Prove that a finite dimensional vector space V has dimension n if and only if n is the maximum number of linearly independent vectors in any subset of V . सिद्ध कीजिए कि एक परिमित आयामी सदिश समष्टि V का आयाम \( n \) है यदि और केवल यदि \( n, \mathrm{~V} \) के किसी उपसमुच्चय में रैखिक रूप से स्वतंत्र सदिशों की अधिकतम संख्या है।

To prove that a finite-dimensional vector space \( V \) has dimension \( n \) if and only if \( n \) is the maximum number of linearly independent vectors in any subset of \( V \), we will establish both implications: 1. **If \( V \) has dimension \( n \), then no subset of \( V \) contains more than \( n \) linearly independent vectors.** 2. **If the maximum number of linearly independent vectors in any subset of \( V \) is \( n \), then the dimension of \( V \) is \( n \).** Let's proceed with each part of the proof. --- ### Part 1: Dimension \( n \) Implies Maximum of \( n \) Linearly Independent Vectors **Assumption:** \( V \) is a finite-dimensional vector space with dimension \( n \). This means there exists a basis for \( V \) consisting of exactly \( n \) vectors. **To Prove:** Any linearly independent subset of \( V \) contains at most \( n \) vectors. **Proof:** 1. Let \( \{v_1, v_2, \dots, v_k\} \) be a linearly independent subset of \( V \). 2. Consider extending this subset to a basis for \( V \). Since \( V \) has dimension \( n \), any basis of \( V \) has exactly \( n \) vectors. 3. Since \( \{v_1, v_2, \dots, v_k\} \) is linearly independent, it can be extended to a basis by adding \( n - k \) additional vectors, assuming \( k \leq n \). 4. If \( k > n \), then having more than \( n \) linearly independent vectors would contradict the fact that the maximum size of a basis is \( n \). 5. Therefore, \( k \leq n \). Hence, any linearly independent subset of \( V \) contains at most \( n \) vectors. --- ### Part 2: Maximum of \( n \) Linearly Independent Vectors Implies Dimension \( n \) **Assumption:** In the vector space \( V \), the maximum number of linearly independent vectors in any subset is \( n \). **To Prove:** The dimension of \( V \) is \( n \). **Proof:** 1. Since there exists a subset of \( V \) with \( n \) linearly independent vectors (by the assumption that the maximum number is \( n \)), let \( \{v_1, v_2, \dots, v_n\} \) be such a set. 2. We need to show that this set is a basis for \( V \), i.e., that it spans \( V \). 3. Suppose, for contradiction, that \( \{v_1, v_2, \dots, v_n\} \) does not span \( V \). Then, there exists a vector \( v_{n+1} \in V \) that is not in the span of \( \{v_1, v_2, \dots, v_n\} \). 4. This implies that \( \{v_1, v_2, \dots, v_n, v_{n+1}\} \) is a linearly independent set with \( n + 1 \) vectors. 5. However, this contradicts the assumption that the maximum number of linearly independent vectors in any subset of \( V \) is \( n \). 6. Therefore, our assumption is false, and \( \{v_1, v_2, \dots, v_n\} \) must span \( V \). 7. Since the set is both linearly independent and spans \( V \), it is a basis for \( V \). 8. Hence, the dimension of \( V \) is \( n \). --- ### Conclusion We have shown both directions: - If \( V \) has dimension \( n \), then no subset of \( V \) contains more than \( n \) linearly independent vectors. - If no subset of \( V \) contains more than \( n \) linearly independent vectors, then the dimension of \( V \) is \( n \). Therefore, a finite-dimensional vector space \( V \) has dimension \( n \) if and only if \( n \) is the maximum number of linearly independent vectors in any subset of \( V \).