Math 170A Assignment #2
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Math 170A Assignment #2
Due date and time: 11pm, August 20, 2023 (PDT)
1. Consider the following matrix A and vector b
l −4 −4 −4 − 1 」 l −23 」
Find the unit lower triangular matrix L and upper triangular matrix U such that A = LU. Find the vector y such that Ly = b and find x such that Ux = y. Verify that x is a solution to Ax = b. Show all computational steps.
Solution. First let’s do Gaussian elimination by row operations.
4 A → ' 7 [ 7 |
−60 |
4 −51 − 17 |
1 4(1)0 |
」 ' l |
So we have A = LU with L = 7(1) [7 |
0 1 6 |
4
0(0) 4 1l [ 0 |
0 |
''(l) 4 9(4) 1 ''(」)
1
−4l
Then we solve Ly = b with forward substitution. Since each ljj = 1 for j = 1, 2, 3, 4, we have
y1 = b1 = −23,
y2 = b2 − l21y1 = − 114 − 7 · ( −23) = 47,
y3 = b3 − l31y1 − l32y2 = −507 − 7 · ( −23) − ( −6) · 47 = −64,
y4 = b4 − l41y1 − l42y2 − l43y3 = −273 − 7 · ( −23) − 4 · 47 − 5 · ( −64) = 20. So y = [−23, 47, −64, 20]T .
We solve Ux = y with backward substitution.
x4 = y4 /u44 = 20/(−4) = −5,
x3 = (y3 − u34x4 )/u33 = ( −64 − 2 · ( −5))/9 = 6,
x2 = (y2 − u24x4 − u23x3 )/u22 = (47 − 2 · ( −5) − 7 · 6)/(− 15) = − 1,
x1 = (y1 − u14x4 − u13x3 − u12x2 )/u11 = ( −23 − ( − 1) · ( −5) − ( −4) · 6 − ( −4) · ( − 1))/(−4) = 2.
So x = [2, − 1, 6, −5]T .
Finally, we verify that Ax = b.
Ax = ' −28
[ 2(2)8(8)
−88 −45
5(1) = .
2. For the following matrix and vector
l —15(5)
A = ' 5 [
— 15 —5
—51 —51
—51 —224
— 12 —69
—69 ' , b = ' 410 ' ,
—37l [ — 151l
find a unit lower triangular matrix L, a diagonal matrix D such that A = LDLT . Find the vector y such that Ly = b, find w such that Dw = y, and find x such that LTx = w, Verify that x is a solution to Ax = b. Show all computational steps.
Solution. First let’s do Gaussian elimination by row operations to obtain the LU decompo-
sition of A. l —5 A → ' [ |
— 15 —6 —36 — 12 |
—5 0 —36 — 12 —219 —69 —69 —37 |
」 ' → l |
5 |
—5 —36 —3 3 |
0 — 12 3 — 13 |
」 ' → l |
— 6(15) —356 [ 0 2 |
0 — 12 3 — 10 |
」 ' l |
Since A is a symmetric matrix, we have U = DLT, where
l3(1) L = ' 1 [0 |
0 1 6 2 |
0 0 1 — 1 |
0(0) l 5 1l [ 0 |
0 —6 0 0 |
0 0 —3 0 |
0(0) — 10l |
So the above matrices L, D satisfy A = LDLT .
Then we solve Ly = b with forward substitution. Since each ljj = 1 for j = 1, 2, 3, 4, we have y1 = b1 = 25,
y2 = b2 — l21y1 = 3 — 3 · 25 = —72,
y3 = b3 — l31y1 — l32y2 = —410 — 1 · 25 — 6 · ( — 72) = —3,
y4 = b4 — l41y1 — l42y2 — l43y3 = — 151 — 0 · 25 — 2 · ( — 72) — ( — 1) · ( —3) = — 10. So y = [25, —72, —3, — 10]T . Then we solve Dw = y for w.
w1 = y1 /d11 = 25/( — 5) = —5,
w2 = y2 /d22 = — 72/( —6) = 12,
w3 = y3 /d33 = —3/( —3) = 1,
w4 = y4 /d44 = — 10/( — 10) = 1.
We get w = [ —5, 12, 1, 1]T . And we compute for the solution to LTx = w by
x4 = w4 = 1,
x3 = w3 — l43x4 = 1 — ( — 1) · 1 = 2,
x2 = w2 — l42x4 — l32x3 = 12 — 2 · 1 — 6 · 2 = —2,
x4 = w1 — l41x4 — l31x3 — l21x2 = —5 — 0 · 1 — 1 · 2 — 3 · ( — 2) = — 1. Then x = [ — 1, —2, 2, 1]T . Finally we verify that Ax = b with the above x.
l —15(5)
Ax = ' 5 [
— 15 —5
—51 —51
—51 —224
— 12 —69
—012 —2(1) 3(25)
69 ' ' 2 ' = ' 410 ' .
—37l[ 1 l [ — 151l
3. Let A be a symmetric positive definite matrix. Show that there exist a unit lower triangular matrix L and a diagonal matrix D such that A = LDLT, where D has positive diagonal entries.
Proof. Consider the Cholesky decomposition of A: A = RTR, where R is an upper triangle matrix with positive diagonal entries. We can express R = D(ˆ)LT with a diagonal matrix D and a unit lower triangle matrix L. Let D :=D(ˆ)2. The D is diagonal and each diagonal entry
is positive. We have
A = RTR = (D(ˆ)LT)T (D(ˆ)LT) = LDLT .
4. Let A = [0 A22(A11 A12)] be a 2 × 2 block matrix, for submatrices A11 ∈ Rn1 ×n1 , A12 ∈ Rn1 ×n2 ,
A22 ∈ Rn2 ×n2 . The 0 is a zero matrix of dimension n2 × n1 . Suppose A11 = L1U1 and A22 = L2U2 are LU factorizations, where L1, L2 are unit lower triangular and U1, U2 are upper triangular. Use L1, L2 , U1 , U2 , A12 to construct a LU factorization for A.
Solution. Denote the LU factorization of A in form of block matrices,
[0(A11) A(A)2(1)2(2)] = [L(L)2(1)1(1) L22(0)][0(U11) U22(U12)],
石 > 石 >石 >
A L U
where L11, L22 are unit lower triangle matrices and U11, U22 are upper triangle matrices. It implies that
A11 = L11U11, A12 = L11U12 , 0 = L21U11, A22 = L21U12 + L22U22 .
Since the LU factorization of A2, A2 are unique, by comparing the terms of A1 = L1U1 and A22 = L2U2, we have
L11 = L1, L21 = 0, L22 = L2
U11 = U1, U12 = L1(−)1 A12, U22 = U2 .
So the LU factorization of A is
A = [0(L)1
5. For p ≥ 1, define the function
L(0)2][0(U)1
L1 U2(−1 A)12] .
∥x∥p :=^p |x1 |p + ··· + |xn|p.
Show that this is a norm function on Rn and p ∥x∥p = ∥x∥∞ .
Proof. First, we prove that ∥x∥p is a norm function.
(a) For every x ∈ Rn , ∥x∥p = ^p |x1 |p + ··· + |xp|p ≥ |xi| ≥ 0 for all i = 1,...,n. When x = 0, it is clear that ∥x∥p = 0. Conversely, if ∥x∥p = 0, then 0 = ∥x∥p ≥ |xi| ≥ 0 for all
i = i,...,n, so x = 0.
(b) For every c ∈ R and every x ∈ Rn ,
∥cx∥p = ^p |cx1 |p + ··· + |cxn|p = ^p |c|p · (|x1 |p + ··· + |xn|p ) = |c|∥x∥p.
(c) For every x,y ∈ Rn, denote ˆ(x) = x/(∥x∥p+ ∥y∥p) and ˆ(y) = y/(∥x|p+ ∥y∥p), then
∥x + y∥p ≤ ∥x∥p+ ∥y∥p ⇔ ∥ˆ(x) + ˆ(y)∥ ≤ ∥ˆ(x)∥p+ ∥ˆ(y)∥p = 1.
Let λ := ∥ˆ(x)∥p, then ∥ˆ(y)∥p = 1 − λ, and 0 ≤ λ, 1 − λ ≤ 1. We have
"
∥ˆ(x) + ˆ(y)∥p(p) = "λ ·
=
≤
+ (1 − λ) · p(p)
' λ · + (1 − λ) · p
(λ + (1 − λ)p
≤ λ + (1 − λ) ≤ λ + (1 − λ) = 1.
The second last inequality holds due to the convexity of the function f(x) = xp , p ≥ 1, and the last inequality holds since ∥ˆ(x)∥p+ ∥ˆ(y)∥p = 1.
Then we prove that limp→+∞∥x∥p = ∥x∥∞. For every x ∈ Rn , we assume that |x1 | ≥ |xi| for all i = 1,...,n without loss of generality. Then ∥x∥∞ = |x1 | and for all p ≥ 1,
√p 1 ≤ ∥∥x(x)∥ ∞ = 1p 1 + x(x)1(2) p + ··· + || x(x)1(n)|| p ≤ √p n.
Since √pn → 1 asp → ∞ , we get limp→∞∥x∥p = ∥x∥∞ .
6. Suppose A has the Cholesky factorization A = RTR, for a nonsingular upper triangular matrix R. Show that ∥x∥A := √xTAx gives a norm function.
Solution. With the Cholesky factorization A = RTR, we have
∥x∥A = √xTAx = √xTRTRx = 4 ∥Rx∥2(2) = ∥Rx∥2 ,
where ∥·∥2 is a norm function we proved in Question 5. Then let us verify that ∥x∥A = ∥Rx∥2 is also a norm function.
(a) For every x ∈ Rn , ∥x∥A = ∥Rx∥2 ≥ 0 and ∥Rx∥2 = 0 if and only if Rx = 0. Since R is nonsingular, Rx = 0 if and only if x = 0. So ∥x∥A = 0 if and only if x = 0.
(b) For every c ∈ R, x ∈ Rn , ∥cx∥A = ∥R(cx)∥2 = |c|∥Rx∥2 = |c|∥x∥A .
(c) For every x,y ∈ Rn, we have
∥x + y∥A = ∥R(x + y)∥2 = ∥Rx + Ry∥2 ≤ ∥Rx∥2 + ∥Ry∥2 = ∥x∥A + ∥y∥A .
7. For a square matrix A = (aij)1≤i,j≤n, define the function
∥A∥max := max |aij| .
Show that ∥A∥maxis a norm function in the matrix space Rn×n , but
2023-09-11