In [1, p. 35 exercise 2.17] we are asked to verify the alternative expression [1, p. 33 (2.77)] for a hermitian function.
Solution: The standard expression of the hermitian function is given by [1, p. 32 (2.73)] :
g(\mathbf{x})=\left(\mathbf{A}\mathbf{x}-\mathbf{b}\right)^{H}\mathbf{R}^{-1}\left(\mathbf{A}\mathbf{x}-\mathbf{b}\right) (1)
=\left(\mathbf{b}-\mathbf{A}\mathbf{x}\right)^{H}\mathbf{R}^{-1} \left(\mathbf{b}-\mathbf{A}\mathbf{x}\right) (2)

The alternative expression [1, p. 33 (2.77)] is given by
g(\mathbf{x})=(\mathbf{b}-\mathbf{A}\mathbf{x}_{0})^{H}\mathbf{R}^{-1}(\mathbf{b}-\mathbf{A}\mathbf{x}_{0}) +
+ (\mathbf{x}-\mathbf{x}_{0})^{H}\mathbf{A}^{H}\mathbf{R}^{-1}\mathbf{A}(\mathbf{x}-\mathbf{x}_{0}) (3)

with \mathbf{x}_{0} given by [1, p. 33 (2.75)] which is reproduced here:
\mathbf{x}_{0}=\left(\mathbf{A}^{H}\mathbf{R}^{-1} \mathbf{A} \right)^{-1} \mathbf{A}^{H}\mathbf{R}^{-1} \mathbf{b} (4)

Using equation 4 we can compute using the matrix property of invertible matrices \left(\mathbf{A} \mathbf{B} \right)^{-1}= \mathbf{B}^{-1}\mathbf{A}^{-1}
\mathbf{A}\mathbf{x}_{0}=\mathbf{A}\left(\mathbf{A}^{H}\mathbf{R}^{-1} \mathbf{A} \right)^{-1} \mathbf{A}^{H}\mathbf{R}^{-1} \mathbf{b}
=\mathbf{A}\left(\mathbf{A}^{H}\mathbf{R}^{-1} \mathbf{A} \right)^{-1} \mathbf{A}^{H}\mathbf{R}^{-1} \mathbf{b}
=\left(\mathbf{A} \mathbf{A}^{-1}\right) \mathbf{R} \left( \left(\mathbf{A} ^{H}\right)^{-1} \mathbf{A}^{H}\right)\mathbf{R}^{-1} \mathbf{b}
=\mathbf{I} \left(\mathbf{R} \mathbf{I}\mathbf{R}^{-1}\right) \mathbf{b}
=\mathbf{b} (5)

Using (5) in (3) considering the matrix property \left(\mathbf{A}\mathbf{x} \right)^{H}= \mathbf{x}^{H}\mathbf{A}^{H} will result in:
g(\mathbf{x})=(\mathbf{b}-\mathbf{b})^{H}\mathbf{R}^{-1}(\mathbf{b}-\mathbf{b}) +(\mathbf{x}-\mathbf{x}_{0})^{H}\mathbf{A}^{H}\mathbf{R}^{-1}\mathbf{A}(\mathbf{x}-\mathbf{x}_{0})
=(\mathbf{A}\mathbf{x}-\mathbf{A}\mathbf{x}_{0})^{H}\mathbf{R}^{-1}(\mathbf{A}\mathbf{x}-\mathbf{A}\mathbf{x}_{0})
=(\mathbf{A}\mathbf{x}-\mathbf{b})^{H}\mathbf{R}^{-1}(\mathbf{A}\mathbf{x}-\mathbf{b}) (6)

The last result (6) verifies the alternative expression (3) to be equivalent to the standard hermitian function (1). QED.

[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.