I'm working on a question from a final review sheet:

Let $V$ be a $5$-dimensional vector space with $3$-dimensional vector spaces $U$ and $W$ such that $\dim(U \cap W)=1.$ Let $(u_1,u_2,u_3)$ be any basis of $U$ and $(w_1,w_2,w_3)$ any basis of $W.$ Prove that for some $i \ne j \in \{1,2,3\},$ $(u_1,u_2,u_3,w_1,w_2)$ is a basis of $V$.

I find the last part with the index $i \ne j \in \{1,2,3\}$ to be a little confusing since it doesn't show up in the basis $(u_1,u_2,u_3,w_1,w_2)$. Here's my attempt at a proof:

Let $(v)$ be a basis of $\dim(U \cap W)$. This can be extended to a basis $B_U = (v,u_1,u_2)$ of $U$ and a basis $B_W = (v,w_1,w_2)$ of $W$. Take $v \in U \cap W$, $u_1,u_2 \in U$,$ \hspace{0.5mm} w_1,w_2 \in W$ and combine them into the sequence $S = (v,u_1,u_2,w_1,w_2)$. Suppose $$\alpha_1 v + \alpha_2 u_1 + \alpha_3 u_2 + \alpha_4 w_1 + \alpha_5 w_2 = 0.$$ As $v,u_1,u_2$ are linearly independent in $U$ and $v,w_1,w_2$ are linearly independent in $W$, we have $$\alpha_1 v + \alpha_2 u_1 + \alpha_3 u_2=0 \Longrightarrow \alpha_1 = \alpha_2 = \alpha_3 = 0$$ and $$\alpha_1v + \alpha_4 w_1 + \alpha_5 w_2 \Longrightarrow \alpha_1 = \alpha_4 = \alpha_5 = 0.$$ Therefore $\alpha_1 v + \alpha_2 u_1 + \alpha_3 u_2 + \alpha_4 w_1 + \alpha_5 w_2 = 0 \Longrightarrow \alpha_k = 0, \hspace{0.5mm} k \in \{1,2,3,4,5\}.$ And so $S$ is a linearly independent list of five vectors in $V$. But $\dim V = 5$ so it must be that $\text{span}(S) = V$ and so $S$ is a basis for $V$.

Please critique me for any errors in my proof


I believe the text is saying that you can extract a basis of $V$ from the set $\{u_1, u_2, u_3, w_1, w_2, w_3 \}$. Else, it is just false. So I'll solve that exercise:

Let $a \neq 0$, $a \in U \cap W$. Without loss of generality, suppose that $a, u_1, u_2$ is a base of $U$ (it is true for some couple of indexes). This means that $u_1, u_2 \not\in W$, else the intersection would be bigger. Which means $u_1, u_2, w_1, w_2, w_3$ is a basis for $V$ (5 vectors, linear indipendence is guaranteed).

  • $\begingroup$ Your proof, while essentially correct, is a bit unclear in the "linear indipendence" part about the coefficients. $\endgroup$ – AnalysisStudent0414 Nov 26 '15 at 22:33

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