NoteNextra-origin/pages/Math429/Math429_L6.md

Lecture 6

Chapter II Finite Dimensional Subspaces

Span and Linear Independence 2A

Recall

Proposition 2.22

In a vector space V, a spanning list \{\vec{v_1},...,\vec{v_n}\}, and an linearly independent list \{\vec{w_1},...,\vec{w_n}\}. Then m\leq n.

Definition 2.26

A list \{\vec{v_1},...,\vec{v_n}\} is called a basis if it is a linearly independent spanning list.

Proposition 2.ex.1

A subspace of a finite dimensional vector space is finite-dimensional.

Proof: Let V be a finite-dimensional vector space and let W be a subspace of V

• Case 1: W=\{\vec{0}\}

• Case 2: Span\{\vec{v_1},...,\vec{v_{k-1}}\}\subset W where \vec{v_1},...,\vec{v_{k-1}} is linearly independent

  If W=Span\{\vec{v_1},...,\vec{v_{k-1}}\}, done. If not, then there exists \vec{v_{k-1}}\in W and \vec{v_k}\cancel{\in} Span\{\vec{v_1},...,\vec{v_{k-1}}\}. This implies Span\{\vec{v_1},...,\vec{v_k}\}\subset W. and \{\vec{v_1},...,\vec{v_k}\} is linearly independent. Continue until Span\{\vec{v_1},...,\vec{v_n}\}=W\subset V, V has a finite spanning set,whose size \geq n by Prop 2.22

Theorem 2.28

A list \{\vec{v_1},...,\vec{v_n}\} is a basis for V if and only if every vector \vec{v}\in V can be uniquely written as

\vec{v}=a_1\vec{v_1}+a_2\vec{v_2}+...+a_n\vec{v_n}

where a_1,...,a_n\in \mathbb{F}

Proof:

\Leftarrow

If every \vec{v}=a_1\vec{v_1}+a_2\vec{v_2}+...+a_n\vec{v_n} with unique choice of a_1,...,a_n, we will show \{\vec{v_1},...,\vec{v_n}\} is a basis

Since every \vec{v} is a linear combination of \{\vec{v_1},...,\vec{v_n}\}, we deduce V=Span\{\vec{v_1},...,\vec{v_n}\}

And by assumption, \vec{0}=a_1\vec{v_1}+a_2\vec{v_2}+...+a_n\vec{v_n} with unique choice of a_1,...,a_n\in \mathbb{F} (this choice is a_1=...=a_n=0) It implies \{\vec{v_1},...,\vec{v_n}\} is linearly independent.

So the list \{\vec{v_1},...,\vec{v_n}\} is a basis.

\Rightarrow

If \{\vec{v_1},...,\vec{v_n}\} is a basis, we will show that every \vec{v} can be uniquely written as \vec{v}=a_1\vec{v_1}+a_2\vec{v_2}+...+a_n\vec{v_n} with unique choice of a_1,...,a_n\in \mathbb{F}

Since \{\vec{v_1},...,\vec{v_n}\} is a basis, it must spans V with each vector being linearly independent.

Since \{\vec{v_1},...,\vec{v_n}\} spans V, there must be some a_1,...,a_n\in \mathbb{F} such that \vec{v}=a_1\vec{v_1}+a_2\vec{v_2}+...+a_n\vec{v_n}

Then \vec{0}=(a_1-b_1)\vec{v_1}+...+(a_n-b_n)\vec{v_n}

Since \{\vec{v_1},...,\vec{v_n}\} is linearly independent, this implies a_i-b_i=0

Lemma 2.30

Every Spanning set of a vector space can we be reduced into a basis.

ideas of Proof:

If the spanning list is not linearly independent, then use Lemma 2.19 to remove a vector.

Lemma 2.32

Every linearly independent list of vectors in a finite dimensional vector space can be extended with a basis.

ideas of Proof:

If \{\vec{v_1},...,\vec{v_{k-1}}\}, we can always add another vector \vec{v_k} \cancel{\in} Span\{\vec{v_1},...,\vec{v_{k-1}}\} to increase the span.

Theorem 2.31

Every finite dimensional vector space has a basis

Proposition (2.33)

Suppose that V is finite-dimensional and U\subset V is a subspace, then \exists W\subset V such that V= U \oplus W

Proof

Since U is a subspace of V, then U is also finite dimensional. Thus U has a basis \{\vec{u_1},...,\vec{u_k}\} This list is linearly independent. So we can extend it into a basis for V, \{\vec{u_1},..,\vec{u_k},\vec{w_1},...,\vec{w_s}\}. Now let W=Span\{\vec{u_1},..,\vec{u_k},\vec{w_1},...,\vec{w_s}\}

Now we need to prove V=U\oplus W.

Since U\subset V and W\subset V then V+W\subset V because U+W is the smallest vector space containing U and W.

Since \{\vec{u_1},..,\vec{u_k},\vec{w_1},...,\vec{w_s}\} is a basis of V, every \vec{v}\in V, \vec{v}\in Span\{\vec{u_1},..,\vec{u_k},\vec{w_1},...,\vec{w_s}\}

\vec{v}=a_1\vec{u_1}+...+a_k\vec{u_k}+b_1\vec{w_1}+...+b_s\vec{w_s}\\

So \vec{v}\in V+W. V=V+W

If \vec{v}\in U\bigcap W, then \vec{v}=a_1\vec{u_1}+...+a_k\vec{u_k}\in V, \vec{v}=b_1\vec{w_1}+...+b_s\vec{w_s}\in W, but \{\vec{u_1},..,\vec{u_k},\vec{w_1},...,\vec{w_s}\} should be an linearly independent spanning set. this implies a_i,b_j=0 So \vec{v}=0