Math 170A Assignment #3
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: daixieit
Math 170A Assignment #3
Due date and time: 11pm, August 27, 2023 (PDT)
1. A permutation matrix P is a matrix such that there is exactly one 1 in every row and every
colum and all other entries are zeros. Show that every permutation matrix is orthogonal.
Proof. Let ei denote the vector where the ithentry is 1 and all other entries are zeros. Then the permutation matrix P can be represented as
P = [ei1 ei2 ··· ein] .
To make P have exactly on e nonzero entry in every row and every column, P must be a square matrix, i.e., P e Rn 根n. Then
l e(e)」 l0(1) 1(0) ···(···) 0(0)」
PT P = ' i2 ' [ei1 ei2 ··· ein ] = ' .(.) .(.) .(.) ' .
[el [0(.) 0(.) ··· 1(.)l
So P is an orthogonal matrix.
2. Let Q e Rm 根n be a matrix. If〈Qx,Qy)=〈x,y)for all x,y e Rn , show that Q is isometric, i.e., show that QT Q = In.
Proof. 〈Qx,Qy)= (Qx)T (Qy) = xT QT Qy and〈x,y)= xTy. If〈Qx,Qy)=〈x,y)for all x,y e Rn, then
xT QT Qx = xTy 今 xT (QT Q — In)y = 0, Ax,y e Rn.
Simply choose x = ei, y = ej for all i,j = 1, 2,...,n, we have every entry of (QT Q — In) equal zero, so QT Q — In = 0.
3. Find a rotation matrix Q such that
Q [ 1(—)2(5)] = [ —013] .
Solution. Let Q = [ csin(os)θ(θ) os(si)nθθ], we can solve from the linear system that cos θ = — 12/13 and sinθ = 5/13. So Q = [ —5(1)/(2)1(/)3(13) —12/(5/1)13(2)].
4. Find real scalars c,s with c2 + s2 = 1 such that
'l0(1) c(0)
[0 s
for all real scalars a,b e R.
0s」' 'l 12(a)」' 'l 13(a)」'
c l(0') 5 l(b ') = 0(b) l(')
Solution. Note that
0(1) c(0)
[0 s
s''(」) ''(l)''(」) = ''(l)12c 5s''(」) .
0 c l[ 5 l [12s + 5cl
We can solve from 12c − 5s = 13, 12s + 5c = 0 for.
c = 1(1)3(2), s = − 13(5).
5. Let u ∈ Rn be with ∥u∥2 = 1 and let Q = In − 2uuT . Show that
QT = Q, Q−1 = QT , Q−1 = Q,
Qu = −u, Qx = x ⇐⇒ ⟨u,x⟩ = 0.
Proof. Since ∥u∥2 = √uTu = 1, we have uTu = 1. Then
()u(I)n= 0 = QT ,
6. Show that if τ = ±∥x∥2 is chosen such that τ and x1 have the same sign, show that every entry ui of the vector
l x2(1) 」 l−τ」
u := τ(x)x1(y) = ' τ x1 ' , y = ' '
[ τ x(n)1 l [0(.) l
satisfies |ui| ≤ 1 and show that 1 ≤ ∥u∥2 ≤ 2.
Proof. When i = 1, ui = 1. When i = 2,...,n,
|ui| = = ≤ ≤ 1.
Since u1 = 1, it is clear that ∥u∥2 ≥ 1. Note that
^2∥x∥2(2) + 2
=
∥x∥2 +
7. Find a reflector matrix Q such that
'l4(3)」' τ」'
Q ' 1 ' = ' 0 ' .
[1l [ 0 l
Solution. Denote this linear system as Qx = y. Since x1 > 0, we choose τ = IxI2 = ^32 + 42 + 1 + 32 + 1 = 6. Then
l4(9)」
u = x — y = 1 ' 1 ' , γ = τ + x1 = 3
[1l
1 ——3(2)6(7) 8(3)6 ——4(9) ——1(2)2(7) ——4(9)
Q = I5 — γuuT = ' —9 —4 53 —3 — 1 ' .
54 ' —27 — 12 —3 45 —3 '
[ —9 —4 — 1 —3 53 l
8. Let A = [ 13 2]. Find a reflector Q and an upper triangular R such that A = QR.
Solution.
τ = Ia1 I2 = ^10, γ = , u = l — .
Q = I — 2uuT = l —^0 — 0(0)] , R = QA = l^010 .
9. Find the least square solution to the overdetermined system
[(l) 1l(」) [x2(x1)] = [(l)l(」) .
Solution. Denote the above system as Ax = b. Apply the Gram-Schmidt process to get a QR decomposition of A.
r11 = Ia1 I2 = ^22 + 32 + 62, u1 = a1 = 1 l3(2)」
Ia1 I2 7 [6l ,
1
7
(2 · ( — 1) + 3 · 9 + 6 · 4) = 7,
r22 = Ia2 — r12u1 I = " l 36」" = ^32 + ( —6)2 + 22 = 7
u2 =
a2 — r12u1 |
Ia2 — r12u1 I2 |
l 36」.
[ 2 l
So we have Q = [(l) 6l(」) and R = [0(7) 7(7)]. The least square solution can be computed by
x = R-1 QT b = 0(1) 1][3(2)
36 2(6)] [(l)l(」) = 103(6)] .
10. Let u,v be the vectors
l 22」 l 2(1) 」
u = ' -2 ' , v = ' -3 ' .
[ 2 l [-4l
Apply the Gram-Schmidt process to find an orthonormal basis for the subspace span{u,v}. Use this to find an isometric matrix Q E R4根2 and an upper triangular R E R2根2 such that [u v] = QR.
Solution.
1 l 22」
r11 = IuI2 = 4, q1 = 4 [('')2l('') ,
r12 =〈q1 , v)= -1, r22 = Iv - r12 q1 I2 = ^29, q2 = 1 3(3)
2^29 -- 7l(7 ') .
So we have
1
Q = ' - -2 , R = [0(4) .] .
-
2023-09-21