# Second order derivation of Quadratic form

I would like to find the second order derivative of a Quadratic form. Assume we have a random complex column vector $x$ and a real constant value $C$. I am interested in computing the following: $$\frac{\partial^2}{\partial x^H \partial x}(x^H C x) = ~?$$

The differentiation of the quadratic form is: $$d(x^H C x) = 2 \mathcal{R}( x^H C dx)$$ where $\mathcal{R}(\cdot)$ denotes the real part.

Are the following expressions correct?

$$d^2(x^H C x) = 2\mathcal{R}(C dx^H dx)$$ since $C$ is real and also the outcome of $dx^H dx$ is real, the operator $\mathcal{R}(\cdot)$ can be ignored and therefore:

$$\frac{\partial^2}{\partial x^H \partial x}(x^H C x) = 2 C$$

And how about, if $C$ is a constant matrix with property $C = C^H$?

• What does $x^H$ or $C^H$ mean? The hermitian conjugate? // What does "random" have to do with this? – Willie Wong May 28 '15 at 10:49
• yes, $(\cdot)^H$ means hermitian transpose. By random I just mean a random variable. In the system I am working it is zero mean Gaussian. – Marco May 28 '15 at 11:37

If you want a derivative with respect to a random variable, then have a look to the Ito's theory. Now, we consider $x,x^H$ as standard variables; note that they are linked; yet we can write a polynomial $P(x,y)$ in $x,y$ under the (unique) form of a polynomial $Q(z,\overline{z})$ in $z,\overline{z}$. For instance, $P$ is harmonic iff $\dfrac{\partial ^2Q}{\partial z\partial \overline{z}}=0$.
Here your function $f:(x,x^H))\in M_{n,1}\times M_{1,n}\rightarrow x^HCx$ is linear with respect to $x$ and to $x^H$. Then $\dfrac{\partial ^2f}{\partial x\partial x^H}:(h,k)\in M_{n,1}\times M_{1,n}\rightarrow kCh=f(h,k)$.
Remark. I think that you only have a vague idea about these concepts; your formula (valid when $C$ is a hermitian matrix)
$d(x^H C x) = 2 \mathcal{R}( x^H C dx)$
is a derivative with respect to $x$ only, that is to say that you consider that $x^H$ is a function of $x$ !!!