Basis, Dimensions of Vector Spaces.
$V/F$, a subset $\beta\subseteq V$ is called a basis of $V$ if
- $\beta$ generates $V$, i.e. $\spa{\beta}=V$.
- $\beta$ is linearly independent.
Prove that $\beta=\set{(1,0), (0,1)}$ is a basis of $\R^2$.
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$e_1=(1,0,0,…)$,
$e_2=(0,1,0, …)$,
$e_3=(0,0,1, …)$,
$e_n=(0,0,…,1, 0, …)$.
Is $S= \set{e_1, e_2, e_3, …}$abasis of $\R^\infty$?
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$R^\infty$ does have a basis, although we don’t know what it is.
Every vector space have a basis.
$e_1=(1,0,0,…)$,
$e_2=(0,1,0, …)$,
$e_3=(0,0,1, …)$,
$e_n=(0,0,…,1, …, 0)$.
$S= \set{e_1, e_2, e_3, …, e_n}$ is a basis of $\R^n$.
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$\beta = \set{1, x, x^2, …, x^n}$ is a basis of $P_n(X)$ over $\R$.
Prove that $\beta = \set{1, x, x^2, …}$ is a basis of $\P(x) / \R$.
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need to prove that $\beta$ generates $\P(x) $.
$a_0 + a_1x + a_2x^2 + … + a_{n} x^{n} \in \P(x)$ $\spa{\beta} = \P(x) $ -
need to prove that $\beta$ is linearly independent.
To prove an infinite set is linearly independent, we only need to prove its every finite subset’s only solution is the trivial representation.
The empty set $\emptyset$ is the basis of $V=\set{0}/F$.
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$\beta \subseteq V$,
$\beta$ is a basis of $V \iff $
$\forall v \in V, \exists v_1, v_2, …, v_n \in \beta$ and unique $a_1, a_2, …, a_n \in F$ s.t.
$$v = \sum^n_{i=1}a_iv_i$$
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$V/F, S$ is a finite generating subset of $V$. Then $\exists T \subseteq S, T$ is a basis of $V$.
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Let $V/F$ and $\beta = \set{v_1, v_2, …, v_n}$ be a basis of $V$.
Let $v \in V$ be a non-zero vector and $v=c_1v_1 + c_2v_2 + … + c_jv_j + … + a_nv_n$ s.t. $c_j \neq 0$ Then $\set{v_1, v_2, …, v_{j-1}, v, v_{j+1}, …, v_n}$ is a basis of $V$. i.e. the $j$th vector of $\beta$ can be replaced by $v$.
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$\beta= \set{e_1,e_2,e_3}$, standard basis.
$u = (-1, 0, \frac12) = -1e_1 + 0e_2+ \frac12 e_3$
$\beta' = \set{u, e_2, e_3}$ is a basis.
$V/F, V$ is finite-dimensional,
Then any two basis of $V$ has exactly same number of elements, namely, $\dim V$.
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$V/F$. If $V$ is generated by a finite subset, then $V$ is called a finite-dimensional vector space. In this case the number of vectors in a basis of $V$ is called the dimension of $V$ denoted by $\dim V$.
If $V$ is NOT a finite-dimensional vector space, or alternatively if there exists an infinite linearly independent subset $S$ of $V$, i.e. $\abs{S} = \infty$ and $S$ is linearly independent, Then $V$ is called an infinite-dimensional vector space, and we write $\dim V = \infty$.
$\beta = \set{1, x, x^2, x^3, …}$
$\dim_\R \P(x) = \infty$
$\dim_\R \R^\infty = \infty$
$e_1 = (1, 0, …, 0)$
$e_2 = (0, 1, …, 0)$
$e_n = (0, 1, …, n)$
$\beta = \set{e_1, e_2, …, e_n}, \dim_\R R^n = n$.
$\dim_\R M_{m\times n} = mn$
$W = \set{A \in M_{2\times 2}(\R) | tr(A) = 0}$
$\dim_\R W = 3$
$\dim_\R \R = 1$
In fact,
$\dim_F F = 1$
$\dim_\C \R = 2$
- $V = \C^2, \F = \C, \dim_\C \C^2 = 1$
$\beta_1 = \set{(1, 0), (0, 1)}/\C$ - $V = \C^2, \F = \R, \dim_\R \C^2 = 2$
$\beta_2 = \set{(1, 0), (i, 0), (0, 1), (0, -i)}/\C$ - $V = \C^n, \F = \R, \dim_\R \C^2 = 2n$
$\dim_\Q \R = \infty$(will be useful in MATH 131A)
Let $V$ be a $n$ dimension vector space over a field $F$.
i.e. $\dim V = n$
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(a) Any finite generating set of $V$ must contain at least $n$ elements.
$\forall S \subseteq V, \spa{S} = V \implies \abs{S} \geq n.$
(b) Any finite generating set of $V$ containing exactly $n$ elements is a basis.
$\forall S \subseteq V, \spa{S} = V, \abs{S} = n \implies S$ is a basis of $V$. -
Any linearly independent subset of $V$ containing exactly $n$ elements is a basis.
$\forall S \subseteq V, S$ is linearly independent, $\abs{S} = n \implies S$ is a basis of $V$. -
Every linearly independent subset of $V$ can be extended to a basis of $V$.
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Let $V$ be a finite dimension vector space over a field $F$, and $W$ a subspace of $V$, then $\dim W \leq \dim V$.
Moreover, if $\dim W = \dim V$, then $W = V$.
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