Apply the product from right to left, choosing every Givens rotation from the current first column. First

J (1, 2)

replaces

(a_{1}, a_{2})

(a_{1}^{2} + a_{2}^{2}, 0)

. Next

J (1, 3)

combines that retained entry with

a_{3}

and sets the third entry to zero. Inductively, after eliminating coordinates

2, \dots, k

, the first coordinate is

a_{1}^{2} + \dots + a_{k}^{2},

(1)

and coordinates

2, \dots, k

are zero. Later Givens rotations involve only coordinate one and a new coordinate, so those zeros remain zero. Therefore

Q_{1} a = (∥ a ∥_{2}, 0, \dots, 0)^{T} = ∥ a ∥_{2} e_{1},

(2)

where

e_{1}

is the first standard basis vector. The zero vector is included by choosing the identity matrix for every zero pair. The nonnegative sign follows from the convention fixed in part (c)(ii). This is the first-column stage of QR decomposition by Givens rotations.

ii

 0  0

Solution

 0  0

Put

C = Q_{1} A

and write its second column as

v

. For

n \geq 3

, take

Q_{2} = J (2, n) J (2, n - 1) \dots J (2, 3),

(1)

where each angle is computed from the current second column, in the right-to-left application order. These Givens rotations eliminate its entries in rows

3, \dots, n

by accumulating their Euclidean norm in row two.

Each rotation acts only on rows with indices at least two. The first column of

C

∥ a ∥_{2} e_{1}

, whose entries in all those rows are zero. Mixing two zeros leaves them zero, and row one is untouched. Thus the first column is unchanged. The resulting matrix has the form

Q_{2} Q_{1} A = ∥ a ∥_{2} 00 ⋮ 0 v_{1} v_{2}^{2} + \dots + v_{n}^{2} 0 ⋮ 0 (n \geq 3) .

(2)

This verifies both required properties, including the case of zero second-column tail. When

n = 2

there are no rows to eliminate: take

Q_{2} = I_{2}

, the empty product. No sign change of

v_{2}

is required in that case. Each Givens row-pair update preserves previously eliminated columns, which is the central invariant of QR decomposition by Givens rotations.