Then by following the logic from the direction

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Unformatted text preview: x = 0 ∀x ∈ Cn Using this, we proceed with the proof: (⇒) : If A is normal, then: A∗ A = AA∗ . This implies: ￿(A∗ A − AA∗ )x, x￿ = 0, ∀x ∈ Cn By the bilinearity of the inner product, this implies that: ￿A∗ Ax, x￿ = ￿AA∗ x, x￿, ∀x ∈ Cn By the definition of the adjoint and the fact that A = (A∗ )∗ , this implies: ||Ax||2 = ￿Ax, Ax￿ = ￿A∗ Ax, x￿ = ￿AA∗ x, x￿ = ￿A∗ x, A∗ x￿ = ||A∗ x||2 , ∀x ∈ Cn Hence: ||Ax|| = ||A∗ x||, ∀x ∈ Cn (⇐) : Suppose ||Ax|| = ||A∗ x|| ∀x ∈ Cn . Then by following the logic from the (⇒) direction backwards, we can get back to: ￿(A∗ A − AA∗ )x, x￿ = 0, ∀x ∈ Cn . Now we may use the Polarization Identity along with some rearranging to conclude that : ￿(A∗ A − AA∗ )x, y ￿ = 0, ∀x, y ∈ Cn . Hence: A∗ A − AA∗ = 0 (refer to today’s clarification) Thus, A is normal. To p...
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This note was uploaded on 01/16/2014 for the course MATH 6304 taught by Professor Bernhardbodmann during the Fall '12 term at University of Houston.

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