Here is the problem:
$$\mathbf{y}=\mathbf{Ax}+\mathbf{b}$$
where $\mathbf{y,x,b}$ are vectors, and$\mathbf{A}$ is matrix(generally rectangular, but with full column rank).
The least squre solution is denoted by:
$$\hat{\mathbf{x}}=\mathbf{A^+y} \quad (1)$$.
Now my problem is to prove that the $i$-th component of $\hat{\mathbf{x}}$, $\hat{x}_i$ is given by:
$$\bf \frac{a_i^H(I-A_iA_i^+)}{||(I-A_iA_i^+)a_i||^2} y\quad (2)$$
where $\bf a_i$ is the $i$-th column of $\bf A$, and $\bf A_i$ is $\bf [a_1,a_2,\cdots,a_{i-1},a_{i+1},\cdots,a_n]$, i.e., $\bf A$ with $\bf a_i$ deleted.
(2) is called ``decorrelator'', i.e., it contains contribution from $ x_i$ and $\bf b$, but no contribution from $x_j, (j\neq i)$.
This is asserted in a book, But I still cannot prove it. Thank you!
Assume the partitioning $A:=[B, c]\in\mathbb{C}^{m\times n}$ where $B$ has $n-1$ columns. Since $A$ has full column rank, $A^+=(A^HA)^{-1}A^H$. We have $$ A^HA=\pmatrix{B^HB & B^Hc\\c^HB & c^Hc}. $$ Using the block inversion formula, we have (with $*$ denoting "something unimportant") $$ (A^HA)^{-1}=\pmatrix{*&*\\-\alpha c^HB(B^HB)^{-1}&\alpha}, $$ where $$ \alpha:=\frac{1}{c^Hc-c^HB^H(B^HB)^{-1}Bc}=\frac{1}{\|(I-BB^+)c\|_2^2}. $$ Hence $$ A^+=\pmatrix{*&*\\-\alpha c^HB(B^HB)^{-1}&\alpha}\pmatrix{B^H\\c^H} $$ and the last row of $A^+$ is given by $$ \alpha c^H[I-B(B^HB)^{-1}B^H]=\frac{c^H(I-BB^+)}{\|(I-BB^+)c\|_2^2}. $$ You can see that this last row of $A^+$ has the same form as the fraction in (2) for $i=n$. To prove the general statement, you can consider a permutation matrix $P$ such that $AP$ has $a_i$ in the last column and use the fact that $(AP)^+=P^TA^+$.