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Math429Linear Algebra (Lecture 5)

Lecture 5

Chapter II Finite Dimensional Subspaces

Span and Linear Independence 2A

Definition 2.15

A list of vector v1⃗,...,vm⃗\vec{v_1},...,\vec{v_m} in VV is called linearly independent if the only choice for a1,...,am∈Fa_1,...,a_m\in \mathbb{F} such that a1v1⃗+...+amvm⃗=0⃗a_1\vec{v_1}+...+a_m\vec{v_m}=\vec{0} is a1=...=am=0a_1=...=a_m=0

If {v1⃗,...,vm⃗}\{\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, {v⃗}\{\vec{v}\}, is lienarly independent, if av⃗=0⃗  ⟹  a=0a\vec{v}=\vec{0}\implies a=0. This implication holds when as long as v⃗≠0⃗\vec{v}\neq \vec{0}.

  • Consider V=F3V=\mathbb{F}^3 {(1,2,3),(1,1,1)}\{(1,2,3),(1,1,1)\}, more generally, {v1⃗,v2⃗}\{\vec{v_1},\vec{v_2}\}, by the definition of linear independence, 0⃗=a1v1⃗+a2v2⃗\vec{0}=a_1\vec{v_1}+a_2\vec{v_2}. This is equivalent to a1v1⃗=−a2v2⃗a_1\vec{v_1}=-a_2\vec{v_2}

    • Case 1: if any of the vector is a zero vector v1⃗=0⃗\vec{v_1}=\vec{0} or v2⃗=0⃗\vec{v_2}=\vec{0}, assume ( v2⃗=0⃗\vec{v_2}=\vec{0} ) then for a1=0a_1=0 and any a2a_2, a1v1⃗=−a2v2⃗a_1\vec{v_1}=-a_2\vec{v_2}.

    • Case 2: if v1⃗≠0⃗\vec{v_1}\neq \vec{0} and v2⃗≠0⃗\vec{v_2}\neq \vec{0} a1v1⃗=−a2v2⃗a_1\vec{v_1}=-a_2\vec{v_2} implies that they lie on the same line.

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

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

Lemma (weak version)

A list of {v1⃗,...,vm⃗}\{\vec{v_1},...,\vec{v_m}\} is linearly dependent   ⟺  \iff there is a vk⃗\vec{v_k} satisfying vk⃗=a1v1⃗+...+ak−1vk−1⃗+ak+1vk+1⃗+...+amvm⃗\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} (vk∈Span{v1⃗,...,vk−1⃗,vk+1⃗,...,vk⃗}v_k\in Span\{\vec{v_1},...,\vec{v_{k-1}},\vec{v_{k+1}},...,\vec{v_k}\})

Proof:

{v1⃗,...,vm⃗}\{\vec{v_1},...,\vec{v_m}\} is linearly dependent   ⟺  \iff a1v1⃗+...+amvm⃗=0⃗a_1\vec{v_1}+...+a_m\vec{v_m}=\vec{0} (with at least one ak≠0a_k\neq 0)

If akvk⃗=−(a1v1⃗+...+ak−1vk−1⃗+ak+1vk+1⃗+...+amvm⃗)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 vk⃗=−1ak(a1v1⃗+...+ak−1vk−1⃗+ak+1vk+1⃗+...+amvm⃗)\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 {v1⃗,...,vm⃗}\{\vec{v_1},...,\vec{v_m}\} is linearly dependent, then ∃vk⃗∈Span{v1⃗,...,vk−1⃗}\exists \vec{v_k} \in Span\{\vec{v_1},...,\vec{v_{k-1}}\}. Moreover, Span{v1⃗,...,vm⃗}=Span{v1⃗,...,vk−1⃗,vk+1⃗,...,vk⃗}Span\{\vec{v_1},...,\vec{v_m}\}=Span\{\vec{v_1},...,\vec{v_{k-1}},\vec{v_{k+1}},...,\vec{v_k}\}

Proof:

{v1⃗,...,vm⃗}\{\vec{v_1},...,\vec{v_m}\} is linearly dependent   ⟹  \implies a1v1⃗+...+amvm⃗=0⃗a_1\vec{v_1}+...+a_m\vec{v_m}=\vec{0}. Let kk be the maximal ii such that ai≠0a_i\neq 0

If v⃗=b1v1⃗+...+bmvm⃗\vec{v}=b_1\vec{v_1}+...+b_m\vec{v_m}, then v⃗=b1v1⃗+...+bk−1vk−1⃗+bk(−1ak(a1v1⃗+....+ak−1vk−1⃗))+bk+1vk+1⃗+...+bmvm⃗∈Span{v1⃗,...,vk−1⃗,vk+1⃗,...,vk⃗}\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 {v1⃗,...,vm⃗}\{\vec{v_1},...,\vec{v_m}\} is linearly independent set, and {u1⃗,...,un⃗}\{\vec{u_1},...,\vec{u_n}\} is a Spanning set, then m≤nm\leq n.

Since Span{u1⃗,...,un⃗}=VSpan\{\vec{u_1},...,\vec{u_n}\}=V , for each vi⃗=a1u1⃗+...+anun⃗\vec{v_i}=a_1\vec{u_1}+...+a_n\vec{u_n} for some scalar a1,...,ana_1,...,a_n. Consider the equation x1v1⃗+...+xmvm⃗=0⃗x_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 ui⃗\vec{u_i} being replaced by v−j⃗\vec{v-j}‘s

Step 1. Consider set {v1⃗,u1⃗,u2⃗,...,un⃗}=V\{\vec{v_1},\vec{u_1},\vec{u_2},...,\vec{u_n}\}=V. Because v1⃗∈Span{u1⃗,...,un⃗}\vec{v_1}\in Span\{\vec{u_1},...,\vec{u_n}\} then the set is linearly dependent. by lemma 2.19, ∃i\exists i such that ui⃗∈Span{v1⃗,u1⃗,u2⃗,...,un⃗}\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 ui⃗\vec{u_i} such that the set is still a Spanning set V={v1⃗,u1⃗,u2⃗,...,ui−1⃗,ui+1⃗,...,un⃗}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 {v1⃗,...,vk⃗,us⃗,...,ut⃗}=V\{\vec{v_1},...,\vec{v_k},\vec{u_s},...,\vec{u_t}\}=V

Step k-1. Consider set {v1⃗,...,vk−1⃗,vk⃗,us⃗,...,ut⃗}=V\{\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 uj⃗∈Span{v1⃗,...,vk−1⃗,vk⃗,us⃗,...,ur⃗}\vec{u_j}\in Span\{\vec{v_1},...,\vec{v_{k-1}},\vec{v_k},\vec{u_s},...,\vec{u_r}\}. with r<jr<j. Then we remove uj⃗\vec{u_j} and update the set.

Basis 2B

Definition 2.26

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

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