3.1 
det([A_{[m#m]}, B_{[m#n]}; C_{[n#m]}, D_{[n#n]}]) = det(A)*det(DCA^{1}B) =
det(D)*det(ABD^{1}C)
 [A B; C D] = [A 0; C
I] [I A^{1}B; 0
DCA^{1}B] = [I B;
0 D] [ABD^{1}C
0; D^{1}C I]
 The result follows since the determinant of a block triangular matrix is
the product of the determinant of the diagonal blocks.

3.2 
det(I_{[n#n]} +
A_{[m#n]}^{T}B_{[m#n]})
= det(I_{[m#m]} +
AB^{T}) =
det(I_{[n#n]} +
B^{T}A) =
det(I_{[m#m]} +
BA^{T})
 det(I + A^{T}B) = det((I +
A^{T}B)^{T}) = det(I +
B^{T}A) [since det
X^{T} = det X]
 det([I_{[m#m]}, B;
A^{T}, I_{[n#n]}]) =
det(I)*det(I + A^{T}IB) =
det(I)*det(I + BIA^{T})
[3.1]
 Hence det(I + A^{T}B) = det(I +
BA^{T}) = det((I +
BA^{T})^{T}) = det(I +
AB^{T})

3.3 
det(I+xy^{T}) =
1+y^{T}x
 det(I+xy^{T}) =
det(I_{[1#1]}+y^{T}x)
[3.2]
 = 1+y^{T}x

3.4 
det(A+xy^{T}) = det(A) ×
(1+y^{T}A^{1}x)
 det(A+xy^{T}) = det(A) *
det(I+A^{1}xy^{T}) = det(A)
* (1+y^{T}A^{1}x)
[3.3]

3.5 
If M =
[A_{[m#m]}, B; C,
D_{[n#n]}] then M^{1} =
[Q^{1}, Q^{1}BD^{1};
D^{1}CQ^{1},
D^{1}(I+CQ^{1}BD^{1})]
where Q =ABD^{1}C and also M^{1} =
[A^{1}(I+BP^{1}CA^{1}),
A^{1}BP^{1};
P^{1}CA^{1}, P^{1}] where
P_{[n#n]}=DCA^{1}B
 [A, B; C, D] [Q^{1},
Q^{1}BD^{1};
D^{1}CQ^{1},
D^{1}(I+CQ^{1}BD^{1})]
=[AQ^{1}BD^{1}CQ^{1},
AQ^{1}BD^{1}+BD^{1}(I+CQ^{1}BD^{1});
CQ^{1}DD^{1}CQ^{1},
CQ^{1}BD^{1}+DD^{1}(I+CQ^{1}BD^{1})]
=[(ABD^{1}C)Q^{1},
AQ^{1}BD^{1}+BD^{1}+BD^{1}CQ^{1}BD^{1};
CQ^{1}CQ^{1},
CQ^{1}BD^{1}+(I+CQ^{1}BD^{1})]
=[I,
(ABD^{1}C)Q^{1}BD^{1}+BD^{1};
0, I] = [I, IBD^{1}+BD^{1}; 0, I] = [I, 0; 0,
I] = I
 Define W = [0, I_{[n#n]};
I_{[m#m]}, 0] and hence
W^{1} = [0, I_{[m#m]};
I_{[n#n]}, 0] and WMW^{1} =
[D, C; B, A]
 From above WM^{1}W^{1} =
(WMW^{1})^{1} = [P^{1},
P^{1}CA^{1};
A^{1}BP^{1},
A^{1}(I+BP^{1}CA^{1})]
 Hence M^{1} =
W^{1}[P^{1},
P^{1}CA^{1};
A^{1}BP^{1},
A^{1}(I+BP^{1}CA^{1})]W =
[A^{1}(I+BP^{1}CA^{1}),
A^{1}BP^{1};
P^{1}CA^{1}, P^{1}]

3.6 
If S is +ve semidefinite
Hermitian, then a^{H}Sb^{2} <=
a^{H}Sa×b^{H}Sb
for any a, b. Also s_{i,j} <=
sqrt(s_{i,i}s_{j,j})
 [a b]^{H}S[a b] is +ve semidefinite
Hermitian and so its determinant is nonnegative
 det([a b]^{H}S[a b]) =
det(a^{H}Sa a^{H}Sb;
b^{H}Sa b^{H}Sb) =
a^{H}Sa×b^{H}Sb
 a^{H}Sb^{2} >= 0
 If we take a = e_{i} and b =
e_{j}, then we get s_{i,j}^{2}
<= s_{i,i}s_{j,j}

3.7 
If B is +ve definite and A
is +ve semidefinite then B^{1}A is diagonalizable and
has nonnegative eigenvalues.
 If S is the +ve definite hermitian square root of
B^{1} (i.e. S^{2}B=I) then
B^{1}A = S (SAS) S^{1} so
B^{1}A and SAS and so have the same
eignevalues.
 SAS = S^{H}AS and so is +ve
semidefinite and so has nonnegative eigenvalues and, since it is hermitian, is
unitarily diagonalizable.
