# Finding dim Sp$\left \{ v_1,v_2,v_3 \right \}$

Let $V$ be a vector space over $\mathbb{R}$. Given $u_1,u_2,u_3,v_1,v_2,v_3 \in V$ such that $u_1,u_2,u_3 \in U= Sp\left \{ v_1,v_2,v_3 \right \}$ and $B=\left \{ u_1-u_2, u_1+3u_3, 4u_2+5u_3 \right \}$ is linearly independent, find dim$U$

So because I was given a spanning set of $U$ a wild guess would be proving $\left \{ v_1,v_2,v_3 \right \}$ is linearly independent. However $\alpha v_1+\beta v_2 +\gamma v_3=0$ doesn't really give me anything since I don't see the connection to $B$.

1. As $U$ is a linear (sub-)space and contains $u_1,u_2,u_3$, it must also contain all linear combinations $$\lambda_1u_1+\lambda_2u_2+\lambda_3u_3,$$ therefore it also contains all the elements of $B$, and so also $\text{span}(B)$.
2. $\text{span}(B)$ is a $3-$dimensional linear subspace of $V$, as $B$ is made of $3$ linearly independent vectors.
• How do you know $V \subseteq Sp B$? – Theorem Jul 30 '16 at 13:52
• I am saying 1. $\text{span}B\subseteq U$ and 2. $\dim(\text{span}B)=3$ I've edited the question to make it more clear. – b00n heT Jul 30 '16 at 13:54
• I think I'm not getting something here. If we want to use $B$ as the basis of $U$ to determine dim$U$, then we must show it spans $U$. If I understood you showed every linear comb. of $B$ is in $U$ but we need to show every element if $U$ is a linear combination of $B$, don't we? – Theorem Jul 30 '16 at 14:11
• That would be a possibility, but showing it explicitly would be a mess. Instead the idea is to use a dimension argument: we know that $V$ is a subspace of dimension less or equal to $3$, as it is spanned by the $3$ vectors $v_i$. But it also contains a subspace of dimension $3$, hence its dimension must be at least $3$. Thus it must be exactly $3$, i.e the vectors $v_i$ are linearly independent – b00n heT Jul 30 '16 at 15:26
• I need to find dim$U$, not dim$V$. I still don't understand how did you get to that conclusion. – Theorem Jul 30 '16 at 15:38