THEOREM 5.6 bon Proof If T V W and S: V→ W are linear transformations, then so are T + S and cT where c is a scalar. We verify this for T+S and leave the proof for cT as an exercise (Exercise 15). Suppose that u and u are vectors in V and a is a scalar. We have and (T+S) (u + v) = = T(u + v) + S(u + v) = T(u) + T(v) + S(u) + S(v) T(u) + S(u) + T (v) + S(v) = (T+S)(u) +(T+S)(v) (T+S) (av) = T(av) + S(av) = aT (v) +aS(v) = a(T(v) + S(v)) = a(T + S)(v) giving us T + S is a linear transformation. As an immediate consequence of Theorem 5.6, we get that linear combinations of linear transformations from one vector space to another are linear transformations.
THEOREM 5.6 bon Proof If T V W and S: V→ W are linear transformations, then so are T + S and cT where c is a scalar. We verify this for T+S and leave the proof for cT as an exercise (Exercise 15). Suppose that u and u are vectors in V and a is a scalar. We have and (T+S) (u + v) = = T(u + v) + S(u + v) = T(u) + T(v) + S(u) + S(v) T(u) + S(u) + T (v) + S(v) = (T+S)(u) +(T+S)(v) (T+S) (av) = T(av) + S(av) = aT (v) +aS(v) = a(T(v) + S(v)) = a(T + S)(v) giving us T + S is a linear transformation. As an immediate consequence of Theorem 5.6, we get that linear combinations of linear transformations from one vector space to another are linear transformations.
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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![**Theorem 5.6**
If \( T : V \to W \) and \( S : V \to W \) are linear transformations, then so are \( T + S \) and \( cT \) where \( c \) is a scalar.
**Proof**
We verify this for \( T + S \) and leave the proof for \( cT \) as an exercise (Exercise 15). Suppose that \( u \) and \( v \) are vectors in \( V \) and \( a \) is a scalar. We have
\[
(T + S)(u + v) = T(u + v) + S(u + v) = T(u) + T(v) + S(u) + S(v)
\]
\[
= T(u) + S(u) + T(v) + S(v) = (T + S)(u) + (T + S)(v)
\]
and
\[
(T + S)(av) = T(av) + S(av) = aT(v) + aS(v) = a(T(v) + S(v)) = a(T + S)(v)
\]
giving us \( T + S \) is a linear transformation.
As an immediate consequence of Theorem 5.6, we get that linear combinations of linear transformations from one vector space to another are linear transformations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbfa44710-6742-4cf8-9a38-b63fd4325c9b%2F000d3a8b-d867-44a3-9605-f956f56971af%2Fitkpqbd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Theorem 5.6**
If \( T : V \to W \) and \( S : V \to W \) are linear transformations, then so are \( T + S \) and \( cT \) where \( c \) is a scalar.
**Proof**
We verify this for \( T + S \) and leave the proof for \( cT \) as an exercise (Exercise 15). Suppose that \( u \) and \( v \) are vectors in \( V \) and \( a \) is a scalar. We have
\[
(T + S)(u + v) = T(u + v) + S(u + v) = T(u) + T(v) + S(u) + S(v)
\]
\[
= T(u) + S(u) + T(v) + S(v) = (T + S)(u) + (T + S)(v)
\]
and
\[
(T + S)(av) = T(av) + S(av) = aT(v) + aS(v) = a(T(v) + S(v)) = a(T + S)(v)
\]
giving us \( T + S \) is a linear transformation.
As an immediate consequence of Theorem 5.6, we get that linear combinations of linear transformations from one vector space to another are linear transformations.

Transcribed Image Text:15. Complete the proof of Theorem 5.6 by showing that if \( T : V \rightarrow W \) is a linear transformation and \( c \) is a scalar, then \( cT \) is a linear transformation.
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