MAT2611 ASSIGNMENT 8 2023
Problem 27
Let 𝐴 be an 𝑛 × 𝑛 matrix, where 𝑛 ≥ 2, 𝑛 ∈ ℕ
Given that 𝐴 is symmetric and orthogonal.
Since 𝐴 is symmetric then 𝐴𝑇 = 𝐴 (1)
Since 𝐴 is orthogonal then 𝐴𝑇 𝐴 = 𝐼𝑛 (2)
On equation (1) to multiply both sides by 𝐴
𝐴𝑇 = 𝐴
𝐴𝑇 𝐴 = 𝐴𝐴
𝐴𝑇 𝐴 = 𝐴2
𝐴2 = 𝐴𝑇 𝐴
To apply the result on equation (2)
𝐴2 = 𝐼𝑛
Let 𝜆 be an eigenvalue of 𝐴.
Therefore, 𝜆2 is an eigenvalue of 𝐴2
To find the eigenvalues 𝛼 of 𝐴2
𝐴2 𝒗 = 𝛼𝒗 where 𝒗 is an 𝑛 × 1 eigenvector of 𝐴2
Now, 𝐴2 = 𝐼𝑛 , therefore:
𝐼𝑛 𝒗 = 𝛼𝒗
𝒗 = 𝛼𝒗
𝒗 − 𝛼𝒗 = 𝟎
(1 − 𝛼)𝒗 = 𝟎
Eigenvectors may not be zero vectors, therefore:
1−𝛼 =0
𝛼=1
The only possible eigenvalue of 𝐴2 is 1
Problem 27
Let 𝐴 be an 𝑛 × 𝑛 matrix, where 𝑛 ≥ 2, 𝑛 ∈ ℕ
Given that 𝐴 is symmetric and orthogonal.
Since 𝐴 is symmetric then 𝐴𝑇 = 𝐴 (1)
Since 𝐴 is orthogonal then 𝐴𝑇 𝐴 = 𝐼𝑛 (2)
On equation (1) to multiply both sides by 𝐴
𝐴𝑇 = 𝐴
𝐴𝑇 𝐴 = 𝐴𝐴
𝐴𝑇 𝐴 = 𝐴2
𝐴2 = 𝐴𝑇 𝐴
To apply the result on equation (2)
𝐴2 = 𝐼𝑛
Let 𝜆 be an eigenvalue of 𝐴.
Therefore, 𝜆2 is an eigenvalue of 𝐴2
To find the eigenvalues 𝛼 of 𝐴2
𝐴2 𝒗 = 𝛼𝒗 where 𝒗 is an 𝑛 × 1 eigenvector of 𝐴2
Now, 𝐴2 = 𝐼𝑛 , therefore:
𝐼𝑛 𝒗 = 𝛼𝒗
𝒗 = 𝛼𝒗
𝒗 − 𝛼𝒗 = 𝟎
(1 − 𝛼)𝒗 = 𝟎
Eigenvectors may not be zero vectors, therefore:
1−𝛼 =0
𝛼=1
The only possible eigenvalue of 𝐴2 is 1
