Question 2: Let V be a finite dimensional vector space over R, and let T: V → V be a linear map. Prove that if T is injective then T is surjective. a) HINT: Use the Rank-Nullity Theorem.

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Chapter2: Second-order Linear Odes
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Please solve problem 2
**Question 2:**

Let \( V \) be a finite dimensional vector space over \( \mathbb{R} \), and let \( T: V \rightarrow V \) be a linear map.

**a)** Prove that if \( T \) is injective then \( T \) is surjective.

**HINT:** Use the Rank-Nullity Theorem.

---

**b)** Prove that if \( T \) is surjective then \( T \) is injective.
Transcribed Image Text:**Question 2:** Let \( V \) be a finite dimensional vector space over \( \mathbb{R} \), and let \( T: V \rightarrow V \) be a linear map. **a)** Prove that if \( T \) is injective then \( T \) is surjective. **HINT:** Use the Rank-Nullity Theorem. --- **b)** Prove that if \( T \) is surjective then \( T \) is injective.
### Linear Algebra Problem

**Problem Statement:**

Let \( V \) be a vector space over \(\mathbb{R}\), and let \( T, S : V \to V \) be linear maps from \( V \) into \( V \). Suppose \( T \circ S \) is invertible. Prove that \( S \) is injective.

**Explanation:**

In the provided problem, \( V \) denotes a vector space over the real numbers \(\mathbb{R}\). The symbols \( T \) and \( S \) represent linear maps that take elements from the vector space \( V \) and map them back to \( V \). The objective is to prove that if the composition \( T \circ S \) (i.e., performing \( S \) first and then \( T \)) is an invertible linear transformation, then \( S \) is an injective (one-to-one) linear map.

**Solution Plan:**

To prove that \( S \) is injective, we need to show that if \( S(v) = S(w) \) for some vectors \( v, w \in V \), then \( v = w \).

1. **Assumption:**
   \( T \circ S \) is invertible which implies there exists a linear map \( (T \circ S)^{-1} \) such that 
   \[
   (T \circ S) \circ (T \circ S)^{-1} = \text{Id}_V
   \]
   where \(\text{Id}_V\) is the identity map on \( V \).

2. **Injectivity of \( S \):**
   Suppose \( S(v) = S(w) \). We need to show \( v = w \).

3. **Action of \( T \) on \( S(v) \) and \( S(w) \):**
   Given \( S(v) = S(w) \), applying \( T \) on both sides gives:
   \[
   T(S(v)) = T(S(w))
   \]

4. **Using the Invertibility:**
   Since \( T \circ S \) is invertible, we have:
   \[
   (T \circ S) \circ (T \circ S)^{-1} = \text{Id}_V
   \]
   Therefore,
   \[
   (
Transcribed Image Text:### Linear Algebra Problem **Problem Statement:** Let \( V \) be a vector space over \(\mathbb{R}\), and let \( T, S : V \to V \) be linear maps from \( V \) into \( V \). Suppose \( T \circ S \) is invertible. Prove that \( S \) is injective. **Explanation:** In the provided problem, \( V \) denotes a vector space over the real numbers \(\mathbb{R}\). The symbols \( T \) and \( S \) represent linear maps that take elements from the vector space \( V \) and map them back to \( V \). The objective is to prove that if the composition \( T \circ S \) (i.e., performing \( S \) first and then \( T \)) is an invertible linear transformation, then \( S \) is an injective (one-to-one) linear map. **Solution Plan:** To prove that \( S \) is injective, we need to show that if \( S(v) = S(w) \) for some vectors \( v, w \in V \), then \( v = w \). 1. **Assumption:** \( T \circ S \) is invertible which implies there exists a linear map \( (T \circ S)^{-1} \) such that \[ (T \circ S) \circ (T \circ S)^{-1} = \text{Id}_V \] where \(\text{Id}_V\) is the identity map on \( V \). 2. **Injectivity of \( S \):** Suppose \( S(v) = S(w) \). We need to show \( v = w \). 3. **Action of \( T \) on \( S(v) \) and \( S(w) \):** Given \( S(v) = S(w) \), applying \( T \) on both sides gives: \[ T(S(v)) = T(S(w)) \] 4. **Using the Invertibility:** Since \( T \circ S \) is invertible, we have: \[ (T \circ S) \circ (T \circ S)^{-1} = \text{Id}_V \] Therefore, \[ (
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