Question 3: Recall that L(R³, R²) denotes the vector space of all linear maps from R³ into R². Let = {Te L(R³, R²): T((1,2,3)) = (0,0)}. a) Prove that is a subspace of L(R³, R²) b) Find dim(S)
Question 3: Recall that L(R³, R²) denotes the vector space of all linear maps from R³ into R². Let = {Te L(R³, R²): T((1,2,3)) = (0,0)}. a) Prove that is a subspace of L(R³, R²) b) Find dim(S)
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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Solve problem 3
Transcribed Image Text:**Question 3:**
Recall that \(\mathcal{L}(\mathbb{R}^3, \mathbb{R}^2)\) denotes the vector space of all linear maps from \(\mathbb{R}^3\) into \(\mathbb{R}^2\).
Let \(\mathcal{S} = \{ T \in \mathcal{L}(\mathbb{R}^3, \mathbb{R}^2) : T((1,2,3)) = (0,0) \} \).
a) Prove that \(\mathcal{S}\) is a subspace of \(\mathcal{L}(\mathbb{R}^3, \mathbb{R}^2)\).
b) Find \(\dim(\mathcal{S})\).
![### Linear Algebra Problem: Injectivity of \(S\) Given an Invertible Composition \(T \circ S\)
**Problem Statement:**
Let \( V \) be a vector space over \( \mathbb{R} \), and let \( T, S : V \rightarrow V \) be linear maps from \( V \) into \( V \). Suppose \( T \circ S \) is invertible. Prove that \( S \) is injective.
**Solution Outline:**
To prove that \( S \) is injective, we need to show that if \( S(v) = S(w) \) for some \( v, w \in V \), then \( v = w \).
1. **Given that \( T \circ S \) is invertible, there exists a linear map \( (T \circ S)^{-1} \) such that:**
\[
(T \circ S) \circ (T \circ S)^{-1} = I_V
\]
where \( I_V \) is the identity map on \( V \).
2. **Assume \( S(v) = S(w) \).**
3. **Apply \( T \) to both sides of the equation:**
\[
T(S(v)) = T(S(w))
\]
This simplifies to:
\[
(T \circ S)(v) = (T \circ S)(w)
\]
4. **Since \( T \circ S \) is invertible, apply \( (T \circ S)^{-1} \) to both sides:**
\[
(T \circ S)^{-1}((T \circ S)(v)) = (T \circ S)^{-1}((T \circ S)(w))
\]
Which simplifies to:
\[
v = w
\]
Hence, \( v = w \) implies that \( S \) is injective, completing the proof.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3cf4b148-a9ff-447b-bf88-730c6d11afaf%2F1ee24083-af88-4b8f-a7de-287e779ba040%2Fsuao9ji_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Linear Algebra Problem: Injectivity of \(S\) Given an Invertible Composition \(T \circ S\)
**Problem Statement:**
Let \( V \) be a vector space over \( \mathbb{R} \), and let \( T, S : V \rightarrow V \) be linear maps from \( V \) into \( V \). Suppose \( T \circ S \) is invertible. Prove that \( S \) is injective.
**Solution Outline:**
To prove that \( S \) is injective, we need to show that if \( S(v) = S(w) \) for some \( v, w \in V \), then \( v = w \).
1. **Given that \( T \circ S \) is invertible, there exists a linear map \( (T \circ S)^{-1} \) such that:**
\[
(T \circ S) \circ (T \circ S)^{-1} = I_V
\]
where \( I_V \) is the identity map on \( V \).
2. **Assume \( S(v) = S(w) \).**
3. **Apply \( T \) to both sides of the equation:**
\[
T(S(v)) = T(S(w))
\]
This simplifies to:
\[
(T \circ S)(v) = (T \circ S)(w)
\]
4. **Since \( T \circ S \) is invertible, apply \( (T \circ S)^{-1} \) to both sides:**
\[
(T \circ S)^{-1}((T \circ S)(v)) = (T \circ S)^{-1}((T \circ S)(w))
\]
Which simplifies to:
\[
v = w
\]
Hence, \( v = w \) implies that \( S \) is injective, completing the proof.
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