Linear subspace/Sum and intersection/Dimension/Fact/Proof

Proof

Let ${}w_{1},\ldots ,w_{k}$ be a basis of ${}U_{1}\cap U_{2}$. On one hand, we can complete this basis, according to fact, to a basis ${}w_{1},\ldots ,w_{k},u_{1},\ldots ,u_{n}$ of ${}U_{1}$, on the other hand, we can complete it to a basis ${}w_{1},\ldots ,w_{k},v_{1},\ldots ,v_{m}$ of ${}U_{2}$. Then

$w_{1},\ldots ,w_{k},u_{1},\ldots ,u_{n},v_{1},\ldots ,v_{m}$

is a generating system of ${}U_{1}+U_{2}$. We claim that it is even a basis. To see this, let

${}a_{1}w_{1}+\cdots +a_{k}w_{k}+b_{1}u_{1}+\cdots +b_{n}u_{n}+c_{1}v_{1}+\cdots +c_{m}v_{m}=0\,.$

This implies that the element

${}a_{1}w_{1}+\cdots +a_{k}w_{k}+b_{1}u_{1}+\cdots +b_{n}u_{n}=-c_{1}v_{1}-\cdots -c_{m}v_{m}\,$

belongs to ${}U_{1}\cap U_{2}$. From this we get directly ${}b_{i}=0$ for ${}i=1,\ldots ,n$ and ${}c_{j}=0$ for ${}j=1,\ldots ,m$. From this we can infer that also ${}a_{\ell }=0$ for all ${}\ell$. Hence, we have linear independence. This gives altogether

${}{\begin{aligned}\dim _{K}{\left(U_{1}\cap U_{2}\right)}+\dim _{K}{\left(U_{1}+U_{2}\right)}&=k+k+n+m\\&=k+n+k+m\\&=\dim _{K}{\left(U_{1}\right)}+\dim _{K}{\left(U_{2}\right)}.\end{aligned}}$

To fact