Let V be a finite-dimensional inner product space. A linear operator Uon V is called a partial isometry if there exists a subspace W of Vsuch that ||U(x)|| = ||x|| for all x € W and U(x) = 0 for all x W-.Observe that W need not be U-invariant. Suppose that U is such anoperator and {v₁, V2,..., , Uk} is an orthonormal basis for W. Prove thefollowing results. = Vi(e) Let T be the linear operator on V that satisfies T(U(v₁))(1 ≤ i ≤ k) and T(w;) = 0 (1 ≤ i ≤ j). Then T is well defined,and T = U*. Hint: Show that (U(x), y) = (x, T(y)) for all x, y € 3.There are four cases.(f) U* is a partial isometry.

In this solution, we will consider a linear operator U on a finite-dimensional inner product space V…

Answered: = Vi O Let T be the linear operator on…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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= Vi O Let T be the linear operator on V that satisfies T(U(v;)) (1 ≤ i ≤ k) and T(w;) = 0 (1 ≤ i ≤ j). Then T is well defined, and T = U*. Hint: Show that (U(x), y) = (x, T(y)) for all x, y € 3. There are four cases. OU* is a partial isometry.

= Vi O Let T be the linear operator on V that satisfies T(U(v;)) (1 ≤ i ≤ k) and T(w;) = 0 (1 ≤ i ≤ j). Then T is well defined, and T = U. Hint: Show that (U(x), y) = (x, T(y)) for all x, y € 3. There are four cases. OU is a partial isometry.