kernel

let $\text{[math]}$ be a linear transformation. the kernel of $\text{[math]}$ , denoted $\text{[math]}$ , is the set of vectors $\text{[math]}$ such that $\text{[math]}$ .
[cite:;taken from @calc_hubbard_2015 definition 2.5.1]

the kernel of a matrix doesnt change when an elementary row operation is applied to the matrix

if $\text{[math]}$ is a linear transformation, then $\text{[math]}$ is a vector subspace of $\text{[math]}$ and $\text{[math]}$ is a vector subspace of $\text{[math]}$ .
[cite:;taken from @calc_hubbard_2015 proposition 2.5.2]
[cite:;also in @algebra_insel_2019 theorem 2.1]

let $\text{[math]}$ and $\text{[math]}$ be vector spaces, and let $\text{[math]}$ be linear. if $\text{[math]}$ is a basis for $\text{[math]}$ , then
$\text{[math]}$ [cite:;also in @algebra_insel_2019 theorem 2.2]

let $\text{[math]}$ be a linear transformation. the system of linear equations $\text{[math]}$ has

at most one solution for every $\text{[math]}$ if and only if $\text{[math]}$ .
at least one solution for every $\text{[math]}$ if and only if $\text{[math]}$ .

[cite:;taken from @calc_hubbard_2015 proposition 2.5.3]

a basis for the kernel is a set of vectors such that any vector $\text{[math]}$ satisfying $\text{[math]}$ can be expressed as a linear combination of those basis vectors. the basis vectors must be in the kernel, and they must be linearly independent.
if a system of linear equations has a solution, then it has a unique solution for any value you choose of the nonpivotal un-knowns. clearly $\text{[math]}$ has a solution, namely $\text{[math]}$ . so the tactic is to choose the values of the nonpivotal (active) unknowns in a convenient way. we take our inspiration from the standard basis vectors: each has one entry equal to 1, and the others 0. we construct a vector $\text{[math]}$ for each nonpivotal column, by setting the entry corresponding to that nonpivotal unknown to be 1, and the entries corresponding to the other nonpivotal unknowns to be 0. the entries corresponding to the pivotal (passive) unknowns will be whatever they have to be to satisfy the equation $\text{[math]}$ .

let $\text{[math]}$ be the number of nonpivotal columns of $\text{[math]}$ , and $\text{[math]}$ their positions. for each nonpivotal column, form the vector $\text{[math]}$ satisfying $\text{[math]}$ and such that its $\text{[math]}$ th entry is 1, and its $\text{[math]}$ th entries are all 0, for $\text{[math]}$ . the vectors $\text{[math]}$ form a basis of $\text{[math]}$ .
[cite:;taken from @calc_hubbard_2015 theorem 2.5.6]

we consider
$\text{[math]}$ the third and fourth columns of $\text{[math]}$ are nonpivotal, so $\text{[math]}$ and $\text{[math]}$ . the system has a unique solution for any values we choose of the third and fourth unknowns. in particular, there is a unique vector $\text{[math]}$ whose third entry is 1 and fourth entry is 0, such that $\text{[math]}$ . there is another, $\text{[math]}$ , whose fourth entry is 1 and third entry is 0, such that $\text{[math]}$ .
$\text{[math]}$ the first, second, and fifth entries of $\text{[math]}$ and $\text{[math]}$ correspond to the pivotal unknowns. we read their values from the first three rows of $\text{[math]}$ (remembering that a solution for $\text{[math]}$ is also a solution for $\text{[math]}$ ).
$\text{[math]}$ which gives
$\text{[math]}$ so for $\text{[math]}$ , where $\text{[math]}$ and $\text{[math]}$ , the first entry is $\text{[math]}$ , the second is $\text{[math]}$ , and the fifth is 0; the corresponding entries for $\text{[math]}$ are $\text{[math]}$ , and 0:
$\text{[math]}$ these two vectors form a basis of the kernel of $\text{[math]}$ .
[cite:;taken2 from @calc_hubbard_2015 example 2.5.7]

[cite:;taken from @calc_hubbard_2015 chapter 2.5]