y = d (X cdot W + b)y = X cdot dW cdot y $$
But I do not understand how can I get from this the derivative $frac{partial L}{partial W} $
matrices derivatives chain-rule
asked Jan 13 at 18:48
Dmitry DenisovDmitry Denisov
62
$endgroup$
add a comment |
$begingroup$
I'm trying to find a derivative of function:
$$L = f cdot y; f = X cdot W + b$$
Matrices shapes: $X.shape=(1, m), W.shape=(m,10), b.shape=(1, 10), y.shape=(10, 1)$
I'm looking for $frac{partial L}{partial W}$
According to chain-rule:
$$frac{partial L}{partial W} = frac{partial L}{partial f} frac{partial f}{partial W} $$
Separately we can find:
$$ frac{partial L}{partial f} = y$$
$$ frac{partial f}{partial W} = X$$
And the problem is that the derivative's dimension of $frac{partial L}{partial W} $ according to my formula is $(10, m)$. However, the dimension should coincide with dimension of $W$.
Also I was advised to find differential of $L$:
$$ d(L) = d(f cdot y) = d(f) cdot y = d (X cdot W + b)y = X cdot dW cdot y $$
But I do not understand how can I get from this the derivative $frac{partial L}{partial W} $
matrices derivatives chain-rule
asked Jan 13 at 18:48
Dmitry DenisovDmitry Denisov
62
$endgroup$
add a comment |
$begingroup$
I'm trying to find a derivative of function:
$$L = f cdot y; f = X cdot W + b$$
Matrices shapes: $X.shape=(1, m), W.shape=(m,10), b.shape=(1, 10), y.shape=(10, 1)$
I'm looking for $frac{partial L}{partial W}$
According to chain-rule:
$$frac{partial L}{partial W} = frac{partial L}{partial f} frac{partial f}{partial W} $$
Separately we can find:
$$ frac{partial L}{partial f} = y$$
$$ frac{partial f}{partial W} = X$$
And the problem is that the derivative's dimension of $frac{partial L}{partial W} $ according to my formula is $(10, m)$. However, the dimension should coincide with dimension of $W$.
Also I was advised to find differential of $L$:
$$ d(L) = d(f cdot y) = d(f) cdot y = d (X cdot W + b)y = X cdot dW cdot y $$
But I do not understand how can I get from this the derivative $frac{partial L}{partial W} $
matrices derivatives chain-rule
asked Jan 13 at 18:48
Dmitry DenisovDmitry Denisov
62
$endgroup$
I'm trying to find a derivative of function:
$$L = f cdot y; f = X cdot W + b$$
Matrices shapes: $X.shape=(1, m), W.shape=(m,10), b.shape=(1, 10), y.shape=(10, 1)$
I'm looking for $frac{partial L}{partial W}$
According to chain-rule:
$$frac{partial L}{partial W} = frac{partial L}{partial f} frac{partial f}{partial W} $$
Separately we can find:
$$ frac{partial L}{partial f} = y$$
$$ frac{partial f}{partial W} = X$$
And the problem is that the derivative's dimension of $frac{partial L}{partial W} $ according to my formula is $(10, m)$. However, the dimension should coincide with dimension of $W$.
Also I was advised to find differential of $L$:
$$ d(L) = d(f cdot y) = d(f) cdot y = d (X cdot W + b)y = X cdot dW cdot y $$
But I do not understand how can I get from this the derivative $frac{partial L}{partial W} $
matrices derivatives chain-rule
matrices derivatives chain-rule
asked Jan 13 at 18:48
Dmitry DenisovDmitry Denisov
62
asked Jan 13 at 18:48
Dmitry DenisovDmitry Denisov
62
asked Jan 13 at 18:48
Dmitry DenisovDmitry Denisov
62
asked Jan 13 at 18:48
Dmitry DenisovDmitry Denisov
62
asked Jan 13 at 18:48
Dmitry DenisovDmitry Denisov
62
62
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's use a convention where a lowercase Latin letter always represents a column vector, an uppercase Latin is a matrix, and a Greek letter is a scalar.
Using this convention your equations are
$$eqalign{
f &= W^Tx + b cr
lambda &= f^Ty cr
}$$
As you have noted, the differential of the scalar function is
$$eqalign{
dlambda &= df^Ty = (dW^Tx)^Ty = x^TdW,y cr
}$$
Let's develop that a bit further by introducing the Trace function
$$eqalign{
dlambda &= {rm Tr}(x^TdW,y) = {rm Tr}(yx^TdW) cr
}$$
Then, depending on your preferred Layout Convention, the gradient is either
$$eqalign{
frac{partiallambda}{partial W} &=yx^T quad{rm or}quad xy^T cr
}$$
Since you expected the the dimensions of the gradient to be those of $W$, it sounds like your preferred layout is $xy^T$
Also note that $frac{partial f}{partial W}neq X.,$ The gradient is a 3rd order tensor, while $X$ is just a 2nd order tensor (aka a matrix). The presence of these 3rd and 4th order tensors as intermediate quantities in the chain rule can make it difficult/impossible to use in practice.
The differential approach suggested by your advisor is often simpler because the differential of a matrix is just another matrix quantity, which is easy to handle.
answered Jan 14 at 19:41
greggreg
8,2751823
$endgroup$
$begingroup$
Thank you very much for your answer, it became more clear for me now! I have 2 questions about your solution: 1) Do I understand correctly that you introduced Trace function, because dλ is scalar so const=Tr(const) ? 2) $frac{partial f}{partial W}$ is 3rd order tensor. Maybe in this case you know how chain rule works in Neural Networks? Because I get derivative from previous layer and I should multiply it by the derivative of current layer according to chain rule. However, as you mentioned, $frac{partial f}{partial W}$ now is a 3rd order tensor, so how can we apply chain rule?
$endgroup$
– Dmitry Denisov
Jan 14 at 22:13
$begingroup$
And even if X is matrix then f is also a matrix and we should take a derivative: matrix-by-matrix?
$endgroup$
– Dmitry Denisov
Jan 15 at 12:14
$begingroup$
@DmitryDenisov 1) Yes, ${rm Tr}(scalar)=scalar,,$ 2) the gradient really is a 3rd order tensor. The point is you never need to calculate 3rd order (vector-by-matrix) or 4th order (matrix-by-matrix) derivatives, and the programs you write will never calculate such quantities either. These online notes are worth a read.
$endgroup$
– greg
Jan 15 at 18:04
$begingroup$
In this row nabla_w[-1] = np.dot(delta, activations[-2].transpose()) they set $frac{partial L}{partial W}$ is equal to $X^T cdot delta$, so it doesn't seem like chain rule. I.e. in another case they also should calculate the derivative using differential on paper, however they stated that chain rule is a universal approach
$endgroup$
– Dmitry Denisov
Jan 16 at 9:57
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let's use a convention where a lowercase Latin letter always represents a column vector, an uppercase Latin is a matrix, and a Greek letter is a scalar.
Using this convention your equations are
$$eqalign{
f &= W^Tx + b cr
lambda &= f^Ty cr
}$$
As you have noted, the differential of the scalar function is
$$eqalign{
dlambda &= df^Ty = (dW^Tx)^Ty = x^TdW,y cr
}$$
Let's develop that a bit further by introducing the Trace function
$$eqalign{
dlambda &= {rm Tr}(x^TdW,y) = {rm Tr}(yx^TdW) cr
}$$
Then, depending on your preferred Layout Convention, the gradient is either
$$eqalign{
frac{partiallambda}{partial W} &=yx^T quad{rm or}quad xy^T cr
}$$
Since you expected the the dimensions of the gradient to be those of $W$, it sounds like your preferred layout is $xy^T$
Also note that $frac{partial f}{partial W}neq X.,$ The gradient is a 3rd order tensor, while $X$ is just a 2nd order tensor (aka a matrix). The presence of these 3rd and 4th order tensors as intermediate quantities in the chain rule can make it difficult/impossible to use in practice.
The differential approach suggested by your advisor is often simpler because the differential of a matrix is just another matrix quantity, which is easy to handle.
answered Jan 14 at 19:41
greggreg
8,2751823
$endgroup$
$begingroup$
Thank you very much for your answer, it became more clear for me now! I have 2 questions about your solution: 1) Do I understand correctly that you introduced Trace function, because dλ is scalar so const=Tr(const) ? 2) $frac{partial f}{partial W}$ is 3rd order tensor. Maybe in this case you know how chain rule works in Neural Networks? Because I get derivative from previous layer and I should multiply it by the derivative of current layer according to chain rule. However, as you mentioned, $frac{partial f}{partial W}$ now is a 3rd order tensor, so how can we apply chain rule?
$endgroup$
– Dmitry Denisov
Jan 14 at 22:13
$begingroup$
And even if X is matrix then f is also a matrix and we should take a derivative: matrix-by-matrix?
$endgroup$
– Dmitry Denisov
Jan 15 at 12:14
$begingroup$
@DmitryDenisov 1) Yes, ${rm Tr}(scalar)=scalar,,$ 2) the gradient really is a 3rd order tensor. The point is you never need to calculate 3rd order (vector-by-matrix) or 4th order (matrix-by-matrix) derivatives, and the programs you write will never calculate such quantities either. These online notes are worth a read.
$endgroup$
– greg
Jan 15 at 18:04
$begingroup$
In this row nabla_w[-1] = np.dot(delta, activations[-2].transpose()) they set $frac{partial L}{partial W}$ is equal to $X^T cdot delta$, so it doesn't seem like chain rule. I.e. in another case they also should calculate the derivative using differential on paper, however they stated that chain rule is a universal approach
$endgroup$
– Dmitry Denisov
Jan 16 at 9:57
add a comment |
$begingroup$
Let's use a convention where a lowercase Latin letter always represents a column vector, an uppercase Latin is a matrix, and a Greek letter is a scalar.
Using this convention your equations are
$$eqalign{
f &= W^Tx + b cr
lambda &= f^Ty cr
}$$
As you have noted, the differential of the scalar function is
$$eqalign{
dlambda &= df^Ty = (dW^Tx)^Ty = x^TdW,y cr
}$$
Let's develop that a bit further by introducing the Trace function
$$eqalign{
dlambda &= {rm Tr}(x^TdW,y) = {rm Tr}(yx^TdW) cr
}$$
Then, depending on your preferred Layout Convention, the gradient is either
$$eqalign{
frac{partiallambda}{partial W} &=yx^T quad{rm or}quad xy^T cr
}$$
Since you expected the the dimensions of the gradient to be those of $W$, it sounds like your preferred layout is $xy^T$
Also note that $frac{partial f}{partial W}neq X.,$ The gradient is a 3rd order tensor, while $X$ is just a 2nd order tensor (aka a matrix). The presence of these 3rd and 4th order tensors as intermediate quantities in the chain rule can make it difficult/impossible to use in practice.
The differential approach suggested by your advisor is often simpler because the differential of a matrix is just another matrix quantity, which is easy to handle.
answered Jan 14 at 19:41
greggreg
8,2751823
$endgroup$
$begingroup$
Thank you very much for your answer, it became more clear for me now! I have 2 questions about your solution: 1) Do I understand correctly that you introduced Trace function, because dλ is scalar so const=Tr(const) ? 2) $frac{partial f}{partial W}$ is 3rd order tensor. Maybe in this case you know how chain rule works in Neural Networks? Because I get derivative from previous layer and I should multiply it by the derivative of current layer according to chain rule. However, as you mentioned, $frac{partial f}{partial W}$ now is a 3rd order tensor, so how can we apply chain rule?
$endgroup$
– Dmitry Denisov
Jan 14 at 22:13
$begingroup$
And even if X is matrix then f is also a matrix and we should take a derivative: matrix-by-matrix?
$endgroup$
– Dmitry Denisov
Jan 15 at 12:14
$begingroup$
@DmitryDenisov 1) Yes, ${rm Tr}(scalar)=scalar,,$ 2) the gradient really is a 3rd order tensor. The point is you never need to calculate 3rd order (vector-by-matrix) or 4th order (matrix-by-matrix) derivatives, and the programs you write will never calculate such quantities either. These online notes are worth a read.
$endgroup$
– greg
Jan 15 at 18:04
$begingroup$
In this row nabla_w[-1] = np.dot(delta, activations[-2].transpose()) they set $frac{partial L}{partial W}$ is equal to $X^T cdot delta$, so it doesn't seem like chain rule. I.e. in another case they also should calculate the derivative using differential on paper, however they stated that chain rule is a universal approach
$endgroup$
– Dmitry Denisov
Jan 16 at 9:57
add a comment |
$begingroup$
Let's use a convention where a lowercase Latin letter always represents a column vector, an uppercase Latin is a matrix, and a Greek letter is a scalar.
Using this convention your equations are
$$eqalign{
f &= W^Tx + b cr
lambda &= f^Ty cr
}$$
As you have noted, the differential of the scalar function is
$$eqalign{
dlambda &= df^Ty = (dW^Tx)^Ty = x^TdW,y cr
}$$
Let's develop that a bit further by introducing the Trace function
$$eqalign{
dlambda &= {rm Tr}(x^TdW,y) = {rm Tr}(yx^TdW) cr
}$$
Then, depending on your preferred Layout Convention, the gradient is either
$$eqalign{
frac{partiallambda}{partial W} &=yx^T quad{rm or}quad xy^T cr
}$$
Since you expected the the dimensions of the gradient to be those of $W$, it sounds like your preferred layout is $xy^T$
Also note that $frac{partial f}{partial W}neq X.,$ The gradient is a 3rd order tensor, while $X$ is just a 2nd order tensor (aka a matrix). The presence of these 3rd and 4th order tensors as intermediate quantities in the chain rule can make it difficult/impossible to use in practice.
The differential approach suggested by your advisor is often simpler because the differential of a matrix is just another matrix quantity, which is easy to handle.
answered Jan 14 at 19:41
greggreg
8,2751823
$endgroup$
Let's use a convention where a lowercase Latin letter always represents a column vector, an uppercase Latin is a matrix, and a Greek letter is a scalar.
Using this convention your equations are
$$eqalign{
f &= W^Tx + b cr
lambda &= f^Ty cr
}$$
As you have noted, the differential of the scalar function is
$$eqalign{
dlambda &= df^Ty = (dW^Tx)^Ty = x^TdW,y cr
}$$
Let's develop that a bit further by introducing the Trace function
$$eqalign{
dlambda &= {rm Tr}(x^TdW,y) = {rm Tr}(yx^TdW) cr
}$$
Then, depending on your preferred Layout Convention, the gradient is either
$$eqalign{
frac{partiallambda}{partial W} &=yx^T quad{rm or}quad xy^T cr
}$$
Since you expected the the dimensions of the gradient to be those of $W$, it sounds like your preferred layout is $xy^T$
Also note that $frac{partial f}{partial W}neq X.,$ The gradient is a 3rd order tensor, while $X$ is just a 2nd order tensor (aka a matrix). The presence of these 3rd and 4th order tensors as intermediate quantities in the chain rule can make it difficult/impossible to use in practice.
The differential approach suggested by your advisor is often simpler because the differential of a matrix is just another matrix quantity, which is easy to handle.
answered Jan 14 at 19:41
greggreg
8,2751823
edited Jan 14 at 19:53
answered Jan 14 at 19:41
greggreg
8,2751823
answered Jan 14 at 19:41
greggreg
8,2751823
answered Jan 14 at 19:41
greggreg
8,2751823
8,2751823
$begingroup$
Thank you very much for your answer, it became more clear for me now! I have 2 questions about your solution: 1) Do I understand correctly that you introduced Trace function, because dλ is scalar so const=Tr(const) ? 2) $frac{partial f}{partial W}$ is 3rd order tensor. Maybe in this case you know how chain rule works in Neural Networks? Because I get derivative from previous layer and I should multiply it by the derivative of current layer according to chain rule. However, as you mentioned, $frac{partial f}{partial W}$ now is a 3rd order tensor, so how can we apply chain rule?
$endgroup$
– Dmitry Denisov
Jan 14 at 22:13
$begingroup$
And even if X is matrix then f is also a matrix and we should take a derivative: matrix-by-matrix?
$endgroup$
– Dmitry Denisov
Jan 15 at 12:14
$begingroup$
@DmitryDenisov 1) Yes, ${rm Tr}(scalar)=scalar,,$ 2) the gradient really is a 3rd order tensor. The point is you never need to calculate 3rd order (vector-by-matrix) or 4th order (matrix-by-matrix) derivatives, and the programs you write will never calculate such quantities either. These online notes are worth a read.
$endgroup$
– greg
Jan 15 at 18:04
$begingroup$
In this row nabla_w[-1] = np.dot(delta, activations[-2].transpose()) they set $frac{partial L}{partial W}$ is equal to $X^T cdot delta$, so it doesn't seem like chain rule. I.e. in another case they also should calculate the derivative using differential on paper, however they stated that chain rule is a universal approach
$endgroup$
– Dmitry Denisov
Jan 16 at 9:57
add a comment |
$begingroup$
Thank you very much for your answer, it became more clear for me now! I have 2 questions about your solution: 1) Do I understand correctly that you introduced Trace function, because dλ is scalar so const=Tr(const) ? 2) $frac{partial f}{partial W}$ is 3rd order tensor. Maybe in this case you know how chain rule works in Neural Networks? Because I get derivative from previous layer and I should multiply it by the derivative of current layer according to chain rule. However, as you mentioned, $frac{partial f}{partial W}$ now is a 3rd order tensor, so how can we apply chain rule?
$endgroup$
– Dmitry Denisov
Jan 14 at 22:13
$begingroup$
And even if X is matrix then f is also a matrix and we should take a derivative: matrix-by-matrix?
$endgroup$
– Dmitry Denisov
Jan 15 at 12:14
$begingroup$
@DmitryDenisov 1) Yes, ${rm Tr}(scalar)=scalar,,$ 2) the gradient really is a 3rd order tensor. The point is you never need to calculate 3rd order (vector-by-matrix) or 4th order (matrix-by-matrix) derivatives, and the programs you write will never calculate such quantities either. These online notes are worth a read.
$endgroup$
– greg
Jan 15 at 18:04
$begingroup$
In this row nabla_w[-1] = np.dot(delta, activations[-2].transpose()) they set $frac{partial L}{partial W}$ is equal to $X^T cdot delta$, so it doesn't seem like chain rule. I.e. in another case they also should calculate the derivative using differential on paper, however they stated that chain rule is a universal approach
$endgroup$
– Dmitry Denisov
Jan 16 at 9:57
$begingroup$
Thank you very much for your answer, it became more clear for me now! I have 2 questions about your solution: 1) Do I understand correctly that you introduced Trace function, because dλ is scalar so const=Tr(const) ? 2) $frac{partial f}{partial W}$ is 3rd order tensor. Maybe in this case you know how chain rule works in Neural Networks? Because I get derivative from previous layer and I should multiply it by the derivative of current layer according to chain rule. However, as you mentioned, $frac{partial f}{partial W}$ now is a 3rd order tensor, so how can we apply chain rule?
$endgroup$
– Dmitry Denisov
Jan 14 at 22:13
$begingroup$
Thank you very much for your answer, it became more clear for me now! I have 2 questions about your solution: 1) Do I understand correctly that you introduced Trace function, because dλ is scalar so const=Tr(const) ? 2) $frac{partial f}{partial W}$ is 3rd order tensor. Maybe in this case you know how chain rule works in Neural Networks? Because I get derivative from previous layer and I should multiply it by the derivative of current layer according to chain rule. However, as you mentioned, $frac{partial f}{partial W}$ now is a 3rd order tensor, so how can we apply chain rule?
$endgroup$
– Dmitry Denisov
Jan 14 at 22:13
$begingroup$
And even if X is matrix then f is also a matrix and we should take a derivative: matrix-by-matrix?
$endgroup$
– Dmitry Denisov
Jan 15 at 12:14
$begingroup$
And even if X is matrix then f is also a matrix and we should take a derivative: matrix-by-matrix?
$endgroup$
– Dmitry Denisov
Jan 15 at 12:14
$begingroup$
@DmitryDenisov 1) Yes, ${rm Tr}(scalar)=scalar,,$ 2) the gradient really is a 3rd order tensor. The point is you never need to calculate 3rd order (vector-by-matrix) or 4th order (matrix-by-matrix) derivatives, and the programs you write will never calculate such quantities either. These online notes are worth a read.
$endgroup$
– greg
Jan 15 at 18:04
$begingroup$
@DmitryDenisov 1) Yes, ${rm Tr}(scalar)=scalar,,$ 2) the gradient really is a 3rd order tensor. The point is you never need to calculate 3rd order (vector-by-matrix) or 4th order (matrix-by-matrix) derivatives, and the programs you write will never calculate such quantities either. These online notes are worth a read.
$endgroup$
– greg
Jan 15 at 18:04
$begingroup$
In this row nabla_w[-1] = np.dot(delta, activations[-2].transpose()) they set $frac{partial L}{partial W}$ is equal to $X^T cdot delta$, so it doesn't seem like chain rule. I.e. in another case they also should calculate the derivative using differential on paper, however they stated that chain rule is a universal approach
$endgroup$
– Dmitry Denisov
Jan 16 at 9:57
$begingroup$
In this row nabla_w[-1] = np.dot(delta, activations[-2].transpose()) they set $frac{partial L}{partial W}$ is equal to $X^T cdot delta$, so it doesn't seem like chain rule. I.e. in another case they also should calculate the derivative using differential on paper, however they stated that chain rule is a universal approach
$endgroup$
– Dmitry Denisov
Jan 16 at 9:57
add a comment |
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