NoteNextra-origin/content/Math429/Math429_L5.md

Lecture 5

Chapter II Finite Dimensional Subspaces

Span and Linear Independence 2A

Definition 2.15

A list of vector \vec{v_1},...,\vec{v_m} in V is called linearly independent if the only choice for a_1,...,a_m\in \mathbb{F} such that a_1\vec{v_1}+...+a_m\vec{v_m}=\vec{0} is a_1=...=a_m=0

If \{\vec{v_1},...,\vec{v_m}\} is NOT linearly independent then we call them linearly dependent.

Examples:

• The empty list is linearly independent.

• Consider the list with a single vector, \{\vec{v}\}, is lienarly independent, if a\vec{v}=\vec{0}\implies a=0. This implication holds when as long as \vec{v}\neq \vec{0}.

• Consider V=\mathbb{F}^3 \{(1,2,3),(1,1,1)\}, more generally, \{\vec{v_1},\vec{v_2}\}, by the definition of linear independence, \vec{0}=a_1\vec{v_1}+a_2\vec{v_2}. This is equivalent to a_1\vec{v_1}=-a_2\vec{v_2}

  • Case 1: if any of the vector is a zero vector \vec{v_1}=\vec{0} or \vec{v_2}=\vec{0}, assume ( \vec{v_2}=\vec{0} ) then for a_1=0 and any a_2, a_1\vec{v_1}=-a_2\vec{v_2}.

  • Case 2: if \vec{v_1}\neq \vec{0} and \vec{v_2}\neq \vec{0} a_1\vec{v_1}=-a_2\vec{v_2} implies that they lie on the same line.

  \{(1,2,3),(1,1,1)\} is linearly independent.

• Consider the list \{(1,2,3),(1,1,1),(-1,0,1)\}, since we can get \vec{0} from a non-trivial solution (1,2,3)-2(1,1,1)-(-1,0,1)=\vec{0}

Lemma (weak version)

A list of \{\vec{v_1},...,\vec{v_m}\} is linearly dependent \iff there is a \vec{v_k} satisfying \vec{v_k}=a_1\vec{v_1}+...+a_{k-1}\vec{v_{k-1}}+a_{k+1}\vec{v_{k+1}}+...+a_m\vec{v_m} (v_k\in Span\{\vec{v_1},...,\vec{v_{k-1}},\vec{v_{k+1}},...,\vec{v_k}\})

Proof:

\{\vec{v_1},...,\vec{v_m}\} is linearly dependent \iff a_1\vec{v_1}+...+a_m\vec{v_m}=\vec{0} (with at least one a_k\neq 0)

If a_k\vec{v_k}=-(a_1\vec{v_1}+...+a_{k-1}\vec{v_{k-1}}+a_{k+1}\vec{v_{k+1}}+...+a_m\vec{v_m}), then \vec{v_k}=-\frac{1}{a_k}(a_1\vec{v_1}+...+a_{k-1}\vec{v_{k-1}}+a_{k+1}\vec{v_{k+1}}+...+a_m\vec{v_m})

Lemma (2.19) (strong version)

If \{\vec{v_1},...,\vec{v_m}\} is linearly dependent, then \exists \vec{v_k} \in Span\{\vec{v_1},...,\vec{v_{k-1}}\}. Moreover, Span\{\vec{v_1},...,\vec{v_m}\}=Span\{\vec{v_1},...,\vec{v_{k-1}},\vec{v_{k+1}},...,\vec{v_k}\}

Proof:

\{\vec{v_1},...,\vec{v_m}\} is linearly dependent \implies a_1\vec{v_1}+...+a_m\vec{v_m}=\vec{0}. Let k be the maximal i such that a_i\neq 0

If \vec{v}=b_1\vec{v_1}+...+b_m\vec{v_m}, then \vec{v}=b_1\vec{v_1}+...+b_{k-1}\vec{v_{k-1}}+b_{k}(-\frac{1}{a_k}(a_1\vec{v_1}+....+a_{k-1}\vec{v_{k-1}}))+b_{k+1}\vec{v_{k+1}}+...+b_m\vec{v_m}\in Span\{\vec{v_1},...,\vec{v_{k-1}},\vec{v_{k+1}},...,\vec{v_k}\}

Proposition 2.22

In a finite dimensional vector space, if \{\vec{v_1},...,\vec{v_m}\} is linearly independent set, and \{\vec{u_1},...,\vec{u_n}\} is a Spanning set, then m\leq n.

Since Span\{\vec{u_1},...,\vec{u_n}\}=V , for each \vec{v_i}=a_1\vec{u_1}+...+a_n\vec{u_n} for some scalar a_1,...,a_n. Consider the equation x_1\vec{v_1}+...+x_m\vec{v_m}=\vec{0}, (if we write it to the matrix form, it will have more columns than the rows. It is guaranteed to have free variables.)

Proof:

We will construct a new Spanning set with elements \vec{u_i} being replaced by $\vec{v-j}$'s

Step 1. Consider set \{\vec{v_1},\vec{u_1},\vec{u_2},...,\vec{u_n}\}=V. Because \vec{v_1}\in Span\{\vec{u_1},...,\vec{u_n}\} then the set is linearly dependent. by lemma 2.19, \exists i such that \vec{u_i}\in Span\{\vec{v_1},\vec{u_1},\vec{u_2},...,\vec{u_n}\}. The lemma 2.19 also implies that we cna remove \vec{u_i} such that the set is still a Spanning set V=\{\vec{v_1},\vec{u_1},\vec{u_2},...,\vec{u_{i-1}},\vec{u_{i+1}},...,\vec{u_n}\}

Step 2. Consider set \{\vec{v_1},...,\vec{v_k},\vec{u_s},...,\vec{u_t}\}=V

Step k-1. Consider set \{\vec{v_1},...,\vec{v_{k-1}},\vec{v_k},\vec{u_s},...,\vec{u_t}\}=V which is linearly dependent. Apply lemma 2.19 again, we can find there is a \vec{u_j}\in Span\{\vec{v_1},...,\vec{v_{k-1}},\vec{v_k},\vec{u_s},...,\vec{u_r}\}. with r<j. Then we remove \vec{u_j} and update the set.

Basis 2B

Definition 2.26

A linearly independent Spanning set is called a basis. "smallest spanning set"