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Math429
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Math429 L5

Lecture 5

Chapter II Finite Dimensional Subspaces

Span and Linear Independence 2A

Definition 2.15

A list of vector v1โƒ—,...,vmโƒ—\vec{v_1},...,\vec{v_m} in VV is called linearly independent if the only choice for a1,...,amโˆˆFa_1,...,a_m\in \mathbb{F} such that a1v1โƒ—+...+amvmโƒ—=0โƒ—a_1\vec{v_1}+...+a_m\vec{v_m}=\vec{0} is a1=...=am=0a_1=...=a_m=0

If {v1โƒ—,...,vmโƒ—}\{\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, {vโƒ—}\{\vec{v}\}, is lienarly independent, if avโƒ—=0โƒ—โ€…โ€ŠโŸนโ€…โ€Ša=0a\vec{v}=\vec{0}\implies a=0. This implication holds when as long as vโƒ—โ‰ 0โƒ—\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, {v1โƒ—,v2โƒ—}\{\vec{v_1},\vec{v_2}\}, by the definition of linear independence, 0โƒ—=a1v1โƒ—+a2v2โƒ—\vec{0}=a_1\vec{v_1}+a_2\vec{v_2}. This is equivalent to a1v1โƒ—=โˆ’a2v2โƒ—a_1\vec{v_1}=-a_2\vec{v_2}

    • Case 1: if any of the vector is a zero vector v1โƒ—=0โƒ—\vec{v_1}=\vec{0} or v2โƒ—=0โƒ—\vec{v_2}=\vec{0}, assume ( v2โƒ—=0โƒ—\vec{v_2}=\vec{0} ) then for a1=0a_1=0 and any a2a_2, a1v1โƒ—=โˆ’a2v2โƒ—a_1\vec{v_1}=-a_2\vec{v_2}.

    • Case 2: if v1โƒ—โ‰ 0โƒ—\vec{v_1}\neq \vec{0} and v2โƒ—โ‰ 0โƒ—\vec{v_2}\neq \vec{0} a1v1โƒ—=โˆ’a2v2โƒ—a_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),(โˆ’1,0,1)}\{(1,2,3),(1,1,1),(-1,0,1)\}, since we can get 0โƒ—\vec{0} from a non-trivial solution (1,2,3)โˆ’2(1,1,1)โˆ’(โˆ’1,0,1)=0โƒ—(1,2,3)-2(1,1,1)-(-1,0,1)=\vec{0}

Lemma (weak version)

A list of {v1โƒ—,...,vmโƒ—}\{\vec{v_1},...,\vec{v_m}\} is linearly dependent โ€…โ€ŠโŸบโ€…โ€Š\iff there is a vkโƒ—\vec{v_k} satisfying vkโƒ—=a1v1โƒ—+...+akโˆ’1vkโˆ’1โƒ—+ak+1vk+1โƒ—+...+amvmโƒ—\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} (vkโˆˆSpan{v1โƒ—,...,vkโˆ’1โƒ—,vk+1โƒ—,...,vkโƒ—}v_k\in Span\{\vec{v_1},...,\vec{v_{k-1}},\vec{v_{k+1}},...,\vec{v_k}\})

Proof:

{v1โƒ—,...,vmโƒ—}\{\vec{v_1},...,\vec{v_m}\} is linearly dependent โ€…โ€ŠโŸบโ€…โ€Š\iff a1v1โƒ—+...+amvmโƒ—=0โƒ—a_1\vec{v_1}+...+a_m\vec{v_m}=\vec{0} (with at least one akโ‰ 0a_k\neq 0)

If akvkโƒ—=โˆ’(a1v1โƒ—+...+akโˆ’1vkโˆ’1โƒ—+ak+1vk+1โƒ—+...+amvmโƒ—)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 vkโƒ—=โˆ’1ak(a1v1โƒ—+...+akโˆ’1vkโˆ’1โƒ—+ak+1vk+1โƒ—+...+amvmโƒ—)\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 {v1โƒ—,...,vmโƒ—}\{\vec{v_1},...,\vec{v_m}\} is linearly dependent, then โˆƒvkโƒ—โˆˆSpan{v1โƒ—,...,vkโˆ’1โƒ—}\exists \vec{v_k} \in Span\{\vec{v_1},...,\vec{v_{k-1}}\}. Moreover, Span{v1โƒ—,...,vmโƒ—}=Span{v1โƒ—,...,vkโˆ’1โƒ—,vk+1โƒ—,...,vkโƒ—}Span\{\vec{v_1},...,\vec{v_m}\}=Span\{\vec{v_1},...,\vec{v_{k-1}},\vec{v_{k+1}},...,\vec{v_k}\}

Proof:

{v1โƒ—,...,vmโƒ—}\{\vec{v_1},...,\vec{v_m}\} is linearly dependent โ€…โ€ŠโŸนโ€…โ€Š\implies a1v1โƒ—+...+amvmโƒ—=0โƒ—a_1\vec{v_1}+...+a_m\vec{v_m}=\vec{0}. Let kk be the maximal ii such that aiโ‰ 0a_i\neq 0

If vโƒ—=b1v1โƒ—+...+bmvmโƒ—\vec{v}=b_1\vec{v_1}+...+b_m\vec{v_m}, then vโƒ—=b1v1โƒ—+...+bkโˆ’1vkโˆ’1โƒ—+bk(โˆ’1ak(a1v1โƒ—+....+akโˆ’1vkโˆ’1โƒ—))+bk+1vk+1โƒ—+...+bmvmโƒ—โˆˆSpan{v1โƒ—,...,vkโˆ’1โƒ—,vk+1โƒ—,...,vkโƒ—}\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 {v1โƒ—,...,vmโƒ—}\{\vec{v_1},...,\vec{v_m}\} is linearly independent set, and {u1โƒ—,...,unโƒ—}\{\vec{u_1},...,\vec{u_n}\} is a Spanning set, then mโ‰คnm\leq n.

Since Span{u1โƒ—,...,unโƒ—}=VSpan\{\vec{u_1},...,\vec{u_n}\}=V , for each viโƒ—=a1u1โƒ—+...+anunโƒ—\vec{v_i}=a_1\vec{u_1}+...+a_n\vec{u_n} for some scalar a1,...,ana_1,...,a_n. Consider the equation x1v1โƒ—+...+xmvmโƒ—=0โƒ—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 uiโƒ—\vec{u_i} being replaced by vโˆ’jโƒ—\vec{v-j}'s

Step 1. Consider set {v1โƒ—,u1โƒ—,u2โƒ—,...,unโƒ—}=V\{\vec{v_1},\vec{u_1},\vec{u_2},...,\vec{u_n}\}=V. Because v1โƒ—โˆˆSpan{u1โƒ—,...,unโƒ—}\vec{v_1}\in Span\{\vec{u_1},...,\vec{u_n}\} then the set is linearly dependent. by lemma 2.19, โˆƒi\exists i such that uiโƒ—โˆˆSpan{v1โƒ—,u1โƒ—,u2โƒ—,...,unโƒ—}\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 uiโƒ—\vec{u_i} such that the set is still a Spanning set V={v1โƒ—,u1โƒ—,u2โƒ—,...,uiโˆ’1โƒ—,ui+1โƒ—,...,unโƒ—}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 {v1โƒ—,...,vkโƒ—,usโƒ—,...,utโƒ—}=V\{\vec{v_1},...,\vec{v_k},\vec{u_s},...,\vec{u_t}\}=V

Step k-1. Consider set {v1โƒ—,...,vkโˆ’1โƒ—,vkโƒ—,usโƒ—,...,utโƒ—}=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 ujโƒ—โˆˆSpan{v1โƒ—,...,vkโˆ’1โƒ—,vkโƒ—,usโƒ—,...,urโƒ—}\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 ujโƒ—\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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