basis

let $\text{[math]}$ be a vector subspace. an ordered set of vectors $\text{[math]}$ is called a basis of $\text{[math]}$ if it satisfies any of the three equivalent conditions.

the set is a maximal linearly independent set: it is linearly independent, and if you add one more vector, the set will no longer be linearly independent.
the set is a minimal spanning set: it spans $\text{[math]}$ , and if you drop one vector, it will no longer span $\text{[math]}$ .
the set is a linearly independent set spanning $\text{[math]}$ .

[cite:;taken from @calc_hubbard_2015 definition 2.4.11]

let $\text{[math]}$ be a finite-dimensional vector space. an ordered basis for $\text{[math]}$ is a basis for $\text{[math]}$ endowed with a specific corder; that is, an ordered basis for $\text{[math]}$ is a finite sequence of linearly independent vectors in $\text{[math]}$ that generates $\text{[math]}$ .
[cite:;taken from @algebra_insel_2019 chapter 2.2 the matrix representation of a linear transformation]

let $\text{[math]}$ be a basis of $\text{[math]}$ :
$\text{[math]}$ since $\text{[math]}$ is linearly independant then:
$\text{[math]}$ this is necessarily true because $\text{[math]}$ has to be 0 for $\text{[math]}$ to zero out and if it wasnt zero then the output vector would have a non-zero in it and it wouldnt be $\text{[math]}$ anymore

a basis doesnt have to be of this simplified form, we can take the basis $\text{[math]}$ , apply elementary row operations to its vectors and we would get another basis for the given vector space

every finitely generated vector space $\text{[math]}$ has atleast one basis

since $\text{[math]}$ is finitely generated there exists a finite set $\text{[math]}$ such that $\text{[math]}$
according to [[can_drop_vector_keep_span][this lemma]], there exists $\text{[math]}$ where $\text{[math]}$ is linearly independant and $\text{[math]}$ , therefore $\text{[math]}$ is a basis

$\text{[math]}$
$\text{[math]}$
$\text{[math]}$ is linearly independant
$\text{[math]}$ is a basis of $\text{[math]}$

given the vector space $\text{[math]}$ , we find a basis of it:
$\text{[math]}$ so $\text{[math]}$ is a basis of $\text{[math]}$ and $\text{[math]}$ , see dimension

consider the vector space $\text{[math]}$ which represents all the matrices with dimensions $\text{[math]}$ over the field $\text{[math]}$
what we are basically looking for is a set of matrices that is linearly independant so that no matrix is a linear combination of the others, which is accomplished with:
$\text{[math]}$ and:
$\text{[math]}$ we prove that $\text{[math]}$ is linearly independant:
$\text{[math]}$ it must be that $\text{[math]}$ and so $\text{[math]}$ is linearly independant because the only linear combination that gives us $\text{[math]}$ is where the coefficients all are 0

given $\text{[math]}$ is a finite set and $\text{[math]}$ and $\text{[math]}$ , then $\text{[math]}$ is linearly dependant
in other words, if we have a finitely generated vector space, if $\text{[math]}$ is a subset of that space such that the number of elements in $\text{[math]}$ is bigger than the number of elements in $\text{[math]}$ , then $\text{[math]}$ is linearly dependant

let $\text{[math]}$ and $\text{[math]}$ such that $\text{[math]}$
let $\text{[math]}$ denote the coefficients, for every $\text{[math]}$ where $\text{[math]}$ there exist the coefficients:
$\text{[math]}$

let $\text{[math]}$ be a finite set, if $\text{[math]}$ and $\text{[math]}$ is linearly independant then $\text{[math]}$

let $\text{[math]}$ be bases of $\text{[math]}$ then $\text{[math]}$

$\text{[math]}$ is a spanning set and $\text{[math]}$ and $\text{[math]}$ is linearly independant then $\text{[math]}$
$\text{[math]}$ is a spanning set and $\text{[math]}$ and $\text{[math]}$ is linearly independant then $\text{[math]}$
therefore $\text{[math]}$

basis of the cartesian coordinate system

the of the cartesian coordinate system consists of the unit vectors that lie along the x, y, and z axes
the x unit vector is denoted by $\text{[math]}$ , the y unit vector by $\text{[math]}$ , and the z unit vector by $\text{[math]}$
$\text{[math]}$

$\text{[math]}$ we can write any vector in terms of the base vectors
$\text{[math]}$