11.11 Finite-Dimensional Vector Spaces

Recall

A list of vectors in \(V_1\) is a finite sequence \(v_1,...,v_k\) with \(v_i\in V\) and no further restrictions.

It's length is \(k\)

A linear combination of the list \(v_1,...,v_k\) is any vector \(v\in V\) that can be written as: \(v=\lambda_1 v_1+...+\lambda_k v_k,\lambda_i\in \mathbb{F}\)

The span of the list \(v_1,...,v_k\) is the subspace \(\langle v_1,\ldots,v_{k}\rangle\) of all linear combination of \(v_1,...,v_k\)

If \(W=\langle v_1,\ldots,v_{k}\rangle\) we say that \(v_1,...,v_k\) span or generate \(W\). Let's agree on: \(\langle\emptyset\rangle=\left\lbrace0\right\rbrace\)

Finitely generated vector space

Definition

A vector space \(V\) is finitely generated, if there is a list \(v_1,...,v_k\) in \(V\) such that \(V=\langle v_1,\ldots,v_{k}\rangle\)

Example

\(V=\mathbb{F}^n=\lang e_1,...,e_n\rangle\) with \(e_{i}=(0,...,1,...,0)\) \(1\) is the i-th
\(V=\mathbb{F}_{n}\left\lbrack x\right\rbrack=\langle1,x,x^2,...,x^{n}\rangle\)
\(M_{m\times n}\left(\mathbb{F}\right)=\langle E_{ij}:1\leq i\leq m,1\leq j\leq n\rangle\) where \(\left(E_{ij}\right)_{lk}=\delta_{il}\delta_{jk},1\leq l\leq m,1\leq k\leq n\)

Kronecker delta:\(\delta_{ij}=\begin{cases}0, & i\neq j\\ 1, & i=j\end{cases}\)
\(V=\mathbb{F}[x]\) is not finitely generated.

Assume YES. \(V=\langle p_1,p_2,\dots,p_{k}\rangle\Rightarrow p=\lambda_1p_1+\dots+\lambda_{k}p_{k}\Rightarrow\deg p\leq\deg p_1+\dots+\deg p_{k}=M\)

Here \(p(x)=x^{M+1}\notin\langle p_1,p_2,\dots,p_{k}\rangle\).

Example

\(\mathbb{Z}_3^3 = \langle (1,1,1), (1,1,0), (1,0,0) \rangle\)

\(=\langle(a,b,c):a,b,c\in\{0,1,2,3\}\rangle\)
Does \((1,0,2), (0,2,1), (2,0,1)\) generate \(\mathbb{Z}_5^3\)?

\(\mathbb{Z}_5^3 = (a, b, c) = \lambda_1 (1,0,2) + \lambda_2 (0,2,1) + \lambda_3 (2,0,1) \quad ? \quad \text{for } \lambda_1, \lambda_2, \lambda_3\)

\(= (\lambda_1 + 2 \lambda_3, 2 \lambda_2, 2 \lambda_1 + \lambda_2 + \lambda_3)\)

\(\begin{cases}\lambda_1+2\lambda_3=a\\ 2\lambda_2=b\\ 2\lambda_1+\lambda_2+\lambda_3=c\end{cases}\) \(\Rightarrow\begin{pmatrix}\begin{array}{ccc|c}1 & 0 & 2 & a\\ 0 & 2 & 0 & b\\ 2 & 1 & 1 & c\end{array}\end{pmatrix}\Rightarrow\begin{pmatrix}\begin{array}{ccc|c}0 & 0 & 1 & \dfrac{2a}{3}+\dfrac{b}{6}-\dfrac{c}{3}\\ 0 & 1 & 0 & \dfrac{b}{2}\\ 1 & 0 & 0 & \dfrac{2c}{3}-\dfrac{b}{3}-\dfrac{a}{3}\end{array}\end{pmatrix}\)

Linear independence/dependence

Definition

A list \(v_1,...,v_k\) in \(V\) is linearly independent if \(0=\lambda_1v_1+...+\lambda_{k}v_{k}\) implies that \(\lambda_1=...=\lambda_k=0\). It is a linear dependent if it is not independent.

The only option is to write \(0\) is the trivial way

Lemma

Let \(v_1,...,v_k\) be a linear independent list in \(V\). The every \(v\in\lang v_1,...,v_k\rang\) is \(v=\lambda_1v_1+...+\lambda_kv_k\) for unique \(\lambda_1,...,\lambda_k\)

Proof

Assume \(v=\lambda_1v_1+...+\lambda_kv_k\) and \(v=\mu_1v_1+...+\mu_{k}v_{k}\)

Then \(0=\left(\lambda_1-\mu_1\right)v_1+...+\left(\lambda_{k}-\mu_{k}\right)v_{k}\)

Since \(v_1,...,v_k\) is linear independent, then by definition we have \(\lambda_1-\mu_1=0,\ldots,\lambda_{k}-\mu_{k}=0\)

Therefore \(\lambda_1=\mu_1,\ldots,\lambda_{k}=\mu_{k}\)

Example

\(v_1\) is linear independent iff \(v_1\neq 0\)
\(v_1,-v_1\) is linear dependent since \(0=0v_1+0v_2\) and \(0=1\cdot v_1+1\cdot (-v_1)\)
\(v_1,v_2,v_3,v_2+2v_1\) is linear dependent since \(0=1\cdot\left(v_2+2v_1\right)+\left(-1\right)\cdot v_2+\left(-2\right)\cdot v_1+0\cdot v_3\) (non-trivial!!!) and \(0=0\cdot\left(v_2+2v_1\right)+0\cdot v_2+0\cdot v_1+0\cdot v_3\)

Propositions

Let \(v_1,...,v_k\) be a linear dependent list in \(V\). Then

\(\exists j\) such that \(v_{j}\in\langle v_1,\ldots,v_{j-1}\rangle\)
\(\langle v_{1},\ldots,\overset{\wedge}{v_{j}},\ldots,v_{k}\rangle=\langle v_{1},. ..,v_{k}\rangle\)

remove \(v_j\)

Proof

Since \(v_1,\dots,v_{k}\) is linear dependent, there exist \(\lambda_1,\dots,\lambda_{k}\) not all \(=0\) such that

\(0=\lambda_1v_1+\dots+\lambda_{k}v_{k}\). Let \(j\) be the largest index such that \(\lambda_j \neq 0\).

Hence \(-\lambda_j v_j = \lambda_1 v_1 + \dots + \lambda_{j-1} v_{j-1}\) (where \(\lambda_i = 0\) for \(i > j\)).

\(\Rightarrow v_{j}=-\frac{\lambda_1}{\lambda_{j}}v_1+\ldots+\left(-\frac{\lambda_{j-1}}{\lambda_{j}}v_{j-1}\right)\).

(b) \(\langle v_1,\ldots,\overset{\wedge}{v_{j}},\ldots,v_{k}\rangle\subseteq\langle v_1,...,v_{k}\rangle\) is clear.

Given \(v\in\langle v_{j},\dots,v_{n}\rangle\Rightarrow v=\mu_1v_1+\dots+\mu_{j}v_{j}+\dots+\mu_{k}v_{k}\).

\(\Rightarrow v=\mu_1v_1+\dots+\mu_{j}\left(-\frac{\lambda_1}{\lambda_{j}}v_1-\dots-\frac{\lambda_{j-1}}{\lambda_{j}}v_{j-1}\right)+\dots+\mu_{k}v_{k}\)

\(=\left(\mu_1-\frac{\mu_{j}\lambda_1}{\lambda_{j}}\right)v_1+\dots+\left(\mu_{j-1}-\frac{\mu_{j}\lambda_{j-1}}{\lambda_{j}}\right)v_{j-1}+0\cdot v_{j}+\mu_{j+1}v_{j+1}+\dots+\mu_{k}v_{k}\).

Theorem

Let \(v_1,...,v_n\) be a linear independent list in \(V\). If \(w_1,...,w_m\) is a list in \(V\) such that \(V=\langle w_1,\ldots,w_{m}\rangle\), then \(n\leq m\)

Proof

Consider \(v_1,w_1,...,w_m\) which is linear dependent. (not unique way to write \(0\))

Previous proposition implies that \(\exists j_1\geq 1\) such that \(\langle v_1,w_1,\ldots,w_{m}\rangle=\langle v_1,w_1,\ldots,\overset{\wedge}{w_{j1}},\ldots,w_{m}\rangle=V\)

Now consider \(v_2,v_1,w_1,...,w_{m}\) which is linear dependent.

Previous proposition implies that \(\exists j_2\geq1,j_2\neq j_1\) such that \(\langle v_2,v_1,w_1,\ldots,w_{m}\rangle=\langle v_1,w_1,\ldots,\overset{\wedge}{w_{j1}},\overset{\wedge}{w_{j2}},\ldots,w_{m}\rangle=V\)

Inductively, we'll get \(\langle v_{n},\ldots,v_1,w_1,\overset{\wedge}{\ldots},w_{m}\rangle\). The number of \(w\) having \(\wedge\) is \(m-n\).

In particular, \(m\geq n\)

Finite-dimensional vector space

A finite-dimensional vector space is a vector space that has a finite basis

A finite-dimensional space requires that there exists at least one finite spanning set.

Theorem

A subspace \(W\) of a finite-dimensional \(V\), is a finite-dimensional.

Proof

Assume \(V = \langle v_1, \dots, v_m \rangle\). Let \(0 \neq w_1 \in W\), \(w_1\) is linear independent. If \(\langle w_1 \rangle = W\), we are done. If not, let \(0\neq w_2\in W-\langle w_1\rangle\).

Then \(w_1, w_2\) is linearly independent. If \(\langle w_1, w_2 \rangle = W\), we are done. If not, let \(0\neq w_3\in W-\langle w_1,w_2\rangle\).

Iterating, we construct \(w_1, w_2, \dots, w_n\) as a linearly independent list in \(V\).

Hence \(n\leq m\) and \(\lang w_1,...,w_n\rang=W\) for some \(n\)

Bases (the phural of basis)

Definition

A basis of a finite-dimensional vector space \(V\), is a linear independent list in \(V\), \(v_1, \dots, v_n\) that generates \(V\).

Remark

Any \(v \in V\) can be written as a linear combination of \(v_1, \dots, v_n\) in a unique way.

Example

(1) \(e_1, \dots, e_n\) in \(\mathbb{F}^n\) is a basis.

(2) \((1,0,1), (-1,2,3), (0,\frac{1}{2},1), (1,1,1)\) is not a basis of \(\mathbb{Q}^3\) (\(\mathbb{R}^3, \mathbb{C}^3\)).

Remark

If \(v_1, \dots, v_n\) is a basis of \(V\), then \(\forall v \in V\), \(\exists! \lambda_1, \dots, \lambda_n\) such that \(v = \lambda_1 v_1 + \dots + \lambda_n v_n\).

Notation

\((\lambda_1,\dots,\lambda_{n})_{\mathcal{B}}\in\mathbb{F}^{n}\) are the coordinates of \(v\) with respect to the basis \(\mathcal{B}: v_1, \dots, v_n\).

Prop implies that every finite-dimensional vector space has a linearly independent list of generators.

That is a basis!!!

Theorem

Every finite-dimensional vector space has a basis

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