First-order Taylor approximation of matrix function Let $f : \Bbb C^{M \times N} \to \Bbb R_0^+$ be defined by $$ f(X) := \left\| X X^H - R \right\|_F^2 $$ I would like to find the first-order Taylor approximation of $f$.

I am familiar with the vector form but I cannot obtain the approximation for this function that depends on a matrix $ X $. I would like to linearize around a point $ X^0 $ to obtain something like this.
$$ f(X) = f(X^0) + \cdots $$
Update: The gradient of $ f(X) $  w.r.t. $ X \in \mathbb{C}^{M \times N} $ is
$$ \nabla_X f(X) = 2 X X^H X - R X - R^H X $$
 A: $
\def\o{{\tt1}}\def\p{\partial}
\def\LR#1{\left(#1\right)}
\def\op#1{\operatorname{#1}}
\def\trace#1{\op{Tr}\LR{#1}}
\def\real#1{\op{\sf Real}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\c#1{\color{red}{#1}}
\def\gradLR#1#2{\LR{\grad{#1}{#2}}}
\def\rgrad#1#2{\frac{\p #1}{\c{\p #2}}}
$For typing convenience, define the matrix variable
$$\eqalign{
M = &\LR{XX^H-R} \qiq
 &\,dM &= X\;dX^H + dX\:\,X^H  \\
 &\{{\rm conjugate}\}&dM^* &= X^*dX^T + dX^*X^T \\
}$$
and the Frobenius product, which is a wonderfully concise notation for the trace
$$\eqalign{
A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; \trace{A^TB} \\
A^*:A &= \|A\|^2_F \\
}$$
Write the objective function using the above notation.
Then calculate its differential and gradient (in the Wirtinger sense)
$$\eqalign{
\phi &= M^*:M \\
 &= \LR{M^*:dM} + \LR{M:dM^*} \\
 &= \LR{M^*:X\;dX^H + M^*:dX\,\,X^H} + \LR{M:X^*dX^T + M:dX^*X^T} \\
 &= (M^HX:dX^*) + (M^*X^*:\c{dX}) + (M^TX^*:\c{dX}) + (MX:dX^*) \\
 &= \LR{M^*+M^T}X^*:\c{dX} + \LR{M+M^H}X{:dX^*} \\
\rgrad{\phi}{X} &= \LR{M^*+M^T}X^* \\
\grad{\phi}{{X^*}} &= \LR{M+M^H}X \;=\; \gradLR{\phi}{X}^* \\\\
}$$
Therefore the first-order Taylor series can be written as
$$\eqalign{
\phi &= \phi(X), \quad \phi_0 = \phi(X_0) \\
\phi
 &= \phi_0 + \gradLR{\phi}{X}:(X-X_0) + \gradLR{\phi}{X^*}:(X-X_0)^* \\
 &= \phi_0 + 2\cdot\real{\gradLR{\phi}{X}:(X-X_0)} \\\\
}$$

Rules for manipulating Frobenius products are easily derived. Here is a summary
$$\eqalign{
A:B  &= B:A &= B^T:A^T \\
CA:B &= A:C^TB &= C:BA^T \\
}$$
It is also useful to know how hermitian and complex conjugation are related
$$\eqalign{
A^H &= \LR{A^T}^* &= \LR{A^*}^T \\
A^* &= \LR{A^H}^T &= \LR{A^T}^H \\
A^T &= \LR{A^*}^H &= \LR{A^H}^* \\
}$$
A: One can use a Taylor series approximation here, just as in the scalar case.
Follow the solution of
Derivative of squared Frobenius norm of a matrix
to compute the necessary derivative(s).
A: In general, the Taylor series of a scalar function $f：V→$, over a $$-vector space $V$ with $∈\{ℝ, ℂ\}$ can be expressed as
$$\begin{aligned} f(x+∆x) &= f(x) + ⟨f(x)∣∆x⟩_{V} + \tfrac{1}{2}⟨^2f(x)∣∆x^{⊗2}⟩_{V^{⊗2}} \\&\qquad+ \tfrac{1}{3!}⟨^3f(x)∣∆x^{⊗3}⟩_{V^{⊗3}} + …
\\&= ∑_{k=0}^{∞} \tfrac{1}{k!} ⟨^kf(x)∣∆x^{⊗k}⟩_{V^{⊗k}}
\end{aligned}$$
In your case $V=ℂ^n⊗ℂ^m$ and $f(x) = 4XX^HX -2RX-2R^⊤X$, hence:
$$ f(X+∆X) ≈ f(X) + ⟨4XX^HX -2RX-2R^⊤X∣∆X⟩_{ℂ^n⊗ℂ^m}$$
Note that the induced inner product on the tensor product of Hilbert spaces $⟨⋅∣⋅⟩_{ℂ^n⊗ℂ^m}$ is equivalent to the Frobenius inner product.
