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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.

Advanced Engineering Mathematics
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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Let V be a finite-dimensional inner product space. A linear operator U on V is called a partial isometry if there exists a subspace W of V such 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 an operator and {v₁, V2,..., , Uk} is an orthonormal basis for W. Prove the following results.
Transcribed Image Text:Let V be a finite-dimensional inner product space. A linear operator U on V is called a partial isometry if there exists a subspace W of V such 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 an operator and {v₁, V2,..., , Uk} is an orthonormal basis for W. Prove the following 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.
Transcribed Image Text:= 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.
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