MAT3701 Assignment 1. 2023

MAT3701 ASSIGNMENT ONE 2023


Question 1


0 1
𝐴=[ ]
−1 0

𝑖 0
𝐵=[ ]
0 −𝑖

𝑊1 = {𝑋 ∈ 𝑀2×2 (ℂ) ∶ 𝐴𝑋 = 𝑋𝐵}

𝑊2 = {𝑋 ∈ 𝑀2×2 (ℂ) ∶ 𝐴𝑋 = 𝑖𝑋}

(1.1)

𝑊1 = {𝑋 ∈ 𝑀2×2 (ℂ) ∶ 𝐴𝑋 = 𝑋𝐵}

𝑎 𝑏 0 1 𝑎 𝑏 𝑎 𝑏 𝑖 0
𝑊1 = {[ ] ∈ 𝑀2×2 (ℂ) ∶ [ ][ ]=[ ][ ]}
𝑐 𝑑 −1 0 𝑐 𝑑 𝑐 𝑑 0 −𝑖

𝑎 𝑏 𝑐 𝑑 𝑖𝑎 −𝑖𝑏
𝑊1 = {[ ] ∈ 𝑀2×2 (ℂ) ∶ [ ]=[ ]}
𝑐 𝑑 −𝑎 −𝑏 𝑖𝑐 −𝑖𝑑

𝑐 𝑑 𝑖𝑎 −𝑖𝑏
𝑊1 = {[ ]=[ ] ∶ 𝑎, 𝑏, 𝑐, 𝑑 ∈ ℂ}
−𝑎 −𝑏 𝑖𝑐 −𝑖𝑑

𝑐 = 𝑖𝑎

𝑑 = −𝑖𝑏

−𝑎 = 𝑖𝑐

−𝑏 = −𝑖𝑑

Therefore:

𝑐 = 𝑖𝑎

1
𝑑=𝑏
−𝑖

1
− 𝑎=𝑐
𝑖

−𝑏
=𝑑
−𝑖

,Therefore:

𝑐 = 𝑖𝑎

𝑖𝑑 = 𝑏

𝑖𝑎 = 𝑐

−𝑖𝑏 = 𝑑

Therefore:

𝑎=𝑎

𝑏=𝑏

𝑐 = 𝑖𝑎

𝑑 = −𝑖𝑏

𝑎 𝑏
𝑊1 = {[ ] ∈ 𝑀2×2 (ℂ)}
𝑐 𝑑

𝑎 𝑏
𝑊1 = {[ ] ∈ 𝑀2×2 (ℂ)}
𝑖𝑎 −𝑖𝑏

𝑎 0 0 𝑏
𝑊1 = {[ ]+[ ] ∈ 𝑀2×2 (ℂ)}
𝑖𝑎 0 0 −𝑖𝑏

1 0 0 1
𝑊1 = {𝑎 [ ]+𝑏[ ] ∶ 𝑎, 𝑏 ∈ ℂ}
𝑖 0 0 −𝑖

1 0 0 1
𝑊1 = 𝑠𝑝𝑎𝑛 {[ ],[ ]}
𝑖 0 0 −𝑖

1 0 0 1
Basis for 𝑊1 is {[ ],[ ]}
𝑖 0 0 −𝑖

(1.2)

0 1
𝐴=[ ]
−1 0

𝑖 0
𝐵=[ ]
0 −𝑖

𝑊1 = {𝑋 ∈ 𝑀2×2 (ℂ) ∶ 𝐴𝑋 = 𝑋𝐵}

𝑊2 = {𝑋 ∈ 𝑀2×2 (ℂ) ∶ 𝐴𝑋 = 𝑖𝑋}

𝑋 ∈ 𝑊1 ∩ 𝑊2

⇒ 𝑋 ∈ 𝑊1 and 𝑋 ∈ 𝑊2

, 𝑎 𝑏
⇒𝑋=[ ] , 𝑎, 𝑏 ∈ ℂ and 𝑋 ∈ 𝑊2
𝑖𝑎 −𝑖𝑏

𝑎 𝑏
⇒𝑋=[ ] , 𝑎, 𝑏 ∈ ℂ and 𝐴𝑋 = 𝑖𝑋
𝑖𝑎 −𝑖𝑏

0 1 𝑎 𝑏 𝑎 𝑏
⇒[ ][ ] = 𝑖[ ]
−1 0 𝑖𝑎 −𝑖𝑏 𝑖𝑎 −𝑖𝑏

𝑖𝑎 −𝑖𝑏 𝑖𝑎 𝑖𝑏
⇒[ ]=[ ]
−𝑎 −𝑏 −𝑎 𝑏

Therefore:

𝑖𝑎 = 𝑖𝑎
−𝑖𝑏 = 𝑖𝑏
−𝑎 = −𝑎
−𝑏 = 𝑏

Therefore:

𝑎=𝑎
𝑏=0

𝑎 𝑏
𝑋=[ ]
𝑖𝑎 −𝑖𝑏

𝑎 0
𝑋=[ ]
𝑖𝑎 0

1 0
𝑋 = 𝑎[ ]
𝑖 0


1 0
𝑊1 ∩ 𝑊2 = 𝑠𝑝𝑎𝑛 {[ ]}
𝑖 0

1 0
Basis for 𝑊1 ∩ 𝑊2 = {[ ]}
𝑖 0

(1.3)

1 0
Since 𝑊1 ∩ 𝑊2 = 𝑠𝑝𝑎𝑛 {[ ]}
𝑖 0
0 0
then 𝑊1 ∩ 𝑊2 ≠ {[ ]}
0 0

𝑊1 ∩ 𝑊2 ≠ 𝑀2×2 (ℂ)