[proofplan] We prove the result directly from the definition of a [quotient space](/page/Quotient%20Space) and the coordinate expansion in the given basis of $V$. First, every coset $x+U$ is represented by a vector $x\in V$, and expanding $x$ in the extended basis shows that the $u_i$-terms vanish in the quotient. Second, any linear relation among the cosets $v_j+U$ gives a vector in $U$; rewriting that vector in the basis $(u_1,\ldots,u_m)$ and using [linear independence](/page/Linear%20Independence) of the full basis of $V$ forces all coefficients to be zero. [/proofplan] [step:Declare the quotient map and identify the zero coset] Define the quotient map \begin{align*} q:V\to V/U,\qquad q(x)=x+U \end{align*} The zero vector of $V/U$ is the coset $U=0+U$. For each $i\in\{1,\ldots,m\}$, since $u_i\in U$, we have \begin{align*} q(u_i)=u_i+U=U. \end{align*} Thus every vector in the subspace $U$ maps to the zero vector of the quotient. [/step] [step:Show that the cosets of the added basis vectors span the quotient] Let $x+U\in V/U$ be arbitrary, with $x\in V$. Since \begin{align*} (u_1,\ldots,u_m,v_1,\ldots,v_r) \end{align*} is a basis of $V$, there exist scalars $\alpha_1,\ldots,\alpha_m,\beta_1,\ldots,\beta_r\in k$ such that \begin{align*} x=\sum_{i=1}^{m}\alpha_i u_i+\sum_{j=1}^{r}\beta_j v_j. \end{align*} Applying $q$ and using linearity of the quotient operations gives \begin{align*} x+U=q(x)=\sum_{i=1}^{m}\alpha_i q(u_i)+\sum_{j=1}^{r}\beta_j q(v_j). \end{align*} Because $q(u_i)=U$ for every $i$, this becomes \begin{align*} x+U=\sum_{j=1}^{r}\beta_j(v_j+U). \end{align*} Hence $(v_1+U,\ldots,v_r+U)$ spans $V/U$. [guided] We want to prove that the displayed cosets span the quotient. An arbitrary element of $V/U$ has the form $x+U$ for some $x\in V$, so the problem is to express this coset as a linear combination of $v_1+U,\ldots,v_r+U$. Since \begin{align*} (u_1,\ldots,u_m,v_1,\ldots,v_r) \end{align*} is a basis of $V$, the vector $x$ has a unique coordinate expansion in that basis. In particular, there exist scalars $\alpha_1,\ldots,\alpha_m,\beta_1,\ldots,\beta_r\in k$ such that \begin{align*} x=\sum_{i=1}^{m}\alpha_i u_i+\sum_{j=1}^{r}\beta_j v_j. \end{align*} Now pass this equality to the quotient by applying the map \begin{align*} q:V\to V/U,\qquad q(y)=y+U \end{align*} The quotient operations are defined so that addition and scalar multiplication of cosets are inherited from $V$, hence \begin{align*} q(x)=\sum_{i=1}^{m}\alpha_i q(u_i)+\sum_{j=1}^{r}\beta_j q(v_j). \end{align*} For each $i$, the vector $u_i$ lies in $U$, so $u_i+U=U$, the zero vector of $V/U$. Therefore all the $u_i$-terms disappear in the quotient: \begin{align*} x+U=q(x)=\sum_{j=1}^{r}\beta_j(v_j+U). \end{align*} Since $x+U$ was arbitrary, the cosets $v_1+U,\ldots,v_r+U$ span $V/U$. [/guided] [/step] [step:Show that the cosets of the added basis vectors are linearly independent] Let $a_1,\ldots,a_r\in k$ satisfy \begin{align*} \sum_{j=1}^{r}a_j(v_j+U)=U. \end{align*} By the definition of addition and scalar multiplication in $V/U$, this means \begin{align*} \left(\sum_{j=1}^{r}a_j v_j\right)+U=U. \end{align*} Thus \begin{align*} \sum_{j=1}^{r}a_j v_j\in U. \end{align*} Since $(u_1,\ldots,u_m)$ is a basis of $U$, there exist scalars $b_1,\ldots,b_m\in k$ such that \begin{align*} \sum_{j=1}^{r}a_j v_j=\sum_{i=1}^{m}b_i u_i. \end{align*} Rearranging in $V$ gives \begin{align*} \sum_{i=1}^{m}(-b_i)u_i+\sum_{j=1}^{r}a_j v_j=0. \end{align*} The list \begin{align*} (u_1,\ldots,u_m,v_1,\ldots,v_r) \end{align*} is linearly independent because it is a basis of $V$. Therefore every coefficient in this linear relation is zero. In particular, \begin{align*} a_1=\cdots=a_r=0. \end{align*} So $(v_1+U,\ldots,v_r+U)$ is linearly independent in $V/U$. [/step] [step:Conclude that the displayed cosets form a basis] The previous two steps show that $(v_1+U,\ldots,v_r+U)$ spans $V/U$ and is linearly independent in $V/U$. Therefore it is a basis of $V/U$. If $r=0$, the same argument says that the empty list spans $V/U$ and is linearly independent; in that case the extended basis hypothesis forces $U=V$, so $V/U$ is the zero [vector space](/page/Vector%20Space) and the empty list is its basis. [/step]