Linear mapping/Dimension formula/Fact/Proof

Proof

Set ${}n=\dim _{K}{\left(V\right)}$. Let ${}U=\operatorname {kern} \varphi \subseteq V$ denote the kernel of the mapping and let ${}k=\dim _{K}{\left(U\right)}$ denote its dimension (${}k\leq n$). Let

$u_{1},\ldots ,u_{k}$

be a basis of ${}U$. Due to fact, there exist vectors

$v_{1},\ldots ,v_{n-k}$

such that

$u_{1},\ldots ,u_{k},\,v_{1},\ldots ,v_{n-k}$

is a basis of ${}V$. We claim that

$w_{j}=\varphi (v_{j}),\,j=1,\ldots ,n-k,$

is a basis of the image. Let ${}w\in W$ be an element of the image ${}\varphi (V)$. Then there exists a vector ${}v\in V$ such that ${}\varphi (v)=w$. We can write ${}v$ with the basis as

${}v=\sum _{i=1}^{k}s_{i}u_{i}+\sum _{j=1}^{n-k}t_{j}v_{j}\,.$

Then we have

${}{\begin{aligned}w&=\varphi (v)\\&=\varphi {\left(\sum _{i=1}^{k}s_{i}u_{i}+\sum _{j=1}^{n-k}t_{j}v_{j}\right)}\\&=\sum _{i=1}^{k}s_{i}\varphi (u_{i})+\sum _{j=1}^{n-k}t_{j}\varphi (v_{j})\\&=\sum _{j=1}^{n-k}t_{j}w_{j},\end{aligned}}$

which means that ${}w$ is a linear combination in terms of the ${}w_{j}$. In order to prove that the family ${}w_{j}$, ${}j=1,\ldots ,n-k$, is linearly independent, let a representation of zero be given,

${}0=\sum _{j=1}^{n-k}t_{j}w_{j}\,.$

Then

${}\varphi {\left(\sum _{j=1}^{n-k}t_{j}v_{j}\right)}=\sum _{j=1}^{n-k}t_{j}\varphi {\left(v_{j}\right)}=0\,.$

Therefore, ${}\sum _{j=1}^{n-k}t_{j}v_{j}$ belongs to the kernel of the mapping. Hence, we can write

${}\sum _{j=1}^{n-k}t_{j}v_{j}=\sum _{i=1}^{k}s_{i}u_{i}\,.$

Since this is altogether a basis of ${}V$, we can infer that all coefficients are ${}0$, in particular, ${}t_{j}=0$.

To fact