# Prove Direct sum of subsets of basis $V$

I have a task to prove this thing and I think I did but I'm not sure if the proof is good.

Question : Let $$B$$ be basis of $$V$$ such that $$B=B_1∪....∪B_k$$ for $$i=1....k$$

So $$U_i=\operatorname{Span}({B_i})$$ --> Prove $$V=U_1⊕....⊕U_k$$

My answer : (1) $$B$$ is basis so the vectors in it are lineary independent which means $$B_1∪....∪B_k$$ are lineary independent.

so lets assume there is a Vector $$v∈B_1∩....∩B_k$$ which means the vector $$v∈B_1,...,B_k$$

so $$v$$ can be written as linear equation of the vectors in $$B_1,...,B_k$$. ----> We get that we have a common vector in $$B_1$$ and $$B_2$$... which means the basis $$B$$ is dependent and contradicts our assumption. ---> the intersection = $${0_V}$$.

(2) Prove $$V=U_1,...,U_k$$ : it's trivial because we know $$V=U_1+...+U_k$$ by the given question.

Is that right or I proved something wrongly / assumed wrongly? Thank you very much!!

If $$\dim V$$ is finite. Let $$\mathcal{B}=\{v_1, v_2,\ldots, v_n\}$$ a basis of V, where $$n=\dim V$$. So, define $$U_i:=span(v_i).$$

Let $$v\in V$$, then $$v=c_1v_1+c_2v_2+\cdots+c_nv_n,$$ Now, note that for all $$i$$, $$c_iv_i\in span(v_i)$$, So $$V=U_1+U_2+\cdots+U_n$$

Finally, you must to show that $$U_1\cap U_2\cap\cdots\cap U_n=\{0\}$$ (Show that if $$w\in U_1\cap U_2\cap\cdots\cap U_n$$ so $$w=0$$). With all you must conclude that $$V=U_1⊕U_2⊕\cdots ⊕U_n.$$

• Ye the Sum part is not hard , to prove that the intersection is ${0}$ is the hard part I think , that's why i'm asking if my proof of the intersection is good. Thank you! Nov 4, 2020 at 7:11
• I mean - by my assumption I get that $v_1 = (c_2 * v_2)/c_1$ , which means $v_1$ is lineary dependent and it contradicts our assumption Nov 4, 2020 at 7:12