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
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Solve problem 3
**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 \(\dim(\mathcal{S})\).
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.
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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