1. **Stating the problem:** We are given two theorems about vector spaces and bases: a) Any set of more than $n$ vectors in $V$ must be linearly dependent. b) Any set of fewer than $n$ vectors in $V$ cannot span $V$. Also, the Basis Theorem states: If a vector space $V$ has a basis of $n$ vectors, then every basis for $V$ has exactly $n$ vectors. 2. **Key definitions and formulas:** - A **basis** $B = \{\vec{v}_1, \vec{v}_2, \ldots, \vec{v}_n\}$ of a vector space $V$ is a set of vectors that is both **linearly independent** and **spans** $V$. - The **dimension** of $V$ is the number $n$ of vectors in any basis of $V$. 3. **Proof of part (a):** - Suppose we have a set $S$ of more than $n$ vectors in $V$, say $m > n$ vectors. - Since $B$ is a basis with $n$ vectors, any vector in $V$ can be written as a linear combination of vectors in $B$. - By the **Linear Dependence Lemma**, any set of more than $n$ vectors in $V$ must be linearly dependent. - This is because you cannot have more than $n$ linearly independent vectors in an $n$-dimensional space. 4. **Proof of part (b):** - Suppose a set $S$ has fewer than $n$ vectors, say $k < n$. - If $S$ spanned $V$, then $S$ would be a basis or could be extended to a basis. - But since $S$ has fewer than $n$ vectors, it cannot span $V$ because the dimension is $n$. - Therefore, $S$ does not span $V$. 5. **Proof of the Basis Theorem:** - Assume $V$ has a basis $B$ with $n$ vectors. - Suppose there is another basis $B'$ with $m$ vectors. - Since $B'$ is a basis, it is linearly independent and spans $V$. - By part (a), $m$ cannot be greater than $n$ (otherwise $B'$ would be dependent). - By part (b), $m$ cannot be less than $n$ (otherwise $B'$ would not span $V$). - Therefore, $m = n$. **Final conclusion:** - Any set of more than $n$ vectors in $V$ is linearly dependent. - Any set of fewer than $n$ vectors in $V$ cannot span $V$. - Every basis of $V$ has exactly $n$ vectors. This completes the proofs.