In this brief description, we will use the numerator layout [1],
and will tacitly assume that all products are conformable.
The derivative of the linear form 𝒰t𝒱 with respect to the vector
𝒱is given as
and since 𝒰t𝒱 is a scalar, we are facing a particular case
of the derivative of a scalar λ with respect to a vector, e.g.,
∂𝒱λ=(∂𝒱1λ ..... ∂𝒱nλ) and it must also be ∂𝒱(𝒰t𝒱)=∂𝒱(𝒱t𝒰)
. Moreover, it is easy to demonstrate that using the denominator
layout, the derivative would have been ∂𝒱(𝒰t𝒱)=𝒰
If both 𝒰 and 𝒱 vectors are function of a third vector 𝒵, we get
which, in the case 𝒰=𝒱=𝒲 reduces to
Dealing with a linear transform 𝒰=a𝒱, if A is 𝓂×𝓂 we have
and if 𝒱 is a function of a vector 𝒲𝒱 we get
From definition of bilinear form, we obtain, for 𝒰tA𝒱 the
derivative
while, for a quadratic form 𝒰tA𝒱(where A is 𝓃 × 𝓃), we get
so that, if A is a symmetric matrix, say, for A = XtX, then