Derivative of a Function of the Diag function
$begingroup$
Suppose there is a vector $U in mathbb{R}^n$. How would you find the derivative of:
$$
F(U)=traceleft(diag(U) A diag(U) right)
$$
where $A in mathbb{R}^{n times n} succ 0 $ and where $diag(cdot)$ creates a diagonal matrix with $(cdot)$ on the leading diagonal. Where the derivative is taken with respect to the vector $U$, i.e.
$$
{partial F(U) over partial U } \
$$
I am more interested in the method used. Thanks in advance.
derivatives matrix-calculus trace
asked Dec 7 '18 at 9:45
p32fr4p32fr4
3717
$endgroup$
add a comment |
$begingroup$
Suppose there is a vector $U in mathbb{R}^n$. How would you find the derivative of:
$$
F(U)=traceleft(diag(U) A diag(U) right)
$$
where $A in mathbb{R}^{n times n} succ 0 $ and where $diag(cdot)$ creates a diagonal matrix with $(cdot)$ on the leading diagonal. Where the derivative is taken with respect to the vector $U$, i.e.
$$
{partial F(U) over partial U } \
$$
I am more interested in the method used. Thanks in advance.
derivatives matrix-calculus trace
asked Dec 7 '18 at 9:45
p32fr4p32fr4
3717
$endgroup$
add a comment |
$begingroup$
Suppose there is a vector $U in mathbb{R}^n$. How would you find the derivative of:
$$
F(U)=traceleft(diag(U) A diag(U) right)
$$
where $A in mathbb{R}^{n times n} succ 0 $ and where $diag(cdot)$ creates a diagonal matrix with $(cdot)$ on the leading diagonal. Where the derivative is taken with respect to the vector $U$, i.e.
$$
{partial F(U) over partial U } \
$$
I am more interested in the method used. Thanks in advance.
derivatives matrix-calculus trace
asked Dec 7 '18 at 9:45
p32fr4p32fr4
3717
$endgroup$
Suppose there is a vector $U in mathbb{R}^n$. How would you find the derivative of:
$$
F(U)=traceleft(diag(U) A diag(U) right)
$$
where $A in mathbb{R}^{n times n} succ 0 $ and where $diag(cdot)$ creates a diagonal matrix with $(cdot)$ on the leading diagonal. Where the derivative is taken with respect to the vector $U$, i.e.
$$
{partial F(U) over partial U } \
$$
I am more interested in the method used. Thanks in advance.
derivatives matrix-calculus trace
derivatives matrix-calculus trace
asked Dec 7 '18 at 9:45
p32fr4p32fr4
3717
asked Dec 7 '18 at 9:45
p32fr4p32fr4
3717
asked Dec 7 '18 at 9:45
p32fr4p32fr4
3717
asked Dec 7 '18 at 9:45
p32fr4p32fr4
3717
asked Dec 7 '18 at 9:45
p32fr4p32fr4
3717
3717
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Posting the solution I identified.
Due to the trace operator evaluating the above is equivalent to evaluating:
$$
{partial left(sumlimits_{i=1}^{n} u_i A_{(i,i)}u_iright)over partial U }=
left( begin{align}
begin{array}
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}\
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}&\
vdots &\
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_n}&\
end{array}end{align}right) \
$$
which becomes:
$$
{partial left( trace(diag(U) A diag(U)) right)over partial U} = 2 diag( A) U
$$
answered Dec 7 '18 at 10:47
p32fr4p32fr4
3717
$endgroup$
2
$begingroup$
You can also derive this using two identities involving the Hadamard product $${rm Diag}(x),A,{rm Diag}(y)=Aodot xy^T$$ $${rm Tr}(Aodot xy^T)=x^T(Iodot A)y=phi$$ So that $$dphi=y^T(Iodot A)dx+x^T(Iodot A)dy$$ Setting $y=x=u$ this becomes $$dphi=2u^T(Iodot A)du$$ $$frac{partialphi}{partial u}=2(Iodot A)u$$
$endgroup$
– greg
Dec 7 '18 at 23:39
$begingroup$
That is a nice alternative thanks @greg
$endgroup$
– p32fr4
Dec 21 '18 at 10:03
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Posting the solution I identified.
Due to the trace operator evaluating the above is equivalent to evaluating:
$$
{partial left(sumlimits_{i=1}^{n} u_i A_{(i,i)}u_iright)over partial U }=
left( begin{align}
begin{array}
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}\
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}&\
vdots &\
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_n}&\
end{array}end{align}right) \
$$
which becomes:
$$
{partial left( trace(diag(U) A diag(U)) right)over partial U} = 2 diag( A) U
$$
answered Dec 7 '18 at 10:47
p32fr4p32fr4
3717
$endgroup$
2
$begingroup$
You can also derive this using two identities involving the Hadamard product $${rm Diag}(x),A,{rm Diag}(y)=Aodot xy^T$$ $${rm Tr}(Aodot xy^T)=x^T(Iodot A)y=phi$$ So that $$dphi=y^T(Iodot A)dx+x^T(Iodot A)dy$$ Setting $y=x=u$ this becomes $$dphi=2u^T(Iodot A)du$$ $$frac{partialphi}{partial u}=2(Iodot A)u$$
$endgroup$
– greg
Dec 7 '18 at 23:39
$begingroup$
That is a nice alternative thanks @greg
$endgroup$
– p32fr4
Dec 21 '18 at 10:03
add a comment |
$begingroup$
Posting the solution I identified.
Due to the trace operator evaluating the above is equivalent to evaluating:
$$
{partial left(sumlimits_{i=1}^{n} u_i A_{(i,i)}u_iright)over partial U }=
left( begin{align}
begin{array}
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}\
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}&\
vdots &\
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_n}&\
end{array}end{align}right) \
$$
which becomes:
$$
{partial left( trace(diag(U) A diag(U)) right)over partial U} = 2 diag( A) U
$$
answered Dec 7 '18 at 10:47
p32fr4p32fr4
3717
$endgroup$
2
$begingroup$
You can also derive this using two identities involving the Hadamard product $${rm Diag}(x),A,{rm Diag}(y)=Aodot xy^T$$ $${rm Tr}(Aodot xy^T)=x^T(Iodot A)y=phi$$ So that $$dphi=y^T(Iodot A)dx+x^T(Iodot A)dy$$ Setting $y=x=u$ this becomes $$dphi=2u^T(Iodot A)du$$ $$frac{partialphi}{partial u}=2(Iodot A)u$$
$endgroup$
– greg
Dec 7 '18 at 23:39
$begingroup$
That is a nice alternative thanks @greg
$endgroup$
– p32fr4
Dec 21 '18 at 10:03
add a comment |
$begingroup$
Posting the solution I identified.
Due to the trace operator evaluating the above is equivalent to evaluating:
$$
{partial left(sumlimits_{i=1}^{n} u_i A_{(i,i)}u_iright)over partial U }=
left( begin{align}
begin{array}
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}\
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}&\
vdots &\
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_n}&\
end{array}end{align}right) \
$$
which becomes:
$$
{partial left( trace(diag(U) A diag(U)) right)over partial U} = 2 diag( A) U
$$
answered Dec 7 '18 at 10:47
p32fr4p32fr4
3717
$endgroup$
Posting the solution I identified.
Due to the trace operator evaluating the above is equivalent to evaluating:
$$
{partial left(sumlimits_{i=1}^{n} u_i A_{(i,i)}u_iright)over partial U }=
left( begin{align}
begin{array}
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}\
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}&\
vdots &\
{partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_n}&\
end{array}end{align}right) \
$$
which becomes:
$$
{partial left( trace(diag(U) A diag(U)) right)over partial U} = 2 diag( A) U
$$
answered Dec 7 '18 at 10:47
p32fr4p32fr4
3717
answered Dec 7 '18 at 10:47
p32fr4p32fr4
3717
answered Dec 7 '18 at 10:47
p32fr4p32fr4
3717
answered Dec 7 '18 at 10:47
p32fr4p32fr4
3717
3717
2
$begingroup$
You can also derive this using two identities involving the Hadamard product $${rm Diag}(x),A,{rm Diag}(y)=Aodot xy^T$$ $${rm Tr}(Aodot xy^T)=x^T(Iodot A)y=phi$$ So that $$dphi=y^T(Iodot A)dx+x^T(Iodot A)dy$$ Setting $y=x=u$ this becomes $$dphi=2u^T(Iodot A)du$$ $$frac{partialphi}{partial u}=2(Iodot A)u$$
$endgroup$
– greg
Dec 7 '18 at 23:39
$begingroup$
That is a nice alternative thanks @greg
$endgroup$
– p32fr4
Dec 21 '18 at 10:03
add a comment |
2
$begingroup$
You can also derive this using two identities involving the Hadamard product $${rm Diag}(x),A,{rm Diag}(y)=Aodot xy^T$$ $${rm Tr}(Aodot xy^T)=x^T(Iodot A)y=phi$$ So that $$dphi=y^T(Iodot A)dx+x^T(Iodot A)dy$$ Setting $y=x=u$ this becomes $$dphi=2u^T(Iodot A)du$$ $$frac{partialphi}{partial u}=2(Iodot A)u$$
$endgroup$
– greg
Dec 7 '18 at 23:39
$begingroup$
That is a nice alternative thanks @greg
$endgroup$
– p32fr4
Dec 21 '18 at 10:03
2
2
$begingroup$
You can also derive this using two identities involving the Hadamard product $${rm Diag}(x),A,{rm Diag}(y)=Aodot xy^T$$ $${rm Tr}(Aodot xy^T)=x^T(Iodot A)y=phi$$ So that $$dphi=y^T(Iodot A)dx+x^T(Iodot A)dy$$ Setting $y=x=u$ this becomes $$dphi=2u^T(Iodot A)du$$ $$frac{partialphi}{partial u}=2(Iodot A)u$$
$endgroup$
– greg
Dec 7 '18 at 23:39
$begingroup$
You can also derive this using two identities involving the Hadamard product $${rm Diag}(x),A,{rm Diag}(y)=Aodot xy^T$$ $${rm Tr}(Aodot xy^T)=x^T(Iodot A)y=phi$$ So that $$dphi=y^T(Iodot A)dx+x^T(Iodot A)dy$$ Setting $y=x=u$ this becomes $$dphi=2u^T(Iodot A)du$$ $$frac{partialphi}{partial u}=2(Iodot A)u$$
$endgroup$
– greg
Dec 7 '18 at 23:39
$begingroup$
That is a nice alternative thanks @greg
$endgroup$
– p32fr4
Dec 21 '18 at 10:03
$begingroup$
That is a nice alternative thanks @greg
$endgroup$
– p32fr4
Dec 21 '18 at 10:03
add a comment |
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