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!!


1 Answer 1


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 $$V=U_1⊕U_2⊕\cdots ⊕U_n.$$

  • $\begingroup$ 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! $\endgroup$
    – Roach87
    Nov 4, 2020 at 7:11
  • $\begingroup$ 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 $\endgroup$
    – Roach87
    Nov 4, 2020 at 7:12

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