$$\dim (U_1+U_2)=\dim U_1+\dim U_2-\dim (U_1\cap U_2)$$

证明:采用数学归纳法.当$\dim U_2=0$时，易得$U_2=\{0\}$.此时维数定理显然成立. 设$\dim  U_2=n$时，维数定理成立.则当$\dim U_2=n+1$时，令$(v_1,\cdots,v_n,v_{n+1})$为$U_2$的一组基.若$v_{n+1}\in U_1$,则$$\dim (U_1+U_2)=\dim (U_1+Span(v_1,\cdots,v_n))$$$$\dim U_1+\dim U_2=\dim U_1+\dim (Span(v_1,\cdots,v_n))+1$$$$\dim (U_1\cap U_2)=\dim (U_1\cap Span(v_1,\cdots,v_n))+1.$$由于$$\dim (U_1+Span(v_1,\cdots,v_n))=\dim U_1+\dim Span(v_1,\cdots,v_n)-\dim (U_1\cap Span(v_1,\cdots,v_n))$$成立（根据归纳假设）,所以$$\dim (U_1+U_2)=\dim U_1+\dim U_2-\dim (U_1\cap U_2)$$成立.若$v_{n+1}\not\in U_1$,则$$\dim (U_1+U_2)=\dim U_1+\dim Span(v_1,\cdots,v_n)+1$$且$$\dim (U_1\cap U_2)=\dim (U_1\cap Span(v_1,\cdots,v_n))$$且$$\dim U_2=\dim Span(v_1,\cdots,v_n)+1$$再结合归纳假设，所以在这种情况下，命题仍成立.根据数学归纳法，可得对于任何有限维向量空间$U_2$,维数定理成立.