a. Show that the product of two $n times n$ lower triangula Quizlet
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8/11/22, 3:41 AM
a. Show that the product of two $n \times n$ lower triangula | Quizlet
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Question
a. Show that the product of two n × n lower triangular matrices is lower triangular. b. Show
that the product of two n × n upper triangular matrices is upper triangular. c. Show that the
inverse of a nonsingular n × n lower triangular matrix is lower triangular.
Solution
Verified
Step 1
1 of 3
= [lij ] is lower -triangular matrix if lij = 0 for
each i = 1, 2, … , j − 1. Let A and B be two n × n lower triangular
(a) A square matrix L
matrices. So
aij = bij = 0
Now AB
for each i = 1, 2, … , j − 1
n
= [cij ] is given by cij = ∑k=1 aik bkj so if i ≤ j − 1 then
j−1
n
n
∑ aik bkj = ∑ aik bkj + ∑ aik bkj
k=1
k=1
k=j
= 0 since k ≤ j − 1 and on second part of above
sum aik = 0 since i ≤ j − 1 < k . Thus, we have
On first part, bkj
j−1
n
n
∑ aik bkj = ∑ aik bkj + ∑ aik bkj = 0
k=1
i.e. cij
k=1
k=j
= 0 if i = 1, 2, … , j − 1. Thus AB be lower triangular as well.
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Step 2
2 of 3
(b) An upper-triangular n × n matrix U
= [uij ] has, for each j =
1, 2, … , n, the entries
uij = 0,
for each i = j + 1, j + 2, … , n
Suppose A and B be two n × n upper triangular matrices. If A
=
[aij ] and B = [bij ] then
⎡
⎤
AB = (∑ aik bkj )
⎣ k=1
⎦
ij
Now suppose that i
k ≥ j + 1 then bkj
n
≥ j + 1 then if k ≤ j + 1 then aik = 0 and if
= 0 so we get
j
n
n
∑ aik bkj = ∑ aik bkj + ∑ aik bkj = 0 + 0
k=1
k=1
k=j+1
Thus the matrix AB is also upper triangular.
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Step 3
3 of 3
(c) Suppose that A is non-singular lower triangular matrix and if A−1 is
not lower triangular then there exists element a−1
lm , where 1
≤ l, m ≤
n and l ≤ m. Now, take element lk -th element of A−1 A which is
n
k
−1
∑ a−1
lj ajk = ∑ alj ajk
j=1
j=1
= 1, 2, … , n then we have
If we take k for each k
k
∑ a−1
lj ajk
j=1
If we exclude the case that l
1
={
0
if l = k
otherwise
= k then we get get
k
∑ a−1
lj ajk = 0,
for each k = 1, 2, … , l − 1, l + 1, … , n
j=1
Now we know that the columns of A is linearly independent since A is
−1
= 0 with m ≥ l, this contradicts
that columns of A are linearly independent i.e. A is non-singular. Thus,
A−1 must also be lower triangular.
non-singular, and we know that alm
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matrices is upper triangular.
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