# Intersection of two vector spaces

It is given that the sequence of vectors $v_1,\cdots , v_n, u_1, ...., u_{m-1}$ is linearly independent and that $u_1, \cdots, u_m$ is also linearly independent where $u_m$ is in the $\text{span}$ of $v_1, \cdots, v_n$.If $V = \text{span}\{v_1, v_2, \cdots , v_n\}$ and $U =\text{span}\{u_1, ...., u_{m}\}$, we should describe the intersection of U and V and determine $\dim(U \cap V)$

From the given, we can deduce that $u_m$ can be expressed as $$u_m=c_1v_1+ c_2v_2+...c_nv_n$$ Similarlly, we have due to LI,

$$c_1v_1+ c_2v_2+...c_nv_n+d_1u_1+...+d_{m-1}u_{m-1}=0$$

What does it mean be "describe the intersection"? And from there how can one find the dimension. Isn't it by establishing the basis? I would appreciate any help.

Let $U' = \text{Span}(u_1, \dots, u_{m-1})$, and $W = \text{Span}(u_m). \$ We know that $V \cap U' = 0$ or else there would be $t$ in both spaces such that $t = \sum a_i v_i = \sum b_i u_i$, but by linear indep. we must have all $a_i, b_i = 0$.
Clearly $U = U' + W$. Suppose that there is a $v \in V \cap U$, i.e. by the that decomposition, $v = u' + w$, where $u' \in U', w \in W. \$ But then $u' = v - w \in V$, since $u_m \in V$ is given, but since $u' \in U'$, by our first paragraph we have $u' = 0$. Thus $v = w \in W$. We have essentially shown that $V \cap U \subset W$. Conversely, if $v \in W$, then its easy to see that $v \in V \cap U$. Thus we've identified the intersection of the two spaces with $W = \text{Span}(u_m)$, which clearly has dimension $1$.
In the answer by Enjoys Math, he has done a fine job by indicating: $U \cap V = W =$ span$(u_m)$, and since $W$ is spanned by a single vector it is also a basis for $W$. I think the argument can be shortened somewhat by recognizing that $u_m \in U$ and $u_m \in V$, so that $u_m \in U \cap V$ and since it is the only basis vector of $U$ in $V$ (Note that $U-V=$ span$\{u_1,\ldots,u_{m-1}\}$) it must then be the only vector in a basis for $U \cap V$.