11.15 Linear independent and basis

CLARIFICATION: \(\{0\}\) vector space is the only one with only one basis.

\(\times \{1\}\) is NOT a basis for \(\{0\}\).

\(\emptyset\) is the only basis for \(\{0\}\).

1. Suppose \(V\) is finite-dimensional and \(U \subseteq V\) is a subspace of \(V\) such that \(\dim(U) = \dim(V) \implies U = V\).

   Let \(n = \dim(U) = \dim(V)\) and let \(\{u_1, \dots, u_n\}\) be a basis for \(U\).

   Since \(\{u_1, \dots, u_n\}\) is linearly independent and it has the same length as \(\dim(V) = n \implies \{u_1, \dots, u_n\}\) is a basis for \(V\) (theorem).

   We want to show that \(U = V\). Since \(U \subseteq V\), we just need to show \(V \subseteq U\).

   Let \(v \in V\), since \(\{u_1, \dots, u_n\}\) is a basis \(\implies v = a_1 u_1 + \dots + a_n u_n \in U\).

   Therefore \(U = V\).

2. Suppose \(U_1, \dots, U_n\) are finite-dimensional subspaces of \(V\). Prove that \(U_1 + \dots + U_n\) is finite-dimensional and \(\dim(U_1 + \dots + U_n) \leq \dim(U_1) + \dots + \dim(U_n)\).

   Let \(B_{k}=\{v_1^{k},\dots,v_{n}^{k}\}\) be a basis for \(U_k\), \(k = 1, \dots, n\).

   Now consider \(B = B_1 \cup B_2 \cup \dots \cup B_n\), we have if \(u_1 + \dots + u_n \in U_1 + \dots + U_n\)

   Since \(u_k \in U_k\), \(\exists a_i^k \in \mathbb{F}\) such that \(u_{k}=a_1^{k}v_1^{k}+\dots+a_{n}^{k}v_{n}^{k}\).

   So \(u_1+\dots+u_{n}=(a_1^1v_1^1+\dots+a_{n_{}}^1v_{n}^1)+\dots+(a_1^{n}v_1^{n}+\dots+a_{n}^{n}v_{n}^{n})\)
   with each term \(\in B\).

   So \(B\) spans \(U_1 + \dots + U_n\) (but it could not be linearly independent).

   \(\implies\dim(U_1+\dots+U_{n})\leq|B|=\dim(U_1)+\dots+\dim(U_{n})\).

3. Suppose that \(U\) and \(W\) are subspaces of \(\mathbb{F}^8\) such that \(\dim(U) = 3\) and \(\dim(W) = 5\) and \(U + W = \mathbb{F}^8\).

   Prove that \(U \oplus W = \mathbb{F}^8\).

   Remember that \(U \oplus W = \mathbb{F}^8 \iff U + W = \mathbb{F}^8\) and \(U \cap W = \{0\}\).

   We have that
   ​\(\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W)\).

   \(\implies 8 = \dim(\mathbb{F}^8) = \dim(U + W) = 3 + 5 - \dim(U \cap W)\)

   Therefore, \(\dim(U \cap W) = 0\), which implies \(U \cap W = \{0\}\).

   Thus, \(U \oplus W = \mathbb{F}^8\).

4. Let \(U\) and \(W\) be subspaces of \(\mathbb{F}^6\) with \(\dim(U) = \dim(W) = 4\). Prove that there exist 2 vectors in \(U \cap W\) such that neither of these vectors is a scalar multiple of the other.

   (i.e., \(\exists v_1, v_2 \in U \cap W\) such that \(v_1 \neq c v_2\) and \(v_2 \neq d v_1\) for any scalars \(c, d \in \mathbb{F}\), so \(\{v_1, v_2\}\) is linearly independent.)

   We know that
   ​\(\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W)\).

   \(\implies \dim(U \cap W) = \dim(U) + \dim(W) - \dim(U + W)\)

   \(= 4 + 4 - \dim(U + W)\)

   \(= 8 - \dim(U + W)\).

   Since \(U\) and \(W\) are subspaces of \(\mathbb{F}^6 \implies U + W\) is also a subspace of \(\mathbb{F}^6\).

   \(\implies \dim(U + W) \leq \dim(\mathbb{F}^6) = 6\)

   \(\implies \dim(U \cap W) = 8 - \dim(U + W) \geq 8 - 6 = 2\)

   So \(\dim(U \cap W) \geq 2\). Therefore, any basis for \(U \cap W\) has at least 2 elements. We consider a subset \(\{v_1, v_2\}\) of any basis \(\{v_1, v_2, \dots, v_s\}\) (of \(U \cap W\)). Since any subset of a linearly independent set is also linearly independent, \(\{v_1, v_2\}\) is linearly independent.

5. Let \(U_1, U_2, U_3\) be subspaces of a finite-dimensional vector space \(V\).

   We know that \(\dim(U_1 + U_2) = \dim(U_1) + \dim(U_2) - \dim(U_1 \cap U_2)\).

   Is it true that:

   \(\dim(U_1 + U_2 + U_3) = \dim(U_1) + \dim(U_2) + \dim(U_3) - \dim(U_1 \cap U_2) - \dim(U_1 \cap U_3) - \dim(U_2 \cap U_3) + \dim(U_1 \cap U_2 \cap U_3)?\)

   Counterexample: Consider \(U_1 = \mathbb{R} \times \{0\}\), \(U_2 = \{(x,x) \in \mathbb{R}^2 : x \in \mathbb{R}\}\), and \(U_3 = \{0\} \times \mathbb{R}\).

   Note that \(U_1 + U_2 + U_3 = \mathbb{R}^2\), and \(U_1 \cap U_2 = \{(0,0)\}\).