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MAT3701 Assignment 2 2022
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  • MAT3701 Linear Algebra (MAT3701)
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  • University Of South Africa (Unisa)

MAT3701 Linear Algebra Assignment TWO of 2022 solutions. Inner product Linear operator Normal matrix Similar matrix Unitarity Unitarily equivalent Spectral decomposition Eigenvalues Eigenvectors Basis Lagrange polynomial Least squares Rigid motion Translation Reflection Rotation

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  • August 18, 2022
  • 24
  • 2022/2023
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Subjects

  • mat3701
  • linear algebra
  • inner product
  • linear operator
  • normal matrix
  • similar matrix
  • unitarity
  • unitarily equivalent
  • spectral decomposition
  • eigenvalues
  • eigenvectors
  • basis
  • lagrange polynomial
  • least squares

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  • University of South Africa (Unisa)
  • MAT3701 Linear Algebra (MAT3701)
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MAT3701 ASSIGNMENT 2 2022

Question 1

Let 𝑢 = (𝑢1 , 𝑢2 , 𝑢3 ) and 𝑣 = (𝑣1 , 𝑣2 , 𝑣3 ) be elements of ℂ3 .
Standard inner product 〈, 〉: ℂ3 × ℂ3 → ℂ defined as:
〈𝑢, 𝑣〉 = 〈(𝑢1 , 𝑢2 , 𝑢3 ) , (𝑣1 , 𝑣2 , 𝑣3 )〉
〈𝑢, 𝑣〉 = 𝑢1 ̅̅̅
𝑣1 + 𝑢2 ̅̅̅
𝑣2 + 𝑢3 ̅̅̅ 𝑣3

1 0
𝑇: ℂ3 → ℂ3 is the orthogonal projection on 𝑉 = 𝑠𝑝𝑎𝑛 {[1] , [1]}.
0 1

(1.1) Basis for 𝑉 ⊥ .

𝑉 ⊥ is a set containing all the vectors in ℂ3 that are orthogonal to all the vectors in 𝑉.

𝑉 ⊥ = {𝑢 ∈ ℂ3 : 〈𝑢, 𝑣〉 = 0, ∀ 𝑣 ∈ 𝑉 }

𝑉 ⊥ = {(𝑢1 , 𝑢2 , 𝑢3 ) ∶ 〈(𝑢1 , 𝑢2 , 𝑢3 ), (1,1, 0)〉 = 0 𝑎𝑛𝑑 〈(𝑢1 , 𝑢2 , 𝑢3 ), (0,1, 1)〉 = 0}

𝑉 ⊥ = {(𝑢1 , 𝑢2 , 𝑢3 ) ∶ 𝑢1 1̅ + 𝑢2 1̅ + 𝑢3 0̅ = 0 𝑎𝑛𝑑 𝑢1 0̅ + 𝑢2 1̅ + 𝑢3 1̅ = 0}

𝑉 ⊥ = {(𝑢1 , 𝑢2 , 𝑢3 ) ∶ 𝑢1 1 + 𝑢2 1 + 𝑢3 0 = 0 𝑎𝑛𝑑 𝑢1 0 + 𝑢2 1 + 𝑢3 1 = 0}

𝑉 ⊥ = {(𝑢1 , 𝑢2 , 𝑢3 ) ∶ 𝑢1 + 𝑢2 = 0 𝑎𝑛𝑑 𝑢2 + 𝑢3 = 0}


𝑉 ⊥ = {(𝑢1 , 𝑢2 , 𝑢3 ) ∶ 𝑢1 = −𝑢2 𝑎𝑛𝑑 𝑢3 = −𝑢2 }


𝑉 ⊥ = {(−𝑢2 , 𝑢2 , −𝑢2 ) ∶ 𝑢2 ∈ ℂ}


𝑉 ⊥ = {𝑢2 (−1,1, −1) ∶ 𝑢2 ∈ ℂ}


𝑉 ⊥ = 𝑠𝑝𝑎𝑛{(−1,1 − 1)}


−1
𝑉 ⊥ = 𝑠𝑝𝑎𝑛 {[ 1 ]}
−1



−1
⊥
Basis for 𝑉 is {[ 1 ]}.
−1

,(1.2) Matrix representation of 𝑇 with respect to the canonical basis for ℂ3 .

1 0 0
Canonical basis for ℂ3 , 𝛽 = {[0] , [1] , [0]} where scalars are complex numbers.
0 0 1

1 0
𝑇: ℂ3 → ℂ3 is the orthogonal projection on 𝑉 = 𝑠𝑝𝑎𝑛 {[1] , [1]}.
0 1

Let 𝑣1 = (1,1, 0) 𝑎𝑛𝑑 𝑣2 = (0,1, 1)

〈𝑢, 𝑣〉 = 𝑢1 ̅̅̅
𝑣1 + 𝑢2 ̅̅̅
𝑣2 + 𝑢3 ̅̅̅
𝑣3

𝑇(𝑢) = 〈𝑢, 𝑣1 〉𝑣1 + 〈𝑢, 𝑣2 〉𝑣2

1 0 0
Apply 𝑇 on , 𝛽 = {[0] , [1] , [0]}
0 0 1

𝑇((1,0,0)) = 〈(1,0,0), (1,1, 0)〉(1,1, 0) + 〈(1,0,0), (0,1, 1)〉(0,1, 1)

𝑇((1,0,0)) = [1 × 1 + 0 × 1 + 0 × 0](1,1, 0) + [1 × 0 + 0 × 1 + 0 × 1](0,1, 1)

𝑇((1,0,0)) = [1](1,1, 0) + [0](0,1, 1)

𝑇((1,0,0)) = (1,1, 0)




𝑇((0,1,0)) = 〈(0,1,0), (1,1, 0)〉(1,1, 0) + 〈(0,1,0), (0,1, 1)〉(0,1, 1)

𝑇((0,1,0)) = [0 × 1 + 1 × 1 + 0 × 0](1,1, 0) + [0 × 0 + 1 × 1 + 0 × 1](0,1, 1)

𝑇((0,1,0)) = [1](1,1, 0) + [1](0,1, 1)

𝑇((0,1,0)) = (1,2, 1)




𝑇((0,0,1)) = 〈(0,0,1), (1,1, 0)〉(1,1, 0) + 〈(0,0,1), (0,1, 1)〉(0,1, 1)

𝑇((0,0,1)) = [0 × 1 + 0 × 1 + 1 × 0](1,1, 0) + [0 × 0 + 0 × 1 + 1 × 1](0,1, 1)

𝑇((0,0,1)) = [0](1,1, 0) + [1](0,1, 1)

𝑇((0,0,1)) = (0,1, 1)

, 1 1 0
[ 𝑇 ] 𝛽 = [1 2 1]
0 1 1


Question 2

Let 𝑔 and ℎ be elements of 𝑃2 (ℝ) and 𝑎, 𝑏 and 𝑐 ∈ ℝ are distinct.
Inner product 〈, 〉: 𝑃2 (ℝ) × 𝑃2 (ℝ) → ℝ defined as:
〈𝑔, ℎ〉 = 𝑔(𝑎)ℎ(𝑎) + 𝑔(𝑏)ℎ(𝑏) + 𝑔(𝑐 )ℎ(𝑐 )

The inner product satisfies:
Let 𝑓, 𝑔 and ℎ ∈ 𝑃2 (ℝ) and 𝑘 ∈ ℝ.
IP1 〈𝑓 + ℎ, 𝑔〉 = 〈𝑓, 𝑔〉 + 〈ℎ, 𝑔〉

IP2 〈𝑘𝑓, 𝑔〉 = 𝑘〈𝑓, 𝑔〉

IP3 〈𝑓, 𝑔〉 = 〈𝑔, 𝑓〉

IP4 (a) 〈𝑓, 𝑓〉 ≥ 0
(b) 〈𝑓, 𝑓〉 = 0 if and only if 𝑓 = 0

The set 𝛽 = {𝑓𝑎 , 𝑓𝑏 , 𝑓𝑐 } of Lagrange polynomials associated with 𝑎, 𝑏 and 𝑐 respectively is a basis for 𝑃2 (ℝ).

The Lagrange polynomials satisfy:
1.
𝑓𝑎 (𝑎) = 1
𝑓𝑏 (𝑏) = 1
𝑓𝑐 (𝑐 ) = 1

2.
𝑓𝑎 (𝑏) = 0 = 𝑓𝑎 (𝑐)
𝑓𝑏 (𝑎) = 0 = 𝑓𝑏 (𝑐)
𝑓𝑐 (𝑎) = 0 = 𝑓𝑐 (𝑏)

Let 𝑃: 𝑃2 (ℝ) → 𝑃2 (ℝ) be the orthogonal projection on
1 1
𝑊 = 𝑠𝑝𝑎𝑛 { (𝑓𝑎 + 2𝑓𝑏 + 2𝑓𝑐 ), (2𝑓𝑎 − 2𝑓𝑏 + 𝑓𝑐 )}
3 3

(2.1) Show that 〈𝑎1 𝑓𝑎 + 𝑏1 𝑓𝑏 + 𝑐1 𝑓𝑐 , 𝑎2 𝑓𝑎 + 𝑏2 𝑓𝑏 + 𝑐2 𝑓𝑐 〉 = 𝑎1 𝑎2 + 𝑏1 𝑏2 + 𝑐1 𝑐2 ∀ 𝑎1 , 𝑎2 , 𝑏2 , 𝑏2 , 𝑐1 , 𝑐2 ∈ ℝ.

Solution
After applying IP1 repeatedly, we get:
〈𝑎1 𝑓𝑎 + 𝑏1 𝑓𝑏 + 𝑐1 𝑓𝑐 , 𝑎2 𝑓𝑎 + 𝑏2 𝑓𝑏 + 𝑐2 𝑓𝑐 〉
= 〈𝑎1 𝑓𝑎 , 𝑎2 𝑓𝑎 + 𝑏2 𝑓𝑏 + 𝑐2 𝑓𝑐 〉 + 〈𝑏1 𝑓𝑏 , 𝑎2 𝑓𝑎 + 𝑏2 𝑓𝑏 + 𝑐2 𝑓𝑐 〉 + 〈𝑐1 𝑓𝑐 , 𝑎2 𝑓𝑎 + 𝑏2 𝑓𝑏 + 𝑐2 𝑓𝑐 〉

After applying IP3 on each term on the RHS, we get:
〈𝑎1 𝑓𝑎 + 𝑏1 𝑓𝑏 + 𝑐1 𝑓𝑐 , 𝑎2 𝑓𝑎 + 𝑏2 𝑓𝑏 + 𝑐2 𝑓𝑐 〉
= 〈𝑎2 𝑓𝑎 + 𝑏2 𝑓𝑏 + 𝑐2 𝑓𝑐 , 𝑎1 𝑓𝑎 〉 + 〈𝑎2 𝑓𝑎 + 𝑏2 𝑓𝑏 + 𝑐2 𝑓𝑐 , 𝑏1 𝑓𝑏 〉 + 〈𝑎2 𝑓𝑎 + 𝑏2 𝑓𝑏 + 𝑐2 𝑓𝑐 , 𝑐1 𝑓𝑐 〉

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