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

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

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

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

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

This implies that the element

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

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

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