# $\dim V = \dim V/U+\dim U$

Is the following Proof Correct?

Theorem. Given that $U$ is a subspace of $V$ and $v_1+U,v_2+U,...,v_m+U$ is a basis for $V/U$ and $u_1,u_2,...,u_n$ is a basis for $U$ prove that $v_1,v_2,...,v_m,u_1,u_2,...,u_n$ is a basis for $V$.

Proof. We know that the dimension of a quotient space is determined as follows$$\dim V/U = \dim V-\dim U$$ which implies that $\dim V = \dim V/U+\dim U$ therefore the list $v_1,v_2,...v_m,u_1,u_2,...,u_n$ is of the right length i.e. $m+n$.

To establish that the list is a basis we need only prove that either the list is linearly independent or spans $V$, we choose the latter option.

Let $v$ be an arbitrary vector in $V$, evidently $v+U\in V/U$ and therefore for some $\alpha_1,\alpha_2,...,\alpha_m\in\mathbf{F}$ $$v+U = \sum_{j=1}^{m}\alpha_j(v_j+U) = \left(\sum_{j=1}^{m}\alpha_jv_j\right)+U$$ which may be equivalently stated as follows $$v-\left(\sum_{j=1}^{m}\alpha_jv_j\right)\in U$$ and consequently for some $\beta_1,\beta_2,...,\beta_n\in\mathbf{F}$ we have $$v-\left(\sum_{j=1}^{m}\alpha_jv_j\right) = \sum_{i=1}^{n}\beta_i u_i$$ which implies that $$v = \sum_{j=1}^{m}\alpha_jv_j+\sum_{i=1}^{n}\beta_i u_i$$ since $v$ was arbitrary it follows that $\operatorname{span}(v_1,v_2,...,v_m,u_1,u_2,...,u_n) = V$.

$\blacksquare$

• Are you sure the exercise supposes you know the dimension formula? I'd rather think it's a way to prove it. – Bernard Sep 24 '17 at 11:16
• Yes the formula is proved prior to the excercise – Atif Farooq Sep 24 '17 at 11:29
• In this case, your proof is quite fine. – Bernard Sep 24 '17 at 11:31