Let V be a vector space, and T : V → V a linear transformation such that T(2√1 – 3√₂) = 27₁ + 502 and T(−3v1 + 5√₂) = −4ở1 – 4ʊ2. Then T(v₁) = V₁ + V₂, T(v₂) = v₁+ → T(40₁ − 402) = ₁+ ₂. - V2. V2,

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 vector space, and \( T : V \rightarrow V \) a linear transformation such that \( T(2\vec{v}_1 - 3\vec{v}_2) = 2\vec{v}_1 + 5\vec{v}_2 \) and \( T(-3\vec{v}_1 + 5\vec{v}_2) = -4\vec{v}_1 - 4\vec{v}_2 \). Then

\[
T(\vec{v}_1) = \, \boxed{} \, \vec{v}_1 + \, \boxed{} \, \vec{v}_2,
\]

\[
T(\vec{v}_2) = \, \boxed{} \, \vec{v}_1 + \, \boxed{} \, \vec{v}_2,
\]

\[
T(4\vec{v}_1 - 4\vec{v}_2) = \, \boxed{} \, \vec{v}_1 + \, \boxed{} \, \vec{v}_2.
\]
Transcribed Image Text:Let \( V \) be a vector space, and \( T : V \rightarrow V \) a linear transformation such that \( T(2\vec{v}_1 - 3\vec{v}_2) = 2\vec{v}_1 + 5\vec{v}_2 \) and \( T(-3\vec{v}_1 + 5\vec{v}_2) = -4\vec{v}_1 - 4\vec{v}_2 \). Then \[ T(\vec{v}_1) = \, \boxed{} \, \vec{v}_1 + \, \boxed{} \, \vec{v}_2, \] \[ T(\vec{v}_2) = \, \boxed{} \, \vec{v}_1 + \, \boxed{} \, \vec{v}_2, \] \[ T(4\vec{v}_1 - 4\vec{v}_2) = \, \boxed{} \, \vec{v}_1 + \, \boxed{} \, \vec{v}_2. \]
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