MMSE transceivers with zero-forcing equalizers - Signal Processing and Optimization for Transceiver Systems - page 424

σ q Tr Σ f ( U f Σ h U f ) M − 1

=

σ q Tr ( U f Σ h U f ) M − 1 Σ − f = σ q

M− 1

B kk
σ f,k

= ,

k =0

where ( A ) M denotes the M M leading principal submatrix of A as usual.

In the preceding, B kk > 0 are the diagonal elements of the Hermitian positive
definite matrix

×

B = ( U f Σ h U f ) M − 1 .

(12 . 84)

Here it is assumed that σ f,k > 0 (i.e., that F has rank M ). Let us take a closer
look at the matrix B and its relation to U f . Write the unitary matrix U f in
partitioned form:

MP − M

U f =

.

U 1 U 2

If we replace U f with

MP − M

U f,new =

U 1 U b U 2 (12 . 85)

for some M × M unitary U b , then B in Eq. (12.84) is replaced with (Problem

12.5)

B new = U b BU b . (12 . 86)
So, as far as B is concerned, the freedom to choose U f is equivalent to the
freedom to choose the unitary matrix U b .

12.B.3 Conversion to a majorization problem

To proceed further we will use the ideas of majorization and Schur convexity
reviewed in Chap. 21. If we assume the ordering

σ f, 0 ≤ σ f, 1 ≤ ... ≤ σ f,M− 1 (12 . 87)

then it can be shown that

B 00 ≤ B 11 ≤ ... ≤ B M− 1 ,M− 1 . (12 . 88)

If this is not the case then we can interchange pairs of elements B kk and B mm
to reduce the MSE further (as we did in Sec. 12.4.1); we have the freedom to do
that by interchanging the columns of the matrix U b . So without loss of generality
we assume the above orderings. Now, from Ex. 1 in Sec. 21.4.1 (Chap. 21), we
know that the sum

M− 1

a k x k (12 . 89)

k =0

is Schur-concave in the region

x 0 ≤ x 1 ≤ ... ≤ x M− 1