Second derivative of norm of matrix power $lVert y - A^kx rVert$
$begingroup$
I am looking for the second derivative w.r.t $A$ (and $x$ and $y$, but those are a little easier):
$$
frac{partial^2}{partial A^2} frac{1}{2} lVert y - A^kx rVert_2^2
$$
I can find the first derivative with www.matrixcalculus.org:
$$
frac{partial}{partial A} frac{1}{2} lVert y - A^kx rVert_2^2 = -k cdot diag(y - A^kx)A^{k-1}cdot diag(x)
$$
But handling the "diag" operator wasn't clear. I am attempting/hoping to show that the Lipschitz constant used for gradient descent depends on $k$.
matrices matrix-calculus
asked Dec 11 '18 at 1:03
RobertRobert
11311
$endgroup$
add a comment |
$begingroup$
I am looking for the second derivative w.r.t $A$ (and $x$ and $y$, but those are a little easier):
$$
frac{partial^2}{partial A^2} frac{1}{2} lVert y - A^kx rVert_2^2
$$
I can find the first derivative with www.matrixcalculus.org:
$$
frac{partial}{partial A} frac{1}{2} lVert y - A^kx rVert_2^2 = -k cdot diag(y - A^kx)A^{k-1}cdot diag(x)
$$
But handling the "diag" operator wasn't clear. I am attempting/hoping to show that the Lipschitz constant used for gradient descent depends on $k$.
matrices matrix-calculus
asked Dec 11 '18 at 1:03
RobertRobert
11311
$endgroup$
$begingroup$
what does $k$ represent?
$endgroup$
– user550103
Dec 11 '18 at 11:44
$begingroup$
@user550103 - it's a scalar integer. In my particular use case, it represents "time" over which the discrete linear system described by $A$ is rolled forward. I should have put that in the question, sorry about that.
$endgroup$
– Robert
Dec 11 '18 at 18:38
add a comment |
$begingroup$
I am looking for the second derivative w.r.t $A$ (and $x$ and $y$, but those are a little easier):
$$
frac{partial^2}{partial A^2} frac{1}{2} lVert y - A^kx rVert_2^2
$$
I can find the first derivative with www.matrixcalculus.org:
$$
frac{partial}{partial A} frac{1}{2} lVert y - A^kx rVert_2^2 = -k cdot diag(y - A^kx)A^{k-1}cdot diag(x)
$$
But handling the "diag" operator wasn't clear. I am attempting/hoping to show that the Lipschitz constant used for gradient descent depends on $k$.
matrices matrix-calculus
asked Dec 11 '18 at 1:03
RobertRobert
11311
$endgroup$
I am looking for the second derivative w.r.t $A$ (and $x$ and $y$, but those are a little easier):
$$
frac{partial^2}{partial A^2} frac{1}{2} lVert y - A^kx rVert_2^2
$$
I can find the first derivative with www.matrixcalculus.org:
$$
frac{partial}{partial A} frac{1}{2} lVert y - A^kx rVert_2^2 = -k cdot diag(y - A^kx)A^{k-1}cdot diag(x)
$$
But handling the "diag" operator wasn't clear. I am attempting/hoping to show that the Lipschitz constant used for gradient descent depends on $k$.
matrices matrix-calculus
matrices matrix-calculus
asked Dec 11 '18 at 1:03
RobertRobert
11311
asked Dec 11 '18 at 1:03
RobertRobert
11311
asked Dec 11 '18 at 1:03
RobertRobert
11311
asked Dec 11 '18 at 1:03
RobertRobert
11311
asked Dec 11 '18 at 1:03
RobertRobert
11311
11311
$begingroup$
what does $k$ represent?
$endgroup$
– user550103
Dec 11 '18 at 11:44
$begingroup$
@user550103 - it's a scalar integer. In my particular use case, it represents "time" over which the discrete linear system described by $A$ is rolled forward. I should have put that in the question, sorry about that.
$endgroup$
– Robert
Dec 11 '18 at 18:38
add a comment |
$begingroup$
what does $k$ represent?
$endgroup$
– user550103
Dec 11 '18 at 11:44
$begingroup$
@user550103 - it's a scalar integer. In my particular use case, it represents "time" over which the discrete linear system described by $A$ is rolled forward. I should have put that in the question, sorry about that.
$endgroup$
– Robert
Dec 11 '18 at 18:38
$begingroup$
what does $k$ represent?
$endgroup$
– user550103
Dec 11 '18 at 11:44
$begingroup$
what does $k$ represent?
$endgroup$
– user550103
Dec 11 '18 at 11:44
$begingroup$
@user550103 - it's a scalar integer. In my particular use case, it represents "time" over which the discrete linear system described by $A$ is rolled forward. I should have put that in the question, sorry about that.
$endgroup$
– Robert
Dec 11 '18 at 18:38
$begingroup$
@user550103 - it's a scalar integer. In my particular use case, it represents "time" over which the discrete linear system described by $A$ is rolled forward. I should have put that in the question, sorry about that.
$endgroup$
– Robert
Dec 11 '18 at 18:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The tricky part is the differential
$$eqalign{
dA^k &= sum_{j=0}^{k-1} A^{j},dA,A^{k-j-1}cr
}$$
Define $w = (A^kx-y)$ and $B=A^T,,$ then find the gradient (first derivative)
$$eqalign{
phi &= tfrac{1}{2}|w|^2_2 = w:w cr
dphi &= 2w:dw = 2w:dA^kx = 2wx^T:dA^k cr
&= sum_{j=0}^{k-1} 2wx^T:A^{j},dA,A^{k-j-1} cr
&= sum_{j=0}^{k-1} 2B^{j}wx^TB^{k-j-1}:dA cr
G=frac{partialphi}{partial A} &= sum_{j=0}^{k-1} 2B^{j}wx^TB^{k-j-1} cr
}$$ where a colon denotes the trace/Frobenius product, i.e. $M:P={rm Tr}(M^TP)$
Note that the gradient is a matrix. That means that the Hessian (second derivative) will be a fourth-order tensor -- which is the reason why the website won't try to calculate it.
In order to calculate the Hessian, the $w$ and $B^i=(A^T)^i$ terms must be differentiated leading to a complicated double summation. Start by finding the differential of the gradient.
$$eqalign{
dG
&= 2sum_{j=0}^{k-1} &dB^{j}wx^TB^{k-j-1}+B^{j}dA^kxx^TB^{k-j-1}+B^{j}wx^TdB^{k-j-1} cr
&= 2sum_{j=0}^{k-1}Bigg(&sum_{i=0}^{j-1}B^{i},dB,B^{j-i-1}wx^TB^{k-j-1}
+ sum_{i=0}^{k-1}B^{j}A^{i},dA,A^{k-i-1}xx^TB^{k-j-1} cr
&,&+ sum_{i=0}^{k-j-2}B^{j}wx^TB^{i},dB,B^{j-i-1}
Bigg) cr
}$$
At this point you can use vectorization to flatten the matrices into vectors and calculate the Hessian as the matrix, i.e.
$$g={rm vec}(G),,,,a={rm vec}(A) implies H=frac{partial g}{partial a}$$
Or you can find the Hessian in tensor form using index notation.
Or perhaps the differential expression is sufficient for whatever purpose you have in mind.
answered Dec 11 '18 at 13:50
greggreg
8,5051823
$endgroup$
$begingroup$
This is great! Thank you! I will go through this and try to understand it more deeply. I bet I can formulate the inequality for the Lipschitz constant in terms of the differential expression: $lVert nabla phi(w_0) - nabla phi(w_1) rVert_2 leq L lVert w_0 - w_1 rVert_2$
$endgroup$
– Robert
Dec 11 '18 at 18:45
add a comment |
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1 Answer
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$begingroup$
The tricky part is the differential
$$eqalign{
dA^k &= sum_{j=0}^{k-1} A^{j},dA,A^{k-j-1}cr
}$$
Define $w = (A^kx-y)$ and $B=A^T,,$ then find the gradient (first derivative)
$$eqalign{
phi &= tfrac{1}{2}|w|^2_2 = w:w cr
dphi &= 2w:dw = 2w:dA^kx = 2wx^T:dA^k cr
&= sum_{j=0}^{k-1} 2wx^T:A^{j},dA,A^{k-j-1} cr
&= sum_{j=0}^{k-1} 2B^{j}wx^TB^{k-j-1}:dA cr
G=frac{partialphi}{partial A} &= sum_{j=0}^{k-1} 2B^{j}wx^TB^{k-j-1} cr
}$$ where a colon denotes the trace/Frobenius product, i.e. $M:P={rm Tr}(M^TP)$
Note that the gradient is a matrix. That means that the Hessian (second derivative) will be a fourth-order tensor -- which is the reason why the website won't try to calculate it.
In order to calculate the Hessian, the $w$ and $B^i=(A^T)^i$ terms must be differentiated leading to a complicated double summation. Start by finding the differential of the gradient.
$$eqalign{
dG
&= 2sum_{j=0}^{k-1} &dB^{j}wx^TB^{k-j-1}+B^{j}dA^kxx^TB^{k-j-1}+B^{j}wx^TdB^{k-j-1} cr
&= 2sum_{j=0}^{k-1}Bigg(&sum_{i=0}^{j-1}B^{i},dB,B^{j-i-1}wx^TB^{k-j-1}
+ sum_{i=0}^{k-1}B^{j}A^{i},dA,A^{k-i-1}xx^TB^{k-j-1} cr
&,&+ sum_{i=0}^{k-j-2}B^{j}wx^TB^{i},dB,B^{j-i-1}
Bigg) cr
}$$
At this point you can use vectorization to flatten the matrices into vectors and calculate the Hessian as the matrix, i.e.
$$g={rm vec}(G),,,,a={rm vec}(A) implies H=frac{partial g}{partial a}$$
Or you can find the Hessian in tensor form using index notation.
Or perhaps the differential expression is sufficient for whatever purpose you have in mind.
answered Dec 11 '18 at 13:50
greggreg
8,5051823
$endgroup$
$begingroup$
This is great! Thank you! I will go through this and try to understand it more deeply. I bet I can formulate the inequality for the Lipschitz constant in terms of the differential expression: $lVert nabla phi(w_0) - nabla phi(w_1) rVert_2 leq L lVert w_0 - w_1 rVert_2$
$endgroup$
– Robert
Dec 11 '18 at 18:45
add a comment |
$begingroup$
The tricky part is the differential
$$eqalign{
dA^k &= sum_{j=0}^{k-1} A^{j},dA,A^{k-j-1}cr
}$$
Define $w = (A^kx-y)$ and $B=A^T,,$ then find the gradient (first derivative)
$$eqalign{
phi &= tfrac{1}{2}|w|^2_2 = w:w cr
dphi &= 2w:dw = 2w:dA^kx = 2wx^T:dA^k cr
&= sum_{j=0}^{k-1} 2wx^T:A^{j},dA,A^{k-j-1} cr
&= sum_{j=0}^{k-1} 2B^{j}wx^TB^{k-j-1}:dA cr
G=frac{partialphi}{partial A} &= sum_{j=0}^{k-1} 2B^{j}wx^TB^{k-j-1} cr
}$$ where a colon denotes the trace/Frobenius product, i.e. $M:P={rm Tr}(M^TP)$
Note that the gradient is a matrix. That means that the Hessian (second derivative) will be a fourth-order tensor -- which is the reason why the website won't try to calculate it.
In order to calculate the Hessian, the $w$ and $B^i=(A^T)^i$ terms must be differentiated leading to a complicated double summation. Start by finding the differential of the gradient.
$$eqalign{
dG
&= 2sum_{j=0}^{k-1} &dB^{j}wx^TB^{k-j-1}+B^{j}dA^kxx^TB^{k-j-1}+B^{j}wx^TdB^{k-j-1} cr
&= 2sum_{j=0}^{k-1}Bigg(&sum_{i=0}^{j-1}B^{i},dB,B^{j-i-1}wx^TB^{k-j-1}
+ sum_{i=0}^{k-1}B^{j}A^{i},dA,A^{k-i-1}xx^TB^{k-j-1} cr
&,&+ sum_{i=0}^{k-j-2}B^{j}wx^TB^{i},dB,B^{j-i-1}
Bigg) cr
}$$
At this point you can use vectorization to flatten the matrices into vectors and calculate the Hessian as the matrix, i.e.
$$g={rm vec}(G),,,,a={rm vec}(A) implies H=frac{partial g}{partial a}$$
Or you can find the Hessian in tensor form using index notation.
Or perhaps the differential expression is sufficient for whatever purpose you have in mind.
answered Dec 11 '18 at 13:50
greggreg
8,5051823
$endgroup$
$begingroup$
This is great! Thank you! I will go through this and try to understand it more deeply. I bet I can formulate the inequality for the Lipschitz constant in terms of the differential expression: $lVert nabla phi(w_0) - nabla phi(w_1) rVert_2 leq L lVert w_0 - w_1 rVert_2$
$endgroup$
– Robert
Dec 11 '18 at 18:45
add a comment |
$begingroup$
The tricky part is the differential
$$eqalign{
dA^k &= sum_{j=0}^{k-1} A^{j},dA,A^{k-j-1}cr
}$$
Define $w = (A^kx-y)$ and $B=A^T,,$ then find the gradient (first derivative)
$$eqalign{
phi &= tfrac{1}{2}|w|^2_2 = w:w cr
dphi &= 2w:dw = 2w:dA^kx = 2wx^T:dA^k cr
&= sum_{j=0}^{k-1} 2wx^T:A^{j},dA,A^{k-j-1} cr
&= sum_{j=0}^{k-1} 2B^{j}wx^TB^{k-j-1}:dA cr
G=frac{partialphi}{partial A} &= sum_{j=0}^{k-1} 2B^{j}wx^TB^{k-j-1} cr
}$$ where a colon denotes the trace/Frobenius product, i.e. $M:P={rm Tr}(M^TP)$
Note that the gradient is a matrix. That means that the Hessian (second derivative) will be a fourth-order tensor -- which is the reason why the website won't try to calculate it.
In order to calculate the Hessian, the $w$ and $B^i=(A^T)^i$ terms must be differentiated leading to a complicated double summation. Start by finding the differential of the gradient.
$$eqalign{
dG
&= 2sum_{j=0}^{k-1} &dB^{j}wx^TB^{k-j-1}+B^{j}dA^kxx^TB^{k-j-1}+B^{j}wx^TdB^{k-j-1} cr
&= 2sum_{j=0}^{k-1}Bigg(&sum_{i=0}^{j-1}B^{i},dB,B^{j-i-1}wx^TB^{k-j-1}
+ sum_{i=0}^{k-1}B^{j}A^{i},dA,A^{k-i-1}xx^TB^{k-j-1} cr
&,&+ sum_{i=0}^{k-j-2}B^{j}wx^TB^{i},dB,B^{j-i-1}
Bigg) cr
}$$
At this point you can use vectorization to flatten the matrices into vectors and calculate the Hessian as the matrix, i.e.
$$g={rm vec}(G),,,,a={rm vec}(A) implies H=frac{partial g}{partial a}$$
Or you can find the Hessian in tensor form using index notation.
Or perhaps the differential expression is sufficient for whatever purpose you have in mind.
answered Dec 11 '18 at 13:50
greggreg
8,5051823
$endgroup$
The tricky part is the differential
$$eqalign{
dA^k &= sum_{j=0}^{k-1} A^{j},dA,A^{k-j-1}cr
}$$
Define $w = (A^kx-y)$ and $B=A^T,,$ then find the gradient (first derivative)
$$eqalign{
phi &= tfrac{1}{2}|w|^2_2 = w:w cr
dphi &= 2w:dw = 2w:dA^kx = 2wx^T:dA^k cr
&= sum_{j=0}^{k-1} 2wx^T:A^{j},dA,A^{k-j-1} cr
&= sum_{j=0}^{k-1} 2B^{j}wx^TB^{k-j-1}:dA cr
G=frac{partialphi}{partial A} &= sum_{j=0}^{k-1} 2B^{j}wx^TB^{k-j-1} cr
}$$ where a colon denotes the trace/Frobenius product, i.e. $M:P={rm Tr}(M^TP)$
Note that the gradient is a matrix. That means that the Hessian (second derivative) will be a fourth-order tensor -- which is the reason why the website won't try to calculate it.
In order to calculate the Hessian, the $w$ and $B^i=(A^T)^i$ terms must be differentiated leading to a complicated double summation. Start by finding the differential of the gradient.
$$eqalign{
dG
&= 2sum_{j=0}^{k-1} &dB^{j}wx^TB^{k-j-1}+B^{j}dA^kxx^TB^{k-j-1}+B^{j}wx^TdB^{k-j-1} cr
&= 2sum_{j=0}^{k-1}Bigg(&sum_{i=0}^{j-1}B^{i},dB,B^{j-i-1}wx^TB^{k-j-1}
+ sum_{i=0}^{k-1}B^{j}A^{i},dA,A^{k-i-1}xx^TB^{k-j-1} cr
&,&+ sum_{i=0}^{k-j-2}B^{j}wx^TB^{i},dB,B^{j-i-1}
Bigg) cr
}$$
At this point you can use vectorization to flatten the matrices into vectors and calculate the Hessian as the matrix, i.e.
$$g={rm vec}(G),,,,a={rm vec}(A) implies H=frac{partial g}{partial a}$$
Or you can find the Hessian in tensor form using index notation.
Or perhaps the differential expression is sufficient for whatever purpose you have in mind.
answered Dec 11 '18 at 13:50
greggreg
8,5051823
edited Dec 11 '18 at 16:13
answered Dec 11 '18 at 13:50
greggreg
8,5051823
answered Dec 11 '18 at 13:50
greggreg
8,5051823
answered Dec 11 '18 at 13:50
greggreg
8,5051823
8,5051823
$begingroup$
This is great! Thank you! I will go through this and try to understand it more deeply. I bet I can formulate the inequality for the Lipschitz constant in terms of the differential expression: $lVert nabla phi(w_0) - nabla phi(w_1) rVert_2 leq L lVert w_0 - w_1 rVert_2$
$endgroup$
– Robert
Dec 11 '18 at 18:45
add a comment |
$begingroup$
This is great! Thank you! I will go through this and try to understand it more deeply. I bet I can formulate the inequality for the Lipschitz constant in terms of the differential expression: $lVert nabla phi(w_0) - nabla phi(w_1) rVert_2 leq L lVert w_0 - w_1 rVert_2$
$endgroup$
– Robert
Dec 11 '18 at 18:45
$begingroup$
This is great! Thank you! I will go through this and try to understand it more deeply. I bet I can formulate the inequality for the Lipschitz constant in terms of the differential expression: $lVert nabla phi(w_0) - nabla phi(w_1) rVert_2 leq L lVert w_0 - w_1 rVert_2$
$endgroup$
– Robert
Dec 11 '18 at 18:45
$begingroup$
This is great! Thank you! I will go through this and try to understand it more deeply. I bet I can formulate the inequality for the Lipschitz constant in terms of the differential expression: $lVert nabla phi(w_0) - nabla phi(w_1) rVert_2 leq L lVert w_0 - w_1 rVert_2$
$endgroup$
– Robert
Dec 11 '18 at 18:45
add a comment |
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$begingroup$
what does $k$ represent?
$endgroup$
– user550103
Dec 11 '18 at 11:44
$begingroup$
@user550103 - it's a scalar integer. In my particular use case, it represents "time" over which the discrete linear system described by $A$ is rolled forward. I should have put that in the question, sorry about that.
$endgroup$
– Robert
Dec 11 '18 at 18:38