Differential of quadratic form

Question

I'm having trouble understanding this: if $\phi(x)=x^{T}Ax$, where $A$ is a matrix of constraints. Then, the differential of $\phi$: $\mathrm{d} \phi=(\mathrm{d} x)^{T} A x+x^{T} A \mathrm{~d} x=x^{T} A^{T} \mathrm{d} x+x^{T} A \mathrm{~d} x=x^{T}\left(A+A^{T}\right) \mathrm{d} x$.

I'm having trouble understanding this step: why is $(\mathrm{d} x)^{T} A x=x^{T} A^{T} \mathrm{d} x$?

I find it interesting because based on the rule: $$ (AB)^{T}=B^{T}A^{T} $$ we have $$ (\mathrm{d} x)^{T} (A x)=[(Ax)^{T} \mathrm{d} x]^{T} $$ Therefore $(\mathrm{d} x)^{T} A x=x^{T} A^{T} \mathrm{d} x$ essentially is saying that $[(Ax)^{T} \mathrm{d} x]^{T}=(Ax)^{T} \mathrm{d} x$

Showing me why this is the case can help me understand this question a lot.

I do know thispost, but I don't understand why we need chain rule here, because the chain rule is: Let $x=x(t)$ and $y=y(t)$ be differentiable at $t$ and suppose that $z=f(x, y)$ is differentiable at the point $(x(t), y(t))$. Then $z=f(x(t), y(t))$ is differentiable at $t$ and $$ \frac{d z}{d t}=\frac{\partial z}{\partial x} \frac{d x}{d t}+\frac{\partial z}{\partial y} \frac{d y}{d t} $$

there's no involvement of another variable t at a lower-level than x & y. There is just one level x here.

Thank you all in advance for your answers.

Since it is a scalar its transpose is equal to itself. – obareey Feb 25 '22 at 13:54 — obareey, Feb 25 '22 at 13:54
but A' = A only if A is symmetric, right? – some Feb 25 '22 at 13:57 — some, Feb 25 '22 at 13:57

score 0 · Accepted Answer · answered Feb 25 '22 at 14:30

You can recover it easily from $$ \phi(x)=\sum_{i,j=1}^n A_{ij} x_i x_j $$

$$ \frac{\partial \phi}{\partial x_k} = \sum_{j=1}^n a_{kj} x_j + \sum_{i=1}^n a_{ik} x_i = [(A+A')x ]_k $$

So, the differential becomes $$ (D\phi)(x)(h) = \sum_{i=1}^n \frac{\partial \phi}{\partial x_i} h_i =x' (A+A')h $$

Differential of quadratic form

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