dimension

the dimension of a vector space $\text{[math]}$ is the number of elements in one of its bases. we write $\text{[math]}$ where $\text{[math]}$ is a basis of $\text{[math]}$ .

$\text{[math]}$ .
$\text{[math]}$ .
$\text{[math]}$ .

let $\text{[math]}$ be a linear transformation. then
$\text{[math]}$ [cite:;taken from @calc_hubbard_2015 theorem 2.5.8 dimension formula]
[cite:;also in @algebra_insel_2019 theorem 2.3 dimension theorem]

$\text{[math]}$ number of free variables.

let $\text{[math]}$ be a finitely spanned vector space, let $\text{[math]}$ then:
$\text{[math]}$

let $\text{[math]}$ be a vector space of dimension $\text{[math]}$ , for every subset $\text{[math]}$ that has $\text{[math]}$ elements the following statements are equal:

$\text{[math]}$ is linearly independant,
$\text{[math]}$ .

we prove $\text{[math]}$
given $\text{[math]}$ is linearly independant, we need to prove $\text{[math]}$
let $\text{[math]}$ , $\text{[math]}$ a basis of $\text{[math]}$ , $\text{[math]}$
let $\text{[math]}$
we know according to previous lemmas that $\text{[math]}$ therefore $\text{[math]}$ is linearly dependant
since $\text{[math]}$ is linearly dependant then we have the following non-trivial linear combination that gives us $\text{[math]}$ :
$\text{[math]}$ we know $\text{[math]}$ because otherwise another $\text{[math]}$ would be equal to 0 and $\text{[math]}$ wouldnt be linearly independant
by manipulating the expression we get:
$\text{[math]}$ therefore $\text{[math]}$ therefore $\text{[math]}$

we prove $\text{[math]}$
given $\text{[math]}$
assume in contradiction that $\text{[math]}$ is linearly dependant
according to a previous lemma we there exists $\text{[math]}$ that is linearly independant such that $\text{[math]}$ therefore $\text{[math]}$ is a basis therefore $\text{[math]}$
we arrived at a contradiction becacuse $\text{[math]}$

if $\text{[math]}$ and $\text{[math]}$ , if $\text{[math]}$ or $\text{[math]}$ is linearly independant then $\text{[math]}$ is a basis of $\text{[math]}$

given $\text{[math]}$ is finitely spanned, if $\text{[math]}$ then $\text{[math]}$ .

according to the previous lemma, since $\text{[math]}$ is finitely spanned then $\text{[math]}$ is finitely spanned too which means $\text{[math]}$ has a basis and a dimension

we assume in contradiction that $\text{[math]}$
assume $\text{[math]}$ is a basis of $\text{[math]}$ , then we know $\text{[math]}$
assume $\text{[math]}$ where $\text{[math]}$ is a basis of $\text{[math]}$ , then we know $\text{[math]}$ which means $\text{[math]}$ is linearly independant but it cant be because we have a set where $\text{[math]}$ with less elements that is linearly independant so a set with $\text{[math]}$ cant be linearly independant
[cite:;also in @algebra_insel_2019 theorem 1.11]

given $\text{[math]}$ is finitely spanned, $\text{[math]}$ and $\text{[math]}$ then $\text{[math]}$

let $\text{[math]}$ be a basis of $\text{[math]}$ and so $\text{[math]}$ and $\text{[math]}$ is linearly independant therefore according to [[n_subset_is_basis][this lemma]] $\text{[math]}$ and since it spans $\text{[math]}$ and is of size $\text{[math]}$ then it is a basis of $\text{[math]}$

if $\text{[math]}$ is finitely spanned and $\text{[math]}$ then for every basis $\text{[math]}$ of $\text{[math]}$ there exists a basis $\text{[math]}$ of $\text{[math]}$ such that $\text{[math]}$
we call this operation as complementing the basis $\text{[math]}$ to the basis $\text{[math]}$

let $\text{[math]}$ and $\text{[math]}$ be finite-dimensional vector spaces of equal dimension, and let $\text{[math]}$ be linear. then the following are equivalent.

$\text{[math]}$ is one-to-one.
$\text{[math]}$ is onto.
$\text{[math]}$ .

[cite:;taken from @algebra_insel_2019 theorem 2.5]