In [1, p. 35 exercise 2.16] we are asked to verify the formulas given for the complex gradient of a hermitian and a linear form [1, p. 31 (2.70)]. To do so, we are instructed to decompose the matrices and vectors into their real and imaginary parts as \mathbf{x}=\mathbf{x}_{R}+j\mathbf{x}_{I}, \mathbf{A^{\prime}}= \mathbf{A}_{R}+j\mathbf{A}_{I}, \mathbf{b^{\prime}}=\mathbf{b}_{R}+j\mathbf{b}_{I}
Solution: The formulas given by [1, p. 31 (2.70)] are
\frac{\partial}{\partial \mathbf{x}}(\mathbf{x}^H\mathbf{A}^{\prime}\mathbf{x})=2\mathbf{A}^{\prime}\mathbf{x} (1)

and
\frac{\partial}{\partial \mathbf{x}} (\mathbf{x}^{H}\mathbf{b}^{\prime}+{\mathbf{b}^{\prime}}^{H}\mathbf{x})=2\mathbf{b}^{\prime} (2)

where \mathbf{A}^{\prime} is a complex n \times n matrix with elements a_{ij} and \mathbf{b}^{\prime} is a complex n \times 1 vector with elements b_{i} and \frac{\partial}{\partial \mathbf{x}} denotes the gradient of a real function in respect to \mathbf{x}. Although it is not explicitly stated (neither in the problem nor in the book text) the matrix must be hermitian for the two relations to hold. For Hermitian matrices one can obtain the following relations for the real (3) and imaginary (4) parts of the matrix:
\mathbf{C}^{H}=\mathbf{C}
\mathbf{C}^{T}_{R}-j\mathbf{C}^{T}_{I}=\mathbf{C}_{R}+j\mathbf{C}_{I}
\mathbf{C}^{T}_{R}=\mathbf{C}_{R} (3)
\mathbf{C}^{T}_{I}=-\mathbf{C}_{I} (4)

A further relation which will be useful for the solution of the problem is the derivation of the formula of the gradient of a quadratic form of a real matrix which is not symmetric. We reproduce [2, rel. (4) from the solution of exercise 2.15] the formula for the real matrix \mathbf{A}_{R} =[a_{ij}] and the real vector \mathbf{x}_{R}=[x_{i}]:
\frac{\partial f(\mathbf{x})}{\partial x_{l}} = 2x_{l}a_{ll} +\sum\limits_{j=1, j\neq l}^{n}a_{lj}x_j +\sum\limits_{i=1, i\neq l}^{n}x_{i}a_{il}
=\sum\limits_{j=1}^{n}a_{lj}x_j +\sum\limits_{i=1}^{n}x_{i}a_{il}

And thus
\frac{\partial f(\mathbf{x})}{\partial \mathbf{x}_{R}} = \mathbf{A}_{R}\mathbf{x}_{R} + \mathbf{A}^{T}_{R}\mathbf{x}_{R} \label{ex216:gradreal}

The last relation is a general expression of the gradient of a real matrix. Expanding the term \mathbf{x}^H\mathbf{A}^{\prime}\mathbf{x} by using the real and imaginary parts we obtain:
\mathbf{x}^H\mathbf{A}^{\prime}\mathbf{x}=(\mathbf{x}^{T}_{R}-j\mathbf{x}^{T}_{I})(\mathbf{A}_{R}+j\mathbf{A}_{I})(\mathbf{x}_{R}+j\mathbf{x}_{I})
=(\mathbf{x}^{T}_{R}\mathbf{A}_{R}+j\mathbf{x}^{T}_{R}\mathbf{A}_{I}-j\mathbf{x}^{T}_{I}\mathbf{A}_{R}+\mathbf{x}^{T}_{I}\mathbf{A}_{I} )(\mathbf{x}_{R}+j\mathbf{x}_{I})
=\mathbf{x}^{T}_{R}\mathbf{A}_{R}\mathbf{x}_{R}+j\mathbf{x}^{T}_{R}\mathbf{A}_{I}\mathbf{x}_{R}-j\mathbf{x}^{T}_{I}\mathbf{A}_{R}\mathbf{x}_{R}+\mathbf{x}^{T}_{I}\mathbf{A}_{I}\mathbf{x}_{R}
+ j\mathbf{x}^{T}_{R}\mathbf{A}_{R}\mathbf{x}_{I}-\mathbf{x}^{T}_{R}\mathbf{A}_{I}\mathbf{x}_{I}+\mathbf{x}^{T}_{I}\mathbf{A}_{R}\mathbf{x}_{I}+j\mathbf{x}^{T}_{I}\mathbf{A}_{I}\mathbf{x}_{I}
=\mathbf{x}^{T}_{R}\mathbf{A}_{R}\mathbf{x}_{R}+ \mathbf{x}^{T}_{I}\mathbf{A}_{I}\mathbf{x}_{R}-\mathbf{x}^{T}_{R}\mathbf{A}_{I}\mathbf{x}_{I} +\mathbf{x}^{T}_{I}\mathbf{A}_{R}\mathbf{x}_{I}
+j ( \mathbf{x}^{T}_{R}\mathbf{A}_{I}\mathbf{x}_{R} - \mathbf{x}^{T}_{I}\mathbf{A}_{R}\mathbf{x}_{R} + \mathbf{x}^{T}_{R}\mathbf{A}_{R}\mathbf{x}_{I} +\mathbf{x}^{T}_{I}\mathbf{A}_{I}\mathbf{x}_{I} ) (5)

Using the definition of the complex gradient ([1, p. 32 (2.69)]) \frac{\partial g(\mathbf{x})}{\partial \mathbf{x}} = \frac{\partial g(\mathbf{x})}{\partial \mathbf{x}_{R}}+j\frac{\partial g(\mathbf{x})}{\partial \mathbf{x}_{I}} we have to calculate the partial derivatives in respect to \mathbf{x}_{R} and \mathbf{x}_{I} :
\frac{\partial \mathbf{x}^H\mathbf{A}^{\prime}\mathbf{x}}{\partial \mathbf{x}_{R}}=2\mathbf{A}_{R}\mathbf{x}_{R}+\mathbf{A}^{T}_{I}\mathbf{x}_{I}-\mathbf{A}_{I}\mathbf{x}_{I}+ j(\mathbf{A}_{I}\mathbf{x}_{R}+\mathbf{A}^{T}_{I}\mathbf{x}_{R}
-\mathbf{A}^{T}_{R}\mathbf{x}_{I}+\mathbf{A}_{R}\mathbf{x}_{I})

and because of (3,4) we obtain the relation
\frac{\partial \mathbf{x}^H\mathbf{A}^{\prime}\mathbf{x}}{\partial \mathbf{x}_{R}} = 2\mathbf{A}_{R}\mathbf{x}_{R}-2 \mathbf{A}_{I}\mathbf{x}_{I} (6)

Similar the gradient to the imaginary part of \mathbf{x} is obtained by
\frac{\partial \mathbf{x}^H\mathbf{A}^{\prime}\mathbf{x}}{\partial \mathbf{x}_{I}} = \mathbf{A}_{I}\mathbf{x}_{R}-\mathbf{A}^{T}_{I}\mathbf{x}_{R}+ 2\mathbf{A}_{R}\mathbf{x}_{I}+j(\mathbf{A}^{T}_{R}\mathbf{x}_{R}-\mathbf{A}_{R}\mathbf{x}_{R})

and because of (3,4) we obtain the relation
\frac{\partial \mathbf{x}^H\mathbf{A}^{\prime}\mathbf{x}}{\partial \mathbf{x}_{I}} = 2\mathbf{A}_{I}\mathbf{x}_{R}+ 2\mathbf{A}_{R}\mathbf{x}_{I} (7)

From (6) and (7) we obtain using the definition of the complex gradian:
\frac{\partial l \mathbf{x}^H\mathbf{A}^{\prime}\mathbf{x}}{\partial \mathbf{x}}=\frac{\partial l \mathbf{x}^H\mathbf{A}^{\prime}\mathbf{x}}{\partial \mathbf{x}_{R}}+j\frac{\partial \mathbf{x}^H\mathbf{A}^{\prime}\mathbf{x}}{\partial \mathbf{x}_{I}}
=2\mathbf{A}_{R}\mathbf{x}_{R}-2 \mathbf{A}_{I}\mathbf{x}_{I}+j( 2\mathbf{A}_{I}\mathbf{x}_{R}+ 2\mathbf{A}_{R}\mathbf{x}_{I} )
=2(\mathbf{A}_{R}+j\mathbf{A}_{I})\mathbf{x}_{R}+ 2 j^{2} \mathbf{A}_{I}\mathbf{x}_{I} +2 j \mathbf{A}_{R}\mathbf{x}_{I}
=2 \mathbf{A}^{\prime}\mathbf{x}_{R}+ 2j ( \mathbf{A}_{R} + j\mathbf{A}_{I}) \mathbf{x}_{I}
=2 \mathbf{A}^{\prime}\mathbf{x}_{R}+ 2j \mathbf{A}^{\prime} \mathbf{x}_{I}
=2 \mathbf{A}^{\prime} (\mathbf{x}_{R}+j \mathbf{x}_{I})
=2 \mathbf{A}^{\prime} \mathbf{x} (11)

Equation 11 proofs the first part of equations [1, p. 31 (2.70)]. The second equation can be solved similar by setting \mathbf{b}^{\prime}=\mathbf{b}_{R}+j\mathbf{b}_{I} and thus:
\mathbf{x}^{H} \mathbf{b}^{\prime}=\mathbf{x}^{T}_{R}\mathbf{b}_{R}+j\mathbf{x}^{T}_{R}\mathbf{b}_{I}-j\mathbf{x}^{T}_{I}\mathbf{b}_{R}+\mathbf{x}^{T}_{I}\mathbf{b}_{I} (8)
\mathbf{b}^{\prime H} \mathbf{x}=\mathbf{b}^{T}_{R} \mathbf{x}_{R}+j\mathbf{b}^{T}_{R} \mathbf{x}_{I}-j\mathbf{b}^{T}_{I} \mathbf{x}_{R}+\mathbf{b}^{T}_{I} \mathbf{x}_{I} (9)

The gradient of \mathbf{x}^{H} \mathbf{b}^{\prime } in respect to the real and imaginary part of \mathbf{x} is obtained by:
\frac{\partial \mathbf{x}^{H} \mathbf{b}^{\prime }}{\partial \mathbf{x}_{R}} = \mathbf{b}_{R} +j\mathbf{b}_{I}
\frac{\partial \mathbf{x}^{H} \mathbf{b}^{\prime }}{\partial \mathbf{x}_{I}} = -j \mathbf{b}_{R} +\mathbf{b}_{I}

while the gradient of \mathbf{b}^{\prime H} \mathbf{x} in respect to the real and imaginary part of \mathbf{x} is given by
\frac{\partial \mathbf{b}^{\prime H} \mathbf{x}}{\partial \mathbf{x}_{R}}= \mathbf{b}_{R} - j\mathbf{b}_{I}
\frac{\partial \mathbf{b}^{\prime H} \mathbf{x}}{\partial \mathbf{x}_{I}}= j\mathbf{b}_{R} +\mathbf{b}_{I}

From the previous relations we obtain:
\frac{\partial \mathbf{x}^{H} \mathbf{b}^{\prime }+\mathbf{b}^{\prime H} \mathbf{x}}{\partial \mathbf{x}_{R} }=2 \mathbf{b}_{R}
\frac{\partial \mathbf{x}^{H} \mathbf{b}^{\prime }+\mathbf{b}^{\prime H} \mathbf{x}}{\partial \mathbf{x}_{I}} = 2 \mathbf{b}_{I}

which can be used to find the complex gradient using its definition ([1, p. 32 (2.69)]) \frac{\partial g(\mathbf{x})}{\partial \mathbf{x}} = \frac{\partial g(\mathbf{x})}{\partial \mathbf{x}_{R}}+j\frac{\partial g(\mathbf{x})}{\partial \mathbf{x}_{I}} :
\frac{\partial}{\partial \mathbf{x}} (\mathbf{x}^{H}\mathbf{b}^{\prime}+{\mathbf{b}^{\prime}}^{H}\mathbf{x})=2 \mathbf{b}_{R}+j2 \mathbf{b}_{I}= 2\mathbf{b}^{\prime} (10)

The result (10) proofs the second part of the equations [1, p. 31 (2.70)]. QED.

[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.
[2] Chatzichrisafis: “Solution of exercise 2.15 from Kay’s Modern Spectral Estimation - Theory and Applications”.