| 3.1 |
det([A[m#m], B[m#n]; C[n#m], D[n#n]]) = det(A)*det(D-CA-1B) =
det(D)*det(A-BD-1C)
- [A B; C D] = [A 0; C
I] [I A-1B; 0
D-CA-1B] = [I B;
0 D] [A-BD-1C
0; D-1C 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]TB[m#n])
= det(I[m#m] +
ABT) =
det(I[n#n] +
BTA) =
det(I[m#m] +
BAT)
- det(I + ATB) = det((I +
ATB)T) = det(I +
BTA) [since det
XT = det X]
- det([I[m#m], -B;
AT, I[n#n]]) =
det(I)*det(I + ATIB) =
det(I)*det(I + BIAT)
[3.1]
- Hence det(I + ATB) = det(I +
BAT) = det((I +
BAT)T) = det(I +
ABT)
|
| 3.3 |
det(I+xyT) =
1+yTx
- det(I+xyT) =
det(I[1#1]+yTx)
[3.2]
- = 1+yTx
|
| 3.4 |
det(A+xyT) = det(A) ×
(1+yTA-1x)
- det(A+xyT) = det(A) *
det(I+A-1xyT) = det(A)
* (1+yTA-1x)
[3.3]
|
| 3.5 |
If M =
[A[m#m], B; C,
D[n#n]] then M-1 =
[Q-1, -Q-1BD-1;
-D-1CQ-1,
D-1(I+CQ-1BD-1)]
where Q =A-BD-1C and also M-1 =
[A-1(I+BP-1CA-1),
-A-1BP-1;
-P-1CA-1, P-1] where
P[n#n]=D-CA-1B
- [A, B; C, D] [Q-1,
-Q-1BD-1;
-D-1CQ-1,
D-1(I+CQ-1BD-1)]
=[AQ-1-BD-1CQ-1,
-AQ-1BD-1+BD-1(I+CQ-1BD-1);
CQ-1-DD-1CQ-1,
-CQ-1BD-1+DD-1(I+CQ-1BD-1)]
=[(A-BD-1C)Q-1,
-AQ-1BD-1+BD-1+BD-1CQ-1BD-1;
CQ-1-CQ-1,
-CQ-1BD-1+(I+CQ-1BD-1)]
=[I,
-(A-BD-1C)Q-1BD-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-1W-1 =
(WMW-1)-1 = [P-1,
-P-1CA-1;
-A-1BP-1,
A-1(I+BP-1CA-1)]
- Hence M-1 =
W-1[P-1,
-P-1CA-1;
-A-1BP-1,
A-1(I+BP-1CA-1)]W =
[A-1(I+BP-1CA-1),
-A-1BP-1;
-P-1CA-1, P-1]
|
| 3.6 |
If S is +ve semidefinite
Hermitian, then |aHSb|2 <=
aHSa×bHSb
for any a, b. Also |si,j| <=
sqrt(si,isj,j)
- [a b]HS[a b] is +ve semidefinite
Hermitian and so its determinant is non-negative
- det([a b]HS[a b]) =
det(aHSa aHSb;
bHSa bHSb) =
aHSa×bHSb
- |aHSb|2 >= 0
- If we take a = ei and b =
ej, then we get |si,j|2
<= si,isj,j
|
| 3.7 |
If B is +ve definite and A
is +ve semidefinite then B-1A is diagonalizable and
has non-negative eigenvalues.
- If S is the +ve definite hermitian square root of
B-1 (i.e. S2B=I) then
B-1A = S (SAS) S-1 so
B-1A and SAS and so have the same
eignevalues.
- SAS = SHAS and so is +ve
semidefinite and so has non-negative eigenvalues and, since it is hermitian, is
unitarily diagonalizable.
|