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EE103 (Winter 2006-07)
3. Triangular matrices
terminology •
forward and backward substitution •
inverse •
3–
Terminology
j > i for = 0^ ija if lower triangular is A a square matrix
= A
00 00 · · ·· · ·^220 a^1121 a^ a
0 nna^1 1 −−,n^1 −n,nnaa · · ·· · ·^2 ,^21 n−ana^1 ,^11 n−^ an a
is lower triangular) TA( j < i for = 0^ ija if upper triangular is A
i for all = 1^ iia upper/lower triangular if unit a triangular matrix is
if the diagonal elements are nonzero nonsingular a triangular matrix is
Forward substitution
lower triangular and nonsingular A with b = Ax solve
... 00.. .· · ·· · · 22 ... 0 a 1121 ... a a
nna · · · 2 na 1 n a
12 ... x^ x
n x
=^
12 ... b^ b
n b
: algorithm
(^11) /a (^1) b := (^1) x
(^22) /a) (^1) x (^21) a − (^2) b( := (^2) x
(^33) /a) (^2) x (^32) a − (^1) x (^31) a − (^3) b( .:= (^3) x
nn/a) 1 −nx 1 − n,na − · · · − (^2) x 2 na − (^1) x 1 na − nb( := n x
flops 2 n 1) = − n + (2 · · · 1 + 3 + 5 + : cost
3–3 Triangular matrices
Recursive formulation
)A matrix n × n (for block matrix formulation
0 11 a [
22 A 21 A
(^1) x ] [
2 X
(^1) b [^ = ]
2 B
]
n × n is^22 A , 1 × 1) − n( is^21 A is scalar,^11 a •
-vector1) − n( is an^2 B is scalar,^1 b -vector,1) − n( is an^2 X is scalar,^1 x •
is nonsingular and lower triangular^22 A , 6 = 0^11 a •
forward substitution
(^11) /a (^1) b := (^1) x 1.
by forward substitution^1 x^21 A −^2 B =^2 X^22 A 2. solve
Left inverse of a triangular matrix
, i.e. ,I = T Y TA as I = Y A to compute a left inverse, write
] ne^ · · ·^2 e^1 e [^ = ] nY^ · · ·^2 Y^1 Y [^ T^ A
)n th unit vector (of sizek is^ ke ; T Y of k is column^ k Y
sets of linear equations n ’s by solvingkY then compute the
ne = nY TA... , , (^2) e = (^2) Y TA , (^1) e = (^1) Y T A
using forward or backward substitution
is triangular and nonsingular, then it has a left inverse A if conclusion:
3–7 Triangular matrices
Inverse of a triangular matrix
if right and left inverse exist, they must be equal:
X = X) Y A ) = (AX( Y = Y
1 −A^ is triangular and nonsingular, then it has an inverse^ A^ : if^ conclusion
I = A 1 −A = 1 − AA
is lower triangular A is lower triangular if 1 −A •
is upper triangular A is upper triangular if 1 −A •
T) 1 −A = ( 1 −) TA(^ •
b 1 −A = x can be expressed as b = Ax solution of •
Summary
is triangular and nonsingular (has nonzero diagonal elements), then: A if
flops 2 n can be solved in b = Ax •
b is solvable for all b = Ax has a full range: A •
= 0 x is = 0 Ax has a zero nullspace: unique solution of A •
b is solvable for all b = x TA has a full range: TA •
= 0 x is = 0 x TA has a zero nullspace: unique solution of TA •
has an inverse A •
has an inverse TA •
3–9 Triangular matrices
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