Exam (elaborations)
COMP 361 or 5611 Assignment 1 complete material update Concordia University
- Course
- COMP 361 or 5611
- Institution
- Concordia University ( )
COMP 361 or 5611 Assignment 1 complete material update Concordia University
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COMP 361 or 5611 Assignment 1 complete material update 2024-2025 ConcordiaUniversity
Each problem is worth equal marks.
Problem 1. Let
1 1 0
A = −2 2 0 .
0 0 3
(a) Find ǁAǁ1 and ǁAǁ∞.
(b) Determine the eigenvalues of AT A.
(c) Find ǁAǁ2.
Solution
(a) ǁAǁ1 = max{1+2+0, 1+2+0, 0+0+3} = 3, ǁAǁ∞ = max{1+1+0, 2+2+0, 0+0+3} = 4.
(b)
1 −2 0 1 1 0 5 −3 0
AT A = 1 2 0 −2 2 0 = −3 5 0
0 0 0 0 3 0 0
The eigenvalues of AT A are the 3solutions, λ, of det(AT A − λI) =90.
5−λ −3 0
det(AT A − λI) = det −3 5−λ 0
0 0 9−λ
= (λ − 8)(λ − 2)(9 − λ)
So the eigenvalues of AT A are {2, 8, 9}.
√ √ √
(c) ǁAǁ2 = max{ 2, 8, 9} = 3.
Problem 2. Let A be an n × n matrix and let C be the maximum absolute column sum of
A, that is,
n
Σ
C = max a| ij .
j
i=1
n
(a) Prove that for any nonzero vector x ∈ R , we
| have
ǁAxǁ1
ǁxǁ1 ≤ C.
1
, (b) Show that we can always find a vector y such that
ǁAyǁ1
ǁyǁ1 = C,
and conclude that ǁAǁ1 = C.
Solution
(a) Let x 0. Then
Σn Σ n
a x|
ǁAxǁ1 | j=1 ij j
i=1
ǁxǁ1 = ǁxǁ1
Σn Σ n |a ||x |
i=1 j=1 ij j
≤
ǁxǁ1
Σn Σni=1 |aij ||xj |
j=1
=
Σn Σ ǁxǁ1
| |aij |
n
j=1 |x i=1
=
j ǁxǁ1
Σn |xj | Cǁxǁ1
j=1
C =
≤
ǁxǁ1 ǁxǁ1
=
C.
(b)
Σ Let kn be the index of the column of A achieving the maximum absolute column sum; so
C = i=1 |aik|. Let y be the vector such that yk = 1 and yi = 0 for all i /= k. Then Ay is
Σn
equal to the kth column of A, so ǁAyǁ1 = i=1 |aik| = C. Also, ǁyǁ1 = 1, so ǁAyǁ 1
ǁyǁ
= C.
1
This proves that ǁAǁ1 ≥ C, which together with part (a) proves that ǁAǁ1 = C.
Problem 3. If x = (x1, . . . , xn) and y = (y1, . . . , yn) are vectors in Rn, then the following
inequality, called the Cauchy-Schwartz inequality, is always true:
n !2 ! !
n n
2Σ 2
xiyi ≤ xi yi .
Σi=1 Σi=1 i=1
Using the Cauchy-Schwartz inequality, prove that
ǁx + yǁ2 ≤ ǁxǁ2 + ǁyǁ2
(this is the triangle inequality for the two-norm).
Solution
2
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