1. **Problem:** Determine whether each statement about basis and dimension is True (T) or False (F). 2. **Recall important definitions and facts:** - A subspace $U$ of a vector space $V$ is a subset of $V$ that is itself a vector space. - The dimension of a vector space is the number of vectors in any basis for that space. - A basis is a linearly independent set that spans the space. 3. **Analyze each statement:** **Statement 1:** If $U$ is a subspace of $V$ with $\dim U = \dim V$, then $U = V$. - Since $U$ is contained in $V$ and both have the same dimension, $U$ must be all of $V$. - **Answer:** T **Statement 2:** If $U$ is a subspace of $V$ and $S$ is a linearly independent set in $U$, then either $S$ is a basis for $U$ or vectors can be added to $S$ to obtain a basis for $U$. - This is a fundamental property of vector spaces: any linearly independent set can be extended to a basis. - **Answer:** T **Statement 3:** If $V$ is a vector space and $v_1, v_2, v_3$ are vectors in $V$ so that $V = \text{Span}\{v_1, v_2, v_3\}$, then $\dim V = 3$. - The span of three vectors can have dimension less than or equal to 3 depending on linear independence. - If $v_1, v_2, v_3$ are linearly dependent, dimension is less than 3. - So this statement is not always true. - **Answer:** F **Statement 4:** If $U$ is a subspace of $V$ and $S$ is a set of vectors in $U$ so that $U = \text{Span } S$, then either $S$ is a basis for $U$ or vectors can be removed from $S$ to obtain a basis for $U$. - This is true because any spanning set can be reduced to a basis by removing dependent vectors. - **Answer:** T **Statement 5:** If $V$ is a vector space with $\dim V = 4$, and $U$ is a subspace of $V$ that is not equal to $V$, then $\dim U \leq 3$. - Since $U \neq V$, dimension of $U$ must be strictly less than 4. - So $\dim U \leq 3$. - **Answer:** T **Final answers:** 1. T 2. T 3. F 4. T 5. T