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THEOREM 2.10. If U is a unitary transformation on the finite-dimensional inner product space X, then each of the eigenvalues of U must have an absolute value equal to 1. Proof. Suppose Ux = 2x, x ±0. By the preceding theorem, then, ||x|| = || Ux|| = ||2x|| = |2| ||x||, which, since |x|| #0, implies that |λ| = 1. Suppose now that A is a normal transformation on a complex finite-dimensional, inner product space. In view of the spectral decomposition theorem, we can write A as A = λ₁E₁ + + λk Ek₂ where the λ, and E, are as in Theorem 2.6. We can also write + λkЕk. A* = λ₁E₁ + If all λ = ₁, then clearly A* = A. Next we note that A*A = |2₁|²E₁ + ··· + |åk|²Ek · Suppose now that each λ¡ (i = 1, 2, ..., k) has absolute value equal to 1. In this case then, clearly, A* A = 1. On the other hand, suppose A*A = 1. This implies that 1 = |2₁|²E₁ + ··· + |2x|²Ek» which, in turn, implies that E₁ = |2₁|²E₁ (1 - |A₂|²) E₂ = 0. Thus, since E, cannot be zero for any i = 1, 2, ..., k, we can conclude that |2|² = 1 for i = 1, 2, ..., k. We summarize these results in Theorem 2.11. or

THEOREM 2.10. If U is a unitary transformation on the finite-dimensional inner product space X, then each of the eigenvalues of U must have an absolute value equal to 1. Proof. Suppose Ux = 2x, x ±0. By the preceding theorem, then, ||x|| = || Ux|| = ||2x|| = |2| ||x||, which, since |x|| #0, implies that |λ| = 1. Suppose now that A is a normal transformation on a complex finite-dimensional, inner product space. In view of the spectral decomposition theorem, we can write A as A = λ₁E₁ + + λk Ek₂ where the λ, and E, are as in Theorem 2.6. We can also write + λkЕk. A* = λ₁E₁ + If all λ = ₁, then clearly A* = A. Next we note that A*A = |2₁|²E₁ + ··· + |åk|²Ek · Suppose now that each λ¡ (i = 1, 2, ..., k) has absolute value equal to 1. In this case then, clearly, A* A = 1. On the other hand, suppose A*A = 1. This implies that 1 = |2₁|²E₁ + ··· + |2x|²Ek» which, in turn, implies that E₁ = |2₁|²E₁ (1 - |A₂|²) E₂ = 0. Thus, since E, cannot be zero for any i = 1, 2, ..., k, we can conclude that |2|² = 1 for i = 1, 2, ..., k. We summarize these results in Theorem 2.11. or

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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THEOREM 2.10. If U is a unitary transformation on the finite-dimensional inner product space X, then each of the eigenvalues of U must have an absolute value equal to 1. Proof. Suppose Ux = λx, x 0. By the preceding theorem, then, ||x|| = || Ux|| = ||2x|| = |2| ||x||, which, since |x|| #0, implies that |2| = 1. Suppose now that A is a normal transformation on a complex finite-dimensional, inner product space. In view of the spectral decomposition theorem, we can write A as A = λ₁E₁ + + λk Ek, where the λ, and E, are as in Theorem 2.6. We can also write + λk Ek. A* = λ₁E₁ + If all λ = ₁, then clearly A* = A. Next we note that A*A = |2₁|²E₁ + · ... + 12x1² Ek. Suppose now that each λ; (i = 1, 2, ..., k) has absolute value equal to 1. In this case then, clearly, A* A = 1. On the other hand, suppose A*A = 1. This implies that 1 = |2₁|²E₂ + ··· + |2x|²Ek » which, in turn, implies that E₁ = |2₁|²E₁ (1 - |A₂|²) E₂ = 0. Thus, since E, cannot be zero for any i = 1, 2, ..., k, we can conclude that |2|² = 1 for i = 1, 2, ..., k. We summarize these results in Theorem 2.11. or
Transcribed Image Text:THEOREM 2.10. If U is a unitary transformation on the finite-dimensional inner product space X, then each of the eigenvalues of U must have an absolute value equal to 1. Proof. Suppose Ux = λx, x 0. By the preceding theorem, then, ||x|| = || Ux|| = ||2x|| = |2| ||x||, which, since |x|| #0, implies that |2| = 1. Suppose now that A is a normal transformation on a complex finite-dimensional, inner product space. In view of the spectral decomposition theorem, we can write A as A = λ₁E₁ + + λk Ek, where the λ, and E, are as in Theorem 2.6. We can also write + λk Ek. A* = λ₁E₁ + If all λ = ₁, then clearly A* = A. Next we note that A*A = |2₁|²E₁ + · ... + 12x1² Ek. Suppose now that each λ; (i = 1, 2, ..., k) has absolute value equal to 1. In this case then, clearly, A* A = 1. On the other hand, suppose A*A = 1. This implies that 1 = |2₁|²E₂ + ··· + |2x|²Ek » which, in turn, implies that E₁ = |2₁|²E₁ (1 - |A₂|²) E₂ = 0. Thus, since E, cannot be zero for any i = 1, 2, ..., k, we can conclude that |2|² = 1 for i = 1, 2, ..., k. We summarize these results in Theorem 2.11. or
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