Mixing
Compute the gradient of $f(x)=|text{diag}(x)|$ with the chain rule
$begingroup$
Consider the function $f:mathbb{R}^ntomathbb{R}$ given by $f(x)=|text{diag}(x)|$, where $text{diag}(x)inmathbb{R}^{ntimes{n}}$ is the diagonal matrix with diagonal entries $x_1,x_2,dots,x_n$, and $|cdot|$ is the spectral norm (matrix 2-norm).
Since the spectral norm of a matrix is its largest singular value, and the singular values of a (square) diagonal matrix are the absolute values of the diagonal entries, we see that $f(x)=|x|_infty$, where $|cdot|_infty$ is the (vector) sup-norm. In this form, it is easier to deduce the properties of $f$--in particular, it is differentiable at any point $xinmathbb{R}^n$ where the largest element of $x$ (in absolute value) is unique. At such a point, the gradient of $f$ is given by
$$
nabla{f(x)}=text{sgn}(x_k)e_k
$$
where $k$ is the index of the (unique) largest entry of $x$ (in absolute value), $e_k$ is the $k^text{th}$ standard basis vector in $mathbb{R}^n$, and $text{sgn}(cdot)$ is the sign function.
I want to deduce the above expression for the gradient using the chain rule applied to $f(x)=(gcirc{h})(x)$, where $g:mathbb{R}^{ntimes{n}}tomathbb{R}$ is given by $g(A)=|A|$, and $h:mathbb{R}^ntomathbb{R}^{ntimes{n}}$ is given by $h(x)=text{diag}(x)$.
The "Jacobian" of $h$ is a three-dimensional object, where
$$
frac{partial[h(x)]_{ij}}{partial{x_k}}=begin{cases}1,&i=j=k,\0,&text{else.}end{cases}
$$
The function $g$ is differentiable at any point $A$ where $A$ has a unique largest singular value, in which case the gradient(?) is given by
$$
nabla{g(A)}=uv^text{T},
$$
where $u$ and $v$ are the left and right singular vectors (respectively) corresponding to the (unique) largest singular value of $A$.
So I essentially have a three-dimensional object and a 2-dimensional object, and I want to apply the chain rule to get the gradient, a 1-dimensional object (i.e. a vector). A straightforward application suggests "multiplying them together" (not sure that concept is even defined), which seems like it would produce a matrix. What simple thing am I missing here?
derivatives matrix-calculus chain-rule
asked Jan 22 at 1:23
David M.David M.
1,856418
$endgroup$
add a comment |
$begingroup$
Consider the function $f:mathbb{R}^ntomathbb{R}$ given by $f(x)=|text{diag}(x)|$, where $text{diag}(x)inmathbb{R}^{ntimes{n}}$ is the diagonal matrix with diagonal entries $x_1,x_2,dots,x_n$, and $|cdot|$ is the spectral norm (matrix 2-norm).
Since the spectral norm of a matrix is its largest singular value, and the singular values of a (square) diagonal matrix are the absolute values of the diagonal entries, we see that $f(x)=|x|_infty$, where $|cdot|_infty$ is the (vector) sup-norm. In this form, it is easier to deduce the properties of $f$--in particular, it is differentiable at any point $xinmathbb{R}^n$ where the largest element of $x$ (in absolute value) is unique. At such a point, the gradient of $f$ is given by
$$
nabla{f(x)}=text{sgn}(x_k)e_k
$$
where $k$ is the index of the (unique) largest entry of $x$ (in absolute value), $e_k$ is the $k^text{th}$ standard basis vector in $mathbb{R}^n$, and $text{sgn}(cdot)$ is the sign function.
I want to deduce the above expression for the gradient using the chain rule applied to $f(x)=(gcirc{h})(x)$, where $g:mathbb{R}^{ntimes{n}}tomathbb{R}$ is given by $g(A)=|A|$, and $h:mathbb{R}^ntomathbb{R}^{ntimes{n}}$ is given by $h(x)=text{diag}(x)$.
The "Jacobian" of $h$ is a three-dimensional object, where
$$
frac{partial[h(x)]_{ij}}{partial{x_k}}=begin{cases}1,&i=j=k,\0,&text{else.}end{cases}
$$
The function $g$ is differentiable at any point $A$ where $A$ has a unique largest singular value, in which case the gradient(?) is given by
$$
nabla{g(A)}=uv^text{T},
$$
where $u$ and $v$ are the left and right singular vectors (respectively) corresponding to the (unique) largest singular value of $A$.
So I essentially have a three-dimensional object and a 2-dimensional object, and I want to apply the chain rule to get the gradient, a 1-dimensional object (i.e. a vector). A straightforward application suggests "multiplying them together" (not sure that concept is even defined), which seems like it would produce a matrix. What simple thing am I missing here?
derivatives matrix-calculus chain-rule
asked Jan 22 at 1:23
David M.David M.
1,856418
$endgroup$
add a comment |
$begingroup$
Consider the function $f:mathbb{R}^ntomathbb{R}$ given by $f(x)=|text{diag}(x)|$, where $text{diag}(x)inmathbb{R}^{ntimes{n}}$ is the diagonal matrix with diagonal entries $x_1,x_2,dots,x_n$, and $|cdot|$ is the spectral norm (matrix 2-norm).
Since the spectral norm of a matrix is its largest singular value, and the singular values of a (square) diagonal matrix are the absolute values of the diagonal entries, we see that $f(x)=|x|_infty$, where $|cdot|_infty$ is the (vector) sup-norm. In this form, it is easier to deduce the properties of $f$--in particular, it is differentiable at any point $xinmathbb{R}^n$ where the largest element of $x$ (in absolute value) is unique. At such a point, the gradient of $f$ is given by
$$
nabla{f(x)}=text{sgn}(x_k)e_k
$$
where $k$ is the index of the (unique) largest entry of $x$ (in absolute value), $e_k$ is the $k^text{th}$ standard basis vector in $mathbb{R}^n$, and $text{sgn}(cdot)$ is the sign function.
I want to deduce the above expression for the gradient using the chain rule applied to $f(x)=(gcirc{h})(x)$, where $g:mathbb{R}^{ntimes{n}}tomathbb{R}$ is given by $g(A)=|A|$, and $h:mathbb{R}^ntomathbb{R}^{ntimes{n}}$ is given by $h(x)=text{diag}(x)$.
The "Jacobian" of $h$ is a three-dimensional object, where
$$
frac{partial[h(x)]_{ij}}{partial{x_k}}=begin{cases}1,&i=j=k,\0,&text{else.}end{cases}
$$
The function $g$ is differentiable at any point $A$ where $A$ has a unique largest singular value, in which case the gradient(?) is given by
$$
nabla{g(A)}=uv^text{T},
$$
where $u$ and $v$ are the left and right singular vectors (respectively) corresponding to the (unique) largest singular value of $A$.
So I essentially have a three-dimensional object and a 2-dimensional object, and I want to apply the chain rule to get the gradient, a 1-dimensional object (i.e. a vector). A straightforward application suggests "multiplying them together" (not sure that concept is even defined), which seems like it would produce a matrix. What simple thing am I missing here?
derivatives matrix-calculus chain-rule
asked Jan 22 at 1:23
David M.David M.
1,856418
$endgroup$
Consider the function $f:mathbb{R}^ntomathbb{R}$ given by $f(x)=|text{diag}(x)|$, where $text{diag}(x)inmathbb{R}^{ntimes{n}}$ is the diagonal matrix with diagonal entries $x_1,x_2,dots,x_n$, and $|cdot|$ is the spectral norm (matrix 2-norm).
Since the spectral norm of a matrix is its largest singular value, and the singular values of a (square) diagonal matrix are the absolute values of the diagonal entries, we see that $f(x)=|x|_infty$, where $|cdot|_infty$ is the (vector) sup-norm. In this form, it is easier to deduce the properties of $f$--in particular, it is differentiable at any point $xinmathbb{R}^n$ where the largest element of $x$ (in absolute value) is unique. At such a point, the gradient of $f$ is given by
$$
nabla{f(x)}=text{sgn}(x_k)e_k
$$
where $k$ is the index of the (unique) largest entry of $x$ (in absolute value), $e_k$ is the $k^text{th}$ standard basis vector in $mathbb{R}^n$, and $text{sgn}(cdot)$ is the sign function.
I want to deduce the above expression for the gradient using the chain rule applied to $f(x)=(gcirc{h})(x)$, where $g:mathbb{R}^{ntimes{n}}tomathbb{R}$ is given by $g(A)=|A|$, and $h:mathbb{R}^ntomathbb{R}^{ntimes{n}}$ is given by $h(x)=text{diag}(x)$.
The "Jacobian" of $h$ is a three-dimensional object, where
$$
frac{partial[h(x)]_{ij}}{partial{x_k}}=begin{cases}1,&i=j=k,\0,&text{else.}end{cases}
$$
The function $g$ is differentiable at any point $A$ where $A$ has a unique largest singular value, in which case the gradient(?) is given by
$$
nabla{g(A)}=uv^text{T},
$$
where $u$ and $v$ are the left and right singular vectors (respectively) corresponding to the (unique) largest singular value of $A$.
So I essentially have a three-dimensional object and a 2-dimensional object, and I want to apply the chain rule to get the gradient, a 1-dimensional object (i.e. a vector). A straightforward application suggests "multiplying them together" (not sure that concept is even defined), which seems like it would produce a matrix. What simple thing am I missing here?
derivatives matrix-calculus chain-rule
derivatives matrix-calculus chain-rule
asked Jan 22 at 1:23
David M.David M.
1,856418
asked Jan 22 at 1:23
David M.David M.
1,856418
asked Jan 22 at 1:23
David M.David M.
1,856418
asked Jan 22 at 1:23
David M.David M.
1,856418
asked Jan 22 at 1:23
David M.David M.
1,856418
1,856418
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A hint rather than a complete answer
It might help to think of $Bbb R^{n times n}$ as $Bbb R^{n^2}$.
Now $h$ is a function from $Bbb R^n$ to $Bbb R^{n^2}$ and $g$ is a function from
$Bbb R^{n^2}$ to $Bbb R$, so the composite goes from $Bbb R^n$ to $Bbb R$. Can you work out the $k$th partial derivative of this?
When you do, you may find out that the result you get is surprisingly closely related to matrix multiplication (of some sort) of the derivatives of $h$ and $g$...or you may not.
Just to get you started, $$nabla g(A)_{ni + j} = u_i v_j,$$
where I'm working with indices that go from $0$ to $n-1$ here.
Can you now work out the $ni + j$th entry of the $k$th partial derivative of $g circ h$?
answered Jan 22 at 1:40
John HughesJohn Hughes
64.4k24191
$endgroup$
$begingroup$
Thought I might have to resort to this--was being lazy and trying to avoid "index counting" from stacking matrices into vectors. Indeed this works!
$endgroup$
– David M.
Jan 22 at 18:53
$begingroup$
And when you were done, did you discover that there was a matrix multiplication hidden in there? (I'm asking seriously...I didn't bother doing it myself, and wonder what the result was...)
$endgroup$
– John Hughes
Jan 22 at 19:24
$begingroup$
Yes it works out following the conventions of the usual multi-dimensional chain rule. The resulting gradient vector is $J_g^text{T}{J_h}$, where $J_ginmathbb{R}^{1times{n^2}}$ is the outer product of $u$ and $v$ stacked columnwise, and $J_hinmathbb{R}^{n^2times{n}}$, where each row of $J_h$ is the gradient of a component of the matrix $text{diag}(x)$ stacked columnwise (lots of words, but basically what you would expect)
$endgroup$
– David M.
Jan 22 at 19:27
add a comment |
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1 Answer
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$begingroup$
A hint rather than a complete answer
It might help to think of $Bbb R^{n times n}$ as $Bbb R^{n^2}$.
Now $h$ is a function from $Bbb R^n$ to $Bbb R^{n^2}$ and $g$ is a function from
$Bbb R^{n^2}$ to $Bbb R$, so the composite goes from $Bbb R^n$ to $Bbb R$. Can you work out the $k$th partial derivative of this?
When you do, you may find out that the result you get is surprisingly closely related to matrix multiplication (of some sort) of the derivatives of $h$ and $g$...or you may not.
Just to get you started, $$nabla g(A)_{ni + j} = u_i v_j,$$
where I'm working with indices that go from $0$ to $n-1$ here.
Can you now work out the $ni + j$th entry of the $k$th partial derivative of $g circ h$?
answered Jan 22 at 1:40
John HughesJohn Hughes
64.4k24191
$endgroup$
$begingroup$
Thought I might have to resort to this--was being lazy and trying to avoid "index counting" from stacking matrices into vectors. Indeed this works!
$endgroup$
– David M.
Jan 22 at 18:53
$begingroup$
And when you were done, did you discover that there was a matrix multiplication hidden in there? (I'm asking seriously...I didn't bother doing it myself, and wonder what the result was...)
$endgroup$
– John Hughes
Jan 22 at 19:24
$begingroup$
Yes it works out following the conventions of the usual multi-dimensional chain rule. The resulting gradient vector is $J_g^text{T}{J_h}$, where $J_ginmathbb{R}^{1times{n^2}}$ is the outer product of $u$ and $v$ stacked columnwise, and $J_hinmathbb{R}^{n^2times{n}}$, where each row of $J_h$ is the gradient of a component of the matrix $text{diag}(x)$ stacked columnwise (lots of words, but basically what you would expect)
$endgroup$
– David M.
Jan 22 at 19:27
add a comment |
$begingroup$
A hint rather than a complete answer
It might help to think of $Bbb R^{n times n}$ as $Bbb R^{n^2}$.
Now $h$ is a function from $Bbb R^n$ to $Bbb R^{n^2}$ and $g$ is a function from
$Bbb R^{n^2}$ to $Bbb R$, so the composite goes from $Bbb R^n$ to $Bbb R$. Can you work out the $k$th partial derivative of this?
When you do, you may find out that the result you get is surprisingly closely related to matrix multiplication (of some sort) of the derivatives of $h$ and $g$...or you may not.
Just to get you started, $$nabla g(A)_{ni + j} = u_i v_j,$$
where I'm working with indices that go from $0$ to $n-1$ here.
Can you now work out the $ni + j$th entry of the $k$th partial derivative of $g circ h$?
answered Jan 22 at 1:40
John HughesJohn Hughes
64.4k24191
$endgroup$
$begingroup$
Thought I might have to resort to this--was being lazy and trying to avoid "index counting" from stacking matrices into vectors. Indeed this works!
$endgroup$
– David M.
Jan 22 at 18:53
$begingroup$
And when you were done, did you discover that there was a matrix multiplication hidden in there? (I'm asking seriously...I didn't bother doing it myself, and wonder what the result was...)
$endgroup$
– John Hughes
Jan 22 at 19:24
$begingroup$
Yes it works out following the conventions of the usual multi-dimensional chain rule. The resulting gradient vector is $J_g^text{T}{J_h}$, where $J_ginmathbb{R}^{1times{n^2}}$ is the outer product of $u$ and $v$ stacked columnwise, and $J_hinmathbb{R}^{n^2times{n}}$, where each row of $J_h$ is the gradient of a component of the matrix $text{diag}(x)$ stacked columnwise (lots of words, but basically what you would expect)
$endgroup$
– David M.
Jan 22 at 19:27
add a comment |
$begingroup$
A hint rather than a complete answer
It might help to think of $Bbb R^{n times n}$ as $Bbb R^{n^2}$.
Now $h$ is a function from $Bbb R^n$ to $Bbb R^{n^2}$ and $g$ is a function from
$Bbb R^{n^2}$ to $Bbb R$, so the composite goes from $Bbb R^n$ to $Bbb R$. Can you work out the $k$th partial derivative of this?
When you do, you may find out that the result you get is surprisingly closely related to matrix multiplication (of some sort) of the derivatives of $h$ and $g$...or you may not.
Just to get you started, $$nabla g(A)_{ni + j} = u_i v_j,$$
where I'm working with indices that go from $0$ to $n-1$ here.
Can you now work out the $ni + j$th entry of the $k$th partial derivative of $g circ h$?
answered Jan 22 at 1:40
John HughesJohn Hughes
64.4k24191
$endgroup$
A hint rather than a complete answer
It might help to think of $Bbb R^{n times n}$ as $Bbb R^{n^2}$.
Now $h$ is a function from $Bbb R^n$ to $Bbb R^{n^2}$ and $g$ is a function from
$Bbb R^{n^2}$ to $Bbb R$, so the composite goes from $Bbb R^n$ to $Bbb R$. Can you work out the $k$th partial derivative of this?
When you do, you may find out that the result you get is surprisingly closely related to matrix multiplication (of some sort) of the derivatives of $h$ and $g$...or you may not.
Just to get you started, $$nabla g(A)_{ni + j} = u_i v_j,$$
where I'm working with indices that go from $0$ to $n-1$ here.
Can you now work out the $ni + j$th entry of the $k$th partial derivative of $g circ h$?
answered Jan 22 at 1:40
John HughesJohn Hughes
64.4k24191
answered Jan 22 at 1:40
John HughesJohn Hughes
64.4k24191
answered Jan 22 at 1:40
John HughesJohn Hughes
64.4k24191
answered Jan 22 at 1:40
John HughesJohn Hughes
64.4k24191
64.4k24191
$begingroup$
Thought I might have to resort to this--was being lazy and trying to avoid "index counting" from stacking matrices into vectors. Indeed this works!
$endgroup$
– David M.
Jan 22 at 18:53
$begingroup$
And when you were done, did you discover that there was a matrix multiplication hidden in there? (I'm asking seriously...I didn't bother doing it myself, and wonder what the result was...)
$endgroup$
– John Hughes
Jan 22 at 19:24
$begingroup$
Yes it works out following the conventions of the usual multi-dimensional chain rule. The resulting gradient vector is $J_g^text{T}{J_h}$, where $J_ginmathbb{R}^{1times{n^2}}$ is the outer product of $u$ and $v$ stacked columnwise, and $J_hinmathbb{R}^{n^2times{n}}$, where each row of $J_h$ is the gradient of a component of the matrix $text{diag}(x)$ stacked columnwise (lots of words, but basically what you would expect)
$endgroup$
– David M.
Jan 22 at 19:27
add a comment |
$begingroup$
Thought I might have to resort to this--was being lazy and trying to avoid "index counting" from stacking matrices into vectors. Indeed this works!
$endgroup$
– David M.
Jan 22 at 18:53
$begingroup$
And when you were done, did you discover that there was a matrix multiplication hidden in there? (I'm asking seriously...I didn't bother doing it myself, and wonder what the result was...)
$endgroup$
– John Hughes
Jan 22 at 19:24
$begingroup$
Yes it works out following the conventions of the usual multi-dimensional chain rule. The resulting gradient vector is $J_g^text{T}{J_h}$, where $J_ginmathbb{R}^{1times{n^2}}$ is the outer product of $u$ and $v$ stacked columnwise, and $J_hinmathbb{R}^{n^2times{n}}$, where each row of $J_h$ is the gradient of a component of the matrix $text{diag}(x)$ stacked columnwise (lots of words, but basically what you would expect)
$endgroup$
– David M.
Jan 22 at 19:27
$begingroup$
Thought I might have to resort to this--was being lazy and trying to avoid "index counting" from stacking matrices into vectors. Indeed this works!
$endgroup$
– David M.
Jan 22 at 18:53
$begingroup$
Thought I might have to resort to this--was being lazy and trying to avoid "index counting" from stacking matrices into vectors. Indeed this works!
$endgroup$
– David M.
Jan 22 at 18:53
$begingroup$
And when you were done, did you discover that there was a matrix multiplication hidden in there? (I'm asking seriously...I didn't bother doing it myself, and wonder what the result was...)
$endgroup$
– John Hughes
Jan 22 at 19:24
$begingroup$
And when you were done, did you discover that there was a matrix multiplication hidden in there? (I'm asking seriously...I didn't bother doing it myself, and wonder what the result was...)
$endgroup$
– John Hughes
Jan 22 at 19:24
$begingroup$
Yes it works out following the conventions of the usual multi-dimensional chain rule. The resulting gradient vector is $J_g^text{T}{J_h}$, where $J_ginmathbb{R}^{1times{n^2}}$ is the outer product of $u$ and $v$ stacked columnwise, and $J_hinmathbb{R}^{n^2times{n}}$, where each row of $J_h$ is the gradient of a component of the matrix $text{diag}(x)$ stacked columnwise (lots of words, but basically what you would expect)
$endgroup$
– David M.
Jan 22 at 19:27
$begingroup$
Yes it works out following the conventions of the usual multi-dimensional chain rule. The resulting gradient vector is $J_g^text{T}{J_h}$, where $J_ginmathbb{R}^{1times{n^2}}$ is the outer product of $u$ and $v$ stacked columnwise, and $J_hinmathbb{R}^{n^2times{n}}$, where each row of $J_h$ is the gradient of a component of the matrix $text{diag}(x)$ stacked columnwise (lots of words, but basically what you would expect)
$endgroup$
– David M.
Jan 22 at 19:27
add a comment |
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