Second derivative of a matrix quartic form












2














I need to compute the second derivative of the following quartic expression: $$x^H A^H x x^H A x$$ where is Hermitian. I have tried to compute the first derivative, and if I am not wrong, it should be: $$(A+A^H) x x^H (A+A^H) x$$
But then, I do not know how to proceed to calculate the second derivative. Could someone sketch the steps I need to follow? Thank you.






linear-algebra matrices derivatives optimization







share|cite|improve this question






















  • Did you compute the first derivative with respect to $x$ or $x^*$?
    – Alex Silva
    Nov 27 at 8:08


















2














I need to compute the second derivative of the following quartic expression: $$x^H A^H x x^H A x$$ where is Hermitian. I have tried to compute the first derivative, and if I am not wrong, it should be: $$(A+A^H) x x^H (A+A^H) x$$
But then, I do not know how to proceed to calculate the second derivative. Could someone sketch the steps I need to follow? Thank you.






linear-algebra matrices derivatives optimization







share|cite|improve this question






















  • Did you compute the first derivative with respect to $x$ or $x^*$?
    – Alex Silva
    Nov 27 at 8:08
















2












2








2







I need to compute the second derivative of the following quartic expression: $$x^H A^H x x^H A x$$ where is Hermitian. I have tried to compute the first derivative, and if I am not wrong, it should be: $$(A+A^H) x x^H (A+A^H) x$$
But then, I do not know how to proceed to calculate the second derivative. Could someone sketch the steps I need to follow? Thank you.






linear-algebra matrices derivatives optimization







share|cite|improve this question













I need to compute the second derivative of the following quartic expression: $$x^H A^H x x^H A x$$ where is Hermitian. I have tried to compute the first derivative, and if I am not wrong, it should be: $$(A+A^H) x x^H (A+A^H) x$$
But then, I do not know how to proceed to calculate the second derivative. Could someone sketch the steps I need to follow? Thank you.






linear-algebra matrices derivatives optimization




linear-algebra matrices derivatives optimization






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asked Nov 27 at 7:49









Fer Nando

164




164












  • Did you compute the first derivative with respect to $x$ or $x^*$?
    – Alex Silva
    Nov 27 at 8:08




















  • Did you compute the first derivative with respect to $x$ or $x^*$?
    – Alex Silva
    Nov 27 at 8:08


















Did you compute the first derivative with respect to $x$ or $x^*$?
– Alex Silva
Nov 27 at 8:08






Did you compute the first derivative with respect to $x$ or $x^*$?
– Alex Silva
Nov 27 at 8:08












2 Answers
2






active

oldest

votes


















3














Define the scalar variables
$$eqalign{
&phi &= x^HAx = (A^Tx^*)^Tx cr
&phi^* &= x^HA^Hx = (A^*x^*)^Tx cr
&psi &= phi^*phi cr
}$$
Find the gradient of your function $(psi)$ with respect to $x$, treating $x^*$ as an independent variable.
$$eqalign{
dphi &= (A^Tx^*)^T,dx cr
dphi^* &= (A^*x^*)^T,dx cr
dpsi
&= phi,dphi^* + phi^*,dphi cr
&= (phi A^*x^* + phi^*A^Tx^*)^Tdx cr
g = frac{partialpsi}{partial x}
&= A^*x^*phi + A^Tx^*phi^* cr
g^* = frac{partialpsi}{partial x^*}
&= Axphi^* + A^Hxphi cr
}$$
Not that we need it, but the last equation is a consequence of the fact that $psi=psi^*,$ (it's real).



Now the Hessian is just the gradient of the gradient, so
$$eqalign{
dg
&= (A^*x^*),dphi + (A^Tx^*),dphi^* cr
&= Big((A^*x^*)(A^Tx^*)^T + (A^Tx^*)(A^*x^*)^TBig),dx cr
H = frac{partial g}{partial x}
&= A^*x^*x^HA + A^Tx^*x^HA^H cr
}$$
Note that the Hessian is symmetric, but it is not Hermitian.






share|cite|improve this answer























  • A question with respect to $dpsi = (phi A^*x^* + phi^*A^Tx^*)^Tdx Rightarrow g = A^*x^*phi + A^Tx^*phi^* $. What rule did you apply that $phi A^*x^*$ became $A^*x^*phi$? Am I misunderstanding something?
    – user550103
    Nov 28 at 8:17












  • Since $phi$ is a scalar, you can move it anywhere you like. It was convenient to move it to the end of the expression, to prepare for the subsequent step of calculating $dg$
    – greg
    Nov 28 at 13:29










  • Oh yes, makes sense... thanks for the clarification, greg!
    – user550103
    Nov 28 at 13:49



















-2














The second derivative of a multi-variable function $f=f(x_1,ldots,x_n)$ is usually expressed as its Hessian matrix
$$
Hf(x_1,ldots,x_n)=left(frac{partial^2 f}{partial x_ipartial x_j}right)_{i,j=1}^n
$$
Here
$$
f(x)=x^tA^txx^tAx=(x^tBx)^2, quad text{where $B=frac{1}{2}(A+A^t)=(b_{ii})$}.
$$
Hence
$$
frac{partial f}{partial x_i}=4(x^tBx)(Bx)_i
$$
and
$$
frac{partial^2 f}{partial x_ipartial x_j}=8(Bx)_i(Bx)_j+4(x^tBx)b_{ij},
$$
and thus
$$
Hf(x)=8Bx(Bx)^t+4(x^tBx)B
$$






share|cite|improve this answer





















  • $x$ is a complex variable.
    – Alex Silva
    Nov 27 at 8:17










  • What is $x^H$? Is it the conjugate of the transpose, or just the transpose?
    – Yiorgos S. Smyrlis
    Nov 27 at 8:19










  • Hermitian = conjugate transpose.
    – Alex Silva
    Nov 27 at 8:22












  • Unfortunately, you CAN NOT differentiate the conjugate. The function $f(z)=overline{z}$ is NOT analytic! Hence, the conjugate applies only to matrix (if at all).
    – Yiorgos S. Smyrlis
    Nov 27 at 8:25






  • 2




    @YiorgosS.Smyrlis I suppose they are talking about Wirtinger derivatives, see en.wikipedia.org/wiki/Wirtinger_derivatives
    – lisyarus
    Nov 27 at 9:22











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














Define the scalar variables
$$eqalign{
&phi &= x^HAx = (A^Tx^*)^Tx cr
&phi^* &= x^HA^Hx = (A^*x^*)^Tx cr
&psi &= phi^*phi cr
}$$
Find the gradient of your function $(psi)$ with respect to $x$, treating $x^*$ as an independent variable.
$$eqalign{
dphi &= (A^Tx^*)^T,dx cr
dphi^* &= (A^*x^*)^T,dx cr
dpsi
&= phi,dphi^* + phi^*,dphi cr
&= (phi A^*x^* + phi^*A^Tx^*)^Tdx cr
g = frac{partialpsi}{partial x}
&= A^*x^*phi + A^Tx^*phi^* cr
g^* = frac{partialpsi}{partial x^*}
&= Axphi^* + A^Hxphi cr
}$$
Not that we need it, but the last equation is a consequence of the fact that $psi=psi^*,$ (it's real).



Now the Hessian is just the gradient of the gradient, so
$$eqalign{
dg
&= (A^*x^*),dphi + (A^Tx^*),dphi^* cr
&= Big((A^*x^*)(A^Tx^*)^T + (A^Tx^*)(A^*x^*)^TBig),dx cr
H = frac{partial g}{partial x}
&= A^*x^*x^HA + A^Tx^*x^HA^H cr
}$$
Note that the Hessian is symmetric, but it is not Hermitian.






share|cite|improve this answer























  • A question with respect to $dpsi = (phi A^*x^* + phi^*A^Tx^*)^Tdx Rightarrow g = A^*x^*phi + A^Tx^*phi^* $. What rule did you apply that $phi A^*x^*$ became $A^*x^*phi$? Am I misunderstanding something?
    – user550103
    Nov 28 at 8:17












  • Since $phi$ is a scalar, you can move it anywhere you like. It was convenient to move it to the end of the expression, to prepare for the subsequent step of calculating $dg$
    – greg
    Nov 28 at 13:29










  • Oh yes, makes sense... thanks for the clarification, greg!
    – user550103
    Nov 28 at 13:49
















3














Define the scalar variables
$$eqalign{
&phi &= x^HAx = (A^Tx^*)^Tx cr
&phi^* &= x^HA^Hx = (A^*x^*)^Tx cr
&psi &= phi^*phi cr
}$$
Find the gradient of your function $(psi)$ with respect to $x$, treating $x^*$ as an independent variable.
$$eqalign{
dphi &= (A^Tx^*)^T,dx cr
dphi^* &= (A^*x^*)^T,dx cr
dpsi
&= phi,dphi^* + phi^*,dphi cr
&= (phi A^*x^* + phi^*A^Tx^*)^Tdx cr
g = frac{partialpsi}{partial x}
&= A^*x^*phi + A^Tx^*phi^* cr
g^* = frac{partialpsi}{partial x^*}
&= Axphi^* + A^Hxphi cr
}$$
Not that we need it, but the last equation is a consequence of the fact that $psi=psi^*,$ (it's real).



Now the Hessian is just the gradient of the gradient, so
$$eqalign{
dg
&= (A^*x^*),dphi + (A^Tx^*),dphi^* cr
&= Big((A^*x^*)(A^Tx^*)^T + (A^Tx^*)(A^*x^*)^TBig),dx cr
H = frac{partial g}{partial x}
&= A^*x^*x^HA + A^Tx^*x^HA^H cr
}$$
Note that the Hessian is symmetric, but it is not Hermitian.






share|cite|improve this answer























  • A question with respect to $dpsi = (phi A^*x^* + phi^*A^Tx^*)^Tdx Rightarrow g = A^*x^*phi + A^Tx^*phi^* $. What rule did you apply that $phi A^*x^*$ became $A^*x^*phi$? Am I misunderstanding something?
    – user550103
    Nov 28 at 8:17












  • Since $phi$ is a scalar, you can move it anywhere you like. It was convenient to move it to the end of the expression, to prepare for the subsequent step of calculating $dg$
    – greg
    Nov 28 at 13:29










  • Oh yes, makes sense... thanks for the clarification, greg!
    – user550103
    Nov 28 at 13:49














3












3








3






Define the scalar variables
$$eqalign{
&phi &= x^HAx = (A^Tx^*)^Tx cr
&phi^* &= x^HA^Hx = (A^*x^*)^Tx cr
&psi &= phi^*phi cr
}$$
Find the gradient of your function $(psi)$ with respect to $x$, treating $x^*$ as an independent variable.
$$eqalign{
dphi &= (A^Tx^*)^T,dx cr
dphi^* &= (A^*x^*)^T,dx cr
dpsi
&= phi,dphi^* + phi^*,dphi cr
&= (phi A^*x^* + phi^*A^Tx^*)^Tdx cr
g = frac{partialpsi}{partial x}
&= A^*x^*phi + A^Tx^*phi^* cr
g^* = frac{partialpsi}{partial x^*}
&= Axphi^* + A^Hxphi cr
}$$
Not that we need it, but the last equation is a consequence of the fact that $psi=psi^*,$ (it's real).



Now the Hessian is just the gradient of the gradient, so
$$eqalign{
dg
&= (A^*x^*),dphi + (A^Tx^*),dphi^* cr
&= Big((A^*x^*)(A^Tx^*)^T + (A^Tx^*)(A^*x^*)^TBig),dx cr
H = frac{partial g}{partial x}
&= A^*x^*x^HA + A^Tx^*x^HA^H cr
}$$
Note that the Hessian is symmetric, but it is not Hermitian.






share|cite|improve this answer














Define the scalar variables
$$eqalign{
&phi &= x^HAx = (A^Tx^*)^Tx cr
&phi^* &= x^HA^Hx = (A^*x^*)^Tx cr
&psi &= phi^*phi cr
}$$
Find the gradient of your function $(psi)$ with respect to $x$, treating $x^*$ as an independent variable.
$$eqalign{
dphi &= (A^Tx^*)^T,dx cr
dphi^* &= (A^*x^*)^T,dx cr
dpsi
&= phi,dphi^* + phi^*,dphi cr
&= (phi A^*x^* + phi^*A^Tx^*)^Tdx cr
g = frac{partialpsi}{partial x}
&= A^*x^*phi + A^Tx^*phi^* cr
g^* = frac{partialpsi}{partial x^*}
&= Axphi^* + A^Hxphi cr
}$$
Not that we need it, but the last equation is a consequence of the fact that $psi=psi^*,$ (it's real).



Now the Hessian is just the gradient of the gradient, so
$$eqalign{
dg
&= (A^*x^*),dphi + (A^Tx^*),dphi^* cr
&= Big((A^*x^*)(A^Tx^*)^T + (A^Tx^*)(A^*x^*)^TBig),dx cr
H = frac{partial g}{partial x}
&= A^*x^*x^HA + A^Tx^*x^HA^H cr
}$$
Note that the Hessian is symmetric, but it is not Hermitian.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 at 5:34

























answered Nov 27 at 20:21









greg

7,5251721




7,5251721












  • A question with respect to $dpsi = (phi A^*x^* + phi^*A^Tx^*)^Tdx Rightarrow g = A^*x^*phi + A^Tx^*phi^* $. What rule did you apply that $phi A^*x^*$ became $A^*x^*phi$? Am I misunderstanding something?
    – user550103
    Nov 28 at 8:17












  • Since $phi$ is a scalar, you can move it anywhere you like. It was convenient to move it to the end of the expression, to prepare for the subsequent step of calculating $dg$
    – greg
    Nov 28 at 13:29










  • Oh yes, makes sense... thanks for the clarification, greg!
    – user550103
    Nov 28 at 13:49


















  • A question with respect to $dpsi = (phi A^*x^* + phi^*A^Tx^*)^Tdx Rightarrow g = A^*x^*phi + A^Tx^*phi^* $. What rule did you apply that $phi A^*x^*$ became $A^*x^*phi$? Am I misunderstanding something?
    – user550103
    Nov 28 at 8:17












  • Since $phi$ is a scalar, you can move it anywhere you like. It was convenient to move it to the end of the expression, to prepare for the subsequent step of calculating $dg$
    – greg
    Nov 28 at 13:29










  • Oh yes, makes sense... thanks for the clarification, greg!
    – user550103
    Nov 28 at 13:49
















A question with respect to $dpsi = (phi A^*x^* + phi^*A^Tx^*)^Tdx Rightarrow g = A^*x^*phi + A^Tx^*phi^* $. What rule did you apply that $phi A^*x^*$ became $A^*x^*phi$? Am I misunderstanding something?
– user550103
Nov 28 at 8:17






A question with respect to $dpsi = (phi A^*x^* + phi^*A^Tx^*)^Tdx Rightarrow g = A^*x^*phi + A^Tx^*phi^* $. What rule did you apply that $phi A^*x^*$ became $A^*x^*phi$? Am I misunderstanding something?
– user550103
Nov 28 at 8:17














Since $phi$ is a scalar, you can move it anywhere you like. It was convenient to move it to the end of the expression, to prepare for the subsequent step of calculating $dg$
– greg
Nov 28 at 13:29




Since $phi$ is a scalar, you can move it anywhere you like. It was convenient to move it to the end of the expression, to prepare for the subsequent step of calculating $dg$
– greg
Nov 28 at 13:29












Oh yes, makes sense... thanks for the clarification, greg!
– user550103
Nov 28 at 13:49




Oh yes, makes sense... thanks for the clarification, greg!
– user550103
Nov 28 at 13:49











-2














The second derivative of a multi-variable function $f=f(x_1,ldots,x_n)$ is usually expressed as its Hessian matrix
$$
Hf(x_1,ldots,x_n)=left(frac{partial^2 f}{partial x_ipartial x_j}right)_{i,j=1}^n
$$
Here
$$
f(x)=x^tA^txx^tAx=(x^tBx)^2, quad text{where $B=frac{1}{2}(A+A^t)=(b_{ii})$}.
$$
Hence
$$
frac{partial f}{partial x_i}=4(x^tBx)(Bx)_i
$$
and
$$
frac{partial^2 f}{partial x_ipartial x_j}=8(Bx)_i(Bx)_j+4(x^tBx)b_{ij},
$$
and thus
$$
Hf(x)=8Bx(Bx)^t+4(x^tBx)B
$$






share|cite|improve this answer





















  • $x$ is a complex variable.
    – Alex Silva
    Nov 27 at 8:17










  • What is $x^H$? Is it the conjugate of the transpose, or just the transpose?
    – Yiorgos S. Smyrlis
    Nov 27 at 8:19










  • Hermitian = conjugate transpose.
    – Alex Silva
    Nov 27 at 8:22












  • Unfortunately, you CAN NOT differentiate the conjugate. The function $f(z)=overline{z}$ is NOT analytic! Hence, the conjugate applies only to matrix (if at all).
    – Yiorgos S. Smyrlis
    Nov 27 at 8:25






  • 2




    @YiorgosS.Smyrlis I suppose they are talking about Wirtinger derivatives, see en.wikipedia.org/wiki/Wirtinger_derivatives
    – lisyarus
    Nov 27 at 9:22
















-2














The second derivative of a multi-variable function $f=f(x_1,ldots,x_n)$ is usually expressed as its Hessian matrix
$$
Hf(x_1,ldots,x_n)=left(frac{partial^2 f}{partial x_ipartial x_j}right)_{i,j=1}^n
$$
Here
$$
f(x)=x^tA^txx^tAx=(x^tBx)^2, quad text{where $B=frac{1}{2}(A+A^t)=(b_{ii})$}.
$$
Hence
$$
frac{partial f}{partial x_i}=4(x^tBx)(Bx)_i
$$
and
$$
frac{partial^2 f}{partial x_ipartial x_j}=8(Bx)_i(Bx)_j+4(x^tBx)b_{ij},
$$
and thus
$$
Hf(x)=8Bx(Bx)^t+4(x^tBx)B
$$






share|cite|improve this answer





















  • $x$ is a complex variable.
    – Alex Silva
    Nov 27 at 8:17










  • What is $x^H$? Is it the conjugate of the transpose, or just the transpose?
    – Yiorgos S. Smyrlis
    Nov 27 at 8:19










  • Hermitian = conjugate transpose.
    – Alex Silva
    Nov 27 at 8:22












  • Unfortunately, you CAN NOT differentiate the conjugate. The function $f(z)=overline{z}$ is NOT analytic! Hence, the conjugate applies only to matrix (if at all).
    – Yiorgos S. Smyrlis
    Nov 27 at 8:25






  • 2




    @YiorgosS.Smyrlis I suppose they are talking about Wirtinger derivatives, see en.wikipedia.org/wiki/Wirtinger_derivatives
    – lisyarus
    Nov 27 at 9:22














-2












-2








-2






The second derivative of a multi-variable function $f=f(x_1,ldots,x_n)$ is usually expressed as its Hessian matrix
$$
Hf(x_1,ldots,x_n)=left(frac{partial^2 f}{partial x_ipartial x_j}right)_{i,j=1}^n
$$
Here
$$
f(x)=x^tA^txx^tAx=(x^tBx)^2, quad text{where $B=frac{1}{2}(A+A^t)=(b_{ii})$}.
$$
Hence
$$
frac{partial f}{partial x_i}=4(x^tBx)(Bx)_i
$$
and
$$
frac{partial^2 f}{partial x_ipartial x_j}=8(Bx)_i(Bx)_j+4(x^tBx)b_{ij},
$$
and thus
$$
Hf(x)=8Bx(Bx)^t+4(x^tBx)B
$$






share|cite|improve this answer












The second derivative of a multi-variable function $f=f(x_1,ldots,x_n)$ is usually expressed as its Hessian matrix
$$
Hf(x_1,ldots,x_n)=left(frac{partial^2 f}{partial x_ipartial x_j}right)_{i,j=1}^n
$$
Here
$$
f(x)=x^tA^txx^tAx=(x^tBx)^2, quad text{where $B=frac{1}{2}(A+A^t)=(b_{ii})$}.
$$
Hence
$$
frac{partial f}{partial x_i}=4(x^tBx)(Bx)_i
$$
and
$$
frac{partial^2 f}{partial x_ipartial x_j}=8(Bx)_i(Bx)_j+4(x^tBx)b_{ij},
$$
and thus
$$
Hf(x)=8Bx(Bx)^t+4(x^tBx)B
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 at 8:13









Yiorgos S. Smyrlis

62.4k1383162




62.4k1383162












  • $x$ is a complex variable.
    – Alex Silva
    Nov 27 at 8:17










  • What is $x^H$? Is it the conjugate of the transpose, or just the transpose?
    – Yiorgos S. Smyrlis
    Nov 27 at 8:19










  • Hermitian = conjugate transpose.
    – Alex Silva
    Nov 27 at 8:22












  • Unfortunately, you CAN NOT differentiate the conjugate. The function $f(z)=overline{z}$ is NOT analytic! Hence, the conjugate applies only to matrix (if at all).
    – Yiorgos S. Smyrlis
    Nov 27 at 8:25






  • 2




    @YiorgosS.Smyrlis I suppose they are talking about Wirtinger derivatives, see en.wikipedia.org/wiki/Wirtinger_derivatives
    – lisyarus
    Nov 27 at 9:22


















  • $x$ is a complex variable.
    – Alex Silva
    Nov 27 at 8:17










  • What is $x^H$? Is it the conjugate of the transpose, or just the transpose?
    – Yiorgos S. Smyrlis
    Nov 27 at 8:19










  • Hermitian = conjugate transpose.
    – Alex Silva
    Nov 27 at 8:22












  • Unfortunately, you CAN NOT differentiate the conjugate. The function $f(z)=overline{z}$ is NOT analytic! Hence, the conjugate applies only to matrix (if at all).
    – Yiorgos S. Smyrlis
    Nov 27 at 8:25






  • 2




    @YiorgosS.Smyrlis I suppose they are talking about Wirtinger derivatives, see en.wikipedia.org/wiki/Wirtinger_derivatives
    – lisyarus
    Nov 27 at 9:22
















$x$ is a complex variable.
– Alex Silva
Nov 27 at 8:17




$x$ is a complex variable.
– Alex Silva
Nov 27 at 8:17












What is $x^H$? Is it the conjugate of the transpose, or just the transpose?
– Yiorgos S. Smyrlis
Nov 27 at 8:19




What is $x^H$? Is it the conjugate of the transpose, or just the transpose?
– Yiorgos S. Smyrlis
Nov 27 at 8:19












Hermitian = conjugate transpose.
– Alex Silva
Nov 27 at 8:22






Hermitian = conjugate transpose.
– Alex Silva
Nov 27 at 8:22














Unfortunately, you CAN NOT differentiate the conjugate. The function $f(z)=overline{z}$ is NOT analytic! Hence, the conjugate applies only to matrix (if at all).
– Yiorgos S. Smyrlis
Nov 27 at 8:25




Unfortunately, you CAN NOT differentiate the conjugate. The function $f(z)=overline{z}$ is NOT analytic! Hence, the conjugate applies only to matrix (if at all).
– Yiorgos S. Smyrlis
Nov 27 at 8:25




2




2




@YiorgosS.Smyrlis I suppose they are talking about Wirtinger derivatives, see en.wikipedia.org/wiki/Wirtinger_derivatives
– lisyarus
Nov 27 at 9:22




@YiorgosS.Smyrlis I suppose they are talking about Wirtinger derivatives, see en.wikipedia.org/wiki/Wirtinger_derivatives
– lisyarus
Nov 27 at 9:22


















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