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

Is the following Proof Correct?

Theorem. Given that $$U$$ is a subspace of $$V$$ and $$\{v_1+U,v_2+U,\ldots,v_m+U\}$$ is a basis for $$V/U$$ and $$\{u_1,u_2,...,u_n\}$$ is a basis for $$U$$ prove that $$\{v_1,v_2,\ldots,v_m,u_1,u_2,\ldots,u_n\}$$ 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 $$v_1,v_2,\ldots v_m,u_1,u_2,\ldots,u_n$$ 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 $$v\in V$$ be arbitrary, evidently $$v+U\in V/U$$ and therefore for some $$\alpha_1,\alpha_2,\ldots,\alpha_m\in\mathbf{F}$$ $$v+U = \sum_{j=1}^{m}\alpha_j(v_j+U) = \left(\sum_{j=1}^{m}\alpha_jv_j\right)+U$$ which may be equivalently stated as follows $$v-\left(\sum_{j=1}^{m}\alpha_jv_j\right)\in U$$ and consequently for some $$\beta_1,\beta_2,\ldots,\beta_n\in\mathbf{F}$$ 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 $$\operatorname{span}\{v_1,v_2,\ldots,v_m,u_1,u_2,\ldots,u_n\} = V$$.

$$\blacksquare$$

• Are you sure the exercise supposes you know the dimension formula? I'd rather think it's a way to prove it. Commented Sep 24, 2017 at 11:16
• Yes the formula is proved prior to the excercise Commented Sep 24, 2017 at 11:29
• In this case, your proof is quite fine. Commented Sep 24, 2017 at 11:31