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Problem 1
Let A =
[
]
. Find a QR-factorization of A.
Solution 1. We first find a suitable orthonormal basis for the span of
the columns of A. Let x 1 =
[
]
. Let x 2 =
[
]
. Next, let
v 1 = x 1 , and
(1) v 2 = x 2 โ
x 2 ยท v 1 v 1 ยท v 1
v 1 =
[
]
[
]
[
]
We now normalize v 1 and v 2. Let u 1 = (^) ||vv^11 || = โ^15
[
]
. Now, ||v 2 || = 11 โ 22 +1^2 5 =^ โ^11
- Let^ u^2 =^
v 2 ||v 2 ||. Thus,
(2) u 2 =
[
]
Let Q =
[
u 1 u 2
]
We now want to find an upper triangular matrix R with positive entries on the diagonal such that A = QR.
Let R =
[
a b 0 c
]
. We compute:
A = QR
[
]
[
] [
a b 0 c
]
[
]
[
a b + 2c 2 a 2 b โ c
]
Hence a =
- Equating the terms in the second columns we obtain the system:
3 =
b + 2c โ 5
โ5 =
2 b โ c โ 5
1
Multiplying the first equation by 2 gives the following system:
2 b + 4c โ 5
โ5 =
2 b โ c โ 5
Subtracting the first equation from the second equation, we get:
11 =
5 c โ 5 11 โ 5
= c.
Plugging this back into the first equation in (5), we get;
b + โ^225 โ 5
3
= b
= b.
A quick inspection shows that R has the desired properties, and that A = QR.
Remark 2. This problem is computationally a bit challenging without a calculator. (I did not use one.) Iโll make sure that the computations are a bit nicer on the actual exam.
Problem 2 Let A be an m ร n matrix such that AT^ A is invertible. Show that the columns of A are linearly independent.
Solution 3. Consider the equation Ax = 0. Weโd like to show that this equation has only the trivial solution, since that is a necessary and sufficient condition for A to have linearly independent columns. We apply AT^ to both sides to get AT^ Ax = AT^0 which is equivalent to AT^ Ax = 0. Since AT^ A is invertible, this last equation has only the trivial solution. Thus, x = 0 , as desired.
Remark 4. This was a homework problem. It is part of the proof of Theorem 14 in section 6.5.
