Matrix Decomposition: Example Solved Problems

MATRIX DECOMPOSITION

WORKED EXAMPLES

Example 1

Find <x,y> if 

Solution:

 <x, y> = 1(4)+2(−5)+3(6) = 4−10+18 = 12

Example 2

Find <x,y> if x = [20, −4, 30, 10] and y = [10,−5,−8,−6].

Solution:

 <x, y> = 20 (10) + (− 4) (− 5) + (30) (−8) + 10 (− 6)

 = 200+20−240−60

 <x, y> = −80

Example 3

Find the magnitude of x = [2, −3, −4]

Solution:

The magnitude is defined as <x, x>

 <x, x> = 2(2)+(−3) (− 3) + (−4) (−4)

 = 4+9+16

 = 29

  The magnitude of x is ||x|| = √29

Example 4

Given that . Find <x, y>, <x, x> and <y,y>.

Solution:

(i) <x,y> = 1(3)+2(4)=3+8=11

(ii) <x, x> = (1)² + (2)² = 5

(iii) <y, y> = (3)² + (4)² = 25

(iv) || x || = √<x, x> = √5

(v) || y || = √<x, y> = √25 = 5

Example 5

Given that x = [10, 20, 30] and y = [5,− 7,3]. Find <x,y>, <x, x> and <y,y>.

Solution:

(i) <x, y> = 10(5) + 20(−7) + 30(3) = 50−140+90=0

(ii) <x, x> = (10)² + (20)² + (30)² = 100 + 400 + 900 = 1400

(iii) <y, y> = (5)² + (−7)² + (3)² = 25+49 +9=83

Example 6

Let  . Verify whether the given vectors are orthogonal or not.

Solution:

Let 

 <x, y> = 1 (1) + 1 (1) + 1 (− 2) = 1+1−2 = 0

 <y, z> = 1 (1) + 1 (− 1) + (− 2) (0) = 1−1+0 = 0

 <z, x> = 1 (1) + (1) (−1) + 1 (0) = 1−1+0 = 0

Since <x, y> = <y, z> = <z, x> = 0 is an orthogonal set because each vector is orthogonal to every other vector.

Example 7

Let  . Verify these vectors are orthogonal or not.

Solution:

 <x,y> = 1(−1) + 1(1) + 1(0) = −1+1=0

 <y, z> = (−1) (1) + (1)(1) + 0(0) = −1+1=0

 <z, x> = 1(1) + 1(1) + 1(0) = 1+1 ≠ 20

These vectors are not orthogonal vectors.

Note: Zero vector is orthogonal to every vector.

Example 8

Verify whether this set of vectors  is orthonormal or not.

Solution:

 <x,x> = (1/√2)² + (1/√2)² + (0)² = 1/2 + 1/2 + 0 = 1

<y,y> = (1/√2)² + (‒1/√2)² + (0)² = 1/2 + 1/2 + 0 = 1

 <z, z> = (0)² + (0)² + (1)² = 0+0+1 = 1

The given set of vectors are orthonormal.

Example 9

Let A = . Calculate the (i) Frobenius norm (ii) L₁ norm and (iii) L_∞ norm.

Solution:

(i) Frobenius norm

 ||A||_F = [ |1|² + |− i|² + |1+i|² + |2− i|² ]^1/2

 =  [ 1 + 1 + (1 + i)(1 − i) + (2 − i)(2 + i) ] ^1/2

 = [1 + 1 + 1 − i + i − i²+4+2i−2i−i² ] ^1/2

 = [1 + 1 + 1 + 1 + 4 + 1]^1/2 = [9]^1⁄2

 ||A||_F = 3

(ii) L₁ − norm

We know that || A ||₁ =

= Max [ |a₁₁| + |a₂₁|, |a₁₂| + |a₂₂| ]

= Max [ |1|+| 1+ i|, |− i│+ |2 − i |

= Max [ √1² + √(1² + 1²), √(−1)² + √(2² + (− 1)²) ]

= Max [1 + √2, 1+ √5]

||A||₁ = 1+√5

(iii) L_∞ norm

We know that || A ||_∞ =

= Max [ |a₁₁| + |a₁₂| , |a₂₁| + |a₂₂| ]

= Max [ |1| + |-i| , |1+i| + |2-i| ]

= Max [ √1² + √(-1)² , √(1²+1²) + √(2²+(‒1)²) ]

 = Max [1 + 1, √2+√5]

= Max [2, √2 + √5]

 ||A||_∞ = √2+√5

Example 10

Determine ||x||_w and ||y||_w for 

Solution:

We know that ||x||_w = √[(x, x)_w] = √[ (wx) (wx) ]

Now wx = 

 = [1+2+0, 0+2+3, 1+0+3]

wx = [3, 5, 4]

 ||x||_w = √[(wx) (wx)] = √[3² +5² +4²] = √[9 + 25 + 16] = √50

  ||y||_w = √(x, y)_w = √[(wy)(wy)]

Now wy = 

= [4+5+0, 0+5+6, 4+0+6]

wy = [9, 11, 10]

|| y ||_w = √[(wy) (wy)] = √[9² + 11² + 10²] = √[81 + 121 + 100] = √302.

Example 11

 If x=  then find (i) ||x||₂ (i) ||x||₁ (iii) ||x||_∞ and (iv) || x ||₅.

Solution:

 (i) ||x||₂ = √(x•x) = √(1.1+2.2 +3.3) = √14

 (ii) ||x||₁ = |1| + |2 | + |3| = 6

 (iii) ||x||_∞ =Max [ |1|, |2|, |3| ] = 3

 (iv) ||x||₅ = [ |1|⁵ + |2|⁵ + |3|⁵ ]^1/5 = (276)^1/5 = 3.077.

Example 12

If  then find (i) ||x||_w, (ii) ||x||₂, (iii) ||x||₁ (iv) ||x||_∞ and (v) ||x||₄.

Solution:

 ||x||_w = √(x, x)_w = √[(wx) (wx)]

Now wx= 

= [ (1(1 − i) + i², (2 + i5)(1 − i) + 0(i) ]

= [1 − i−1, 2−2i + 5i − 5i²]

= [−i, 7+3i]

(i) || x ||_w = √[(− i) (i) + (7 + 3i) (7 − 3i)] = √[ (− i²) + 49 − 21i + 21i – 9i² ]

= √ [ 1 + 49 + 9 ] = √59

(ii) ||x||₂ = √(x, x) = √[(1 − i) (1 + i) + i (−i)] =√[1−i²−i²]

= √(1+1+1) = √3

(iii) ||x||₁ = | 1 − i |+| i | = √[(1 − i) (1 + i)] + √[i (− i)] = √[1² − i²] + √−i²

= √[1 +1] + √1 = √2 + 1

(iv) || x ||_∞ = Max [ |1−i|, | i | ] = Max [√[1+(−1)²], √1² ]

= Max [√2, 1] = √2

(v) ||x||₄ = [ |1−i|⁴ + |i|⁴ ]^1/4 = [(√[1² +(− 1)²)⁴ +  (√1²)⁴ ]^1⁄4

= [ (√2)⁴ + (1)⁴ ]^1/4 = [ 4 + 1 ]^1/4 = 5^1/4

Example 13

Find the distance between x and y with respect to

(i) Euclidean norm

(ii) the inner product norm with respect to w

(iii) l₃ norm for

Solution:

For these vectors, x−y = [1−4, 2−5, 3−6] = [−3, −3, −3]

(i) || x − y ||₂ = √ [ (−3)² + (− 3)² + (− 3)² ] = √27

(ii) w (x − y) = [−6, −6, −6]

 || x − y ||_w = √[ w(x − y) • w (x − y)]

 = √ [ (− 6)² + (− 6)² + (− 6)² ] = √[36 + 36 +36] = √108

(iii) || x − y ||₃ = [ |− 3|³ + |− 3|³ + |− 3|³ ]^1/3

 = [27 +27 +27]^1/3 = (81)^1/3

Example 14

Let   calculate (i) Frobenius norm (ii) L₁ norm (iii) L_∞ and (iv) Spectral norm.

Solution:

(i) ||A||_F = px = [ |4 + i3|² + |−7|² + |3|² + |i4|² ]^1/2

 = √[25 +49 + 9 + 16] = √99

 (ii) || A ||₁ = Max [ | 4+ 3i | + |3| , | −7 | + | i4 | ]

= Max [5 +3, 7+4] = 11

(iii) || A ||_∞ = Max [ |4+3i| + |−7 |, |3|+|i4| ]

= Max [5 +7, 3+4] = 12

(iv) Spectral norm

Which has the characteristic equation

 λ² − 99λ + 337 = 0

By solving this equation, we have

⇒  λ = 95.470 and λ₂=3.530

 || A ||_S = Max [√95.470, √3.530] = √95.470

Example 15

Let A = . Calculate (i) Frobenius norm (ie) ||A||_F (ii) ||A||₁, (iii) ||A||_∞, and (iv) Spectral Norm.

Solution:

(i) ||A||_F = 13.638

(ii) ||A||₁ = Max [9, 10, 9] = 10

(iii) ||A||_∞ = Max [11, 8, 9] = 11

(iv) A^‒TA=A^TA = 

which has eigenvalues 68.5078, 36.4922 and 81.0

|| A ||_S = Max [√68.5078, √36.4922, √81]

= √81 = 9