To illustrate the process, let us consider the KKT conditions
⋱−I−IQkSkAkSk⊤RkBkAk⊤Bk⊤−I−IPk+1λkxkukλk+1xk+1=−ck−1qkrkckpk+1.
Using the last row we can express xk+1 as
xk+1=Pk+1−1λk+1−Pk+1−1pk+1
and consequently eliminate it from the KKT system
⋱−I−IQkSkAkSk⊤RkBkAk⊤Bk⊤−Pk+1−1λkxkukλk+1=−ck−1qkrkck+Pk+1−1pk+1.
In a similar fashion we may eliminate
λk+1=Pk+1(Akxk+Bkuk)+Pk+1ck+pk+1,
attaining
⋱−I−IQk+Ak⊤Pk+1AkSk+Bk⊤Pk+1AkSk⊤+Ak⊤Pk+1BkRk+Bk⊤Pk+1Bkλkxkuk=−ck−1qk+Ak⊤(pk+1+Pk+1ck)rk+Bk⊤(pk+1+Pk+1ck).
Finally, we may also eliminate
uk=−(Rk+Bk⊤Pk+1Bk)−1(rk+Bk⊤(pk+1+Pk+1ck))=−(Rk+Bk⊤Pk+1Bk)−1(Sk+Bk⊤Pk+1Ak)xk
to close the recursion
[⋱−I−IPk][λkxk]=−[ck−1pk]
where
Pkpk=Qk+Ak⊤Pk+1Ak−(Sk+Bk⊤Pk+1Ak)⊤(Rk+Bk⊤Pk+1Bk)−1(Sk+Bk⊤Pk+1Ak)=qk+Ak⊤(pk+1+Pk+1ck)=−(Sk+Bk⊤Pk+1Ak)⊤(Rk+Bk⊤Pk+1Bk)−1(rk+Bk⊤(pk+1+Pk+1ck)).
This process can be repeated for k←k−1.