# MATH 323 Lecture 13

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Brain dump on Vector Space

## Review

1. span: $\mathrm {Span} (v_{1},\ldots ,v_{n})\subseteq V$ 2. linear independence: $c_{1}\,v_{1}+\dots +c_{n}\,v_{n}=0$ 3. if 1 & 2, then $v$ 's define a basis ($w=c_{1}\,v_{1}+\dots +c_{n}\,v_{n}\quad \forall w\in V$ 4. cardinality of basis is dimension
• if $\{v_{i}\}_{i=1}^{n}$ and $\{w_{j}\}_{j=1}^{k}$ are bases for $V$ , then $n=k$ .
• Note: $\dim V=0\iff V=\{0\}$ ## Bases of Subspace

For a vector space $V$ with dimension less than ∞, if $W\subset V$ , then $\dim W<\dim V$ .

## Infinite Dimension

Let $P=\bigcup _{n=1}^{\infty }P_{n}$ represent the space of polynomials. Then $\dim P_{n}=|\{1,x,\ldots ,x^{n-1}\}|=n$ and $\dim P=\infty$ ## Column and Row Space

For a $m\times n$ matrix, $M=(v_{1},\ldots ,v_{n})$ , where $v$ 's are columns.

$\mathrm {Span} (v_{1},\ldots ,v_{n})\subseteq \mathrm {R} ^{m}$ is called the column space of $M$ .

$\mathrm {Span} ({\vec {w}}:{\mbox{rows of }}M)\subseteq \mathrm {R} ^{n}$ is called row space of $M$ .

## Theorem 3.4.4

For vector space $V$ with dimension $\dim V=n>0$ , then

1. no set of fewer than $n$ vectors can span $V$ ;
2. any subset of fewer than $n$ linearly independent vectors can be extended (i.e. vectors can be added) to form a basis for $V$ ; and
3. any spanning set containing more than $n$ vectors can be pared down (i.e. vectors can be removed) to form a basis for $V$ .

## Bases in Euclidean Space

${\vec {v}}_{1},\ldots ,{\vec {v}}_{n}\in \mathbb {R} ^{n}$ is a basis iff $\left|M\right|\neq 0$ , where $M=\left({\vec {v}}_{1},\ldots ,{\vec {v}}_{n}\right)$ is a $n\times n$ matrix.

### Standard Bases

${\hat {\imath }},{\hat {\jmath }},{\hat {k}}$ is standard basis in $\mathbb {R} ^{3}$ .

A standard basis for $\mathbb {R} ^{n}$ is a set $\{{\vec {e}}_{1},\ldots ,{\vec {e}}_{n}\}$ such that $i$ th entry of ${\vec {e}}_{i}$ is $1$ , and all other entries are $0$ Let $[a,b]$ denote an ordered set representation, such that $[a,b]\neq [b,a]$ . $[{\vec {e}}_{1},{\vec {e}}_{2}]$ will denote an ordered basis for $\mathbb {R} ^{2}$ , for example.

## Conversion of Bases

### To Standard Basis

For ${\vec {u}}_{1},{\vec {u}}_{2}\in \mathbb {R} ^{2}$ , $\mathrm {Span} ({\vec {u}}_{1},{\vec {u}}_{2})=\mathbb {R} ^{2}$ , and let $[{\vec {u}}_{1},{\vec {u}}_{2}]$ be a second basis (i.e. not standard).

We can use basis vectors to describe any point ${\vec {x}}=c_{1}\,{\vec {u}}_{1}+c_{2}\,{\vec {u}}_{2}$ in $\mathbb {R} ^{2}$ .

1. Given ${\vec {x}}=\left\langle x_{1},x_{2}\right\rangle$ , find its coordinates w.r.t. $[{\vec {u}}_{1},{\vec {u}}_{2}]$ 2. Given $c_{1}\,{\vec {u}}_{1}+c_{2}\,{\vec {u}}_{2}$ , find its coordinates w.r.t. $[{\vec {e}}_{1},{\vec {e}}_{2}]$ .

### Examples

${\vec {u}}_{1}=\left\langle 3,2\right\rangle =3{\vec {e}}_{1}+2{\vec {e}}_{2}$ and ${\vec {u}}_{2}=\left\langle 1,1\right\rangle ={\vec {e}}_{1}+{\vec {e}}_{2}$ .

Case 2: $c_{1}\,{\vec {u}}_{1}+c_{2}\,{\vec {u}}_{2}=c_{1}(3{\vec {e}}_{1}+2{\vec {e}}_{2})+c_{2}({\vec {e}}_{1}+{\vec {e}}_{2})=(3c_{1}+c_{2}){\vec {e}}_{1}+(2c_{1}+c_{2}){\vec {e}}_{2}$ .

Case 1: ${\vec {x}}=\left\langle 3c_{1}+c_{2},2c_{1}+c_{2}\right\rangle ={\begin{bmatrix}3&1\\2&1\end{bmatrix}}\,{\begin{bmatrix}c_{1}\\c_{2}\end{bmatrix}}$ .

Note: The 2×2 matrix $U=({\vec {u}}_{1},{\vec {u}}_{2})$ is called the transition matrix.

Assuming $|U|\neq 0$ , ${\vec {c}}=U^{-1}\,{\vec {x}}$ ### To Arbitrary Basis

$[{\vec {v}}_{1},{\vec {v}}_{2}]\to [{\vec {u}}_{1},{\vec {u}}_{2}]$ ${\vec {x}}=c_{1}\,{\vec {v}}_{1}+c_{2}\,{\vec {v}}_{2}=d_{1}\,{\vec {u}}_{1}+d_{2}\,{\vec {u}}_{2}$ Let $V=({\vec {v}}_{1},{\vec {v}}_{2})$ and $U=({\vec {u}}_{1}{\vec {u}}_{2})$ . Then (assuming $V$ and $U$ are nonsingular)

{\begin{aligned}V\,{\vec {c}}&=U\,{\vec {d}}\\{\vec {d}}&=U^{-1}\,V\,{\vec {c}}\end{aligned}} The product $S=U^{-1}\,V$ is the tranformation matrix from $[{\vec {v}}_{1},{\vec {v}}_{2}]$ to $[{\vec {u}}_{1},{\vec {u}}_{2}]$ ### Summary

Cases for transformation matrix $S$ 1. $[{\vec {v}}_{1},{\vec {v}}_{2}]\to [{\vec {e}}_{1},{\vec {e}}_{2}]:S=V$ 2. $[{\vec {v}}_{1},{\vec {v}}_{2}]\to [{\vec {u}}_{1},{\vec {u}}_{2}]:S=U^{-1}\,V$ 3. $[{\vec {e}}_{1},{\vec {e}}_{2}]\to [{\vec {u}}_{1},{\vec {u}}_{2}]:S=U^{-1}$ Note that $S$ for standard basis $[{\vec {e}}_{1},{\vec {e}}_{2}]$ is the identity matrix, so $S=U^{-1}I=U^{-1}$ ## Coordinate Vector

For vector space $V$ with dimension $\dim V=n$ , let $E=[v_{1},v_{2},\ldots ,v_{n}]$ be the ordered basis for $V$ . Thus $v=c_{1}\,v_{1}+c_{2}\,v_{2}+\dots +c_{n}\,v_{n}\quad \forall v\in V$ .

The vector ${\vec {c}}=\left\langle c_{1},c_{2},\ldots ,c_{n}\right\rangle \in \mathbb {R} ^{n}$ is called the coordinate vector of $V$ w.r.t. $E$ and is denoted ${\vec {c}}=[V]_{E}$ .

Therefore, the previous section could be rewritten as

$[X]_{F}=S[X]_{E}\,\!$ where $E$ and $F$ are ordered bases for $U$ and $V$ , respectively, and $S=U^{-1}V$ ## Review

Maximum number of points = 50 + 7 pt. bonus question

Look over Determinants, null spaces, and solving linear systems of equations.