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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 $\operatorname{span}\{v_1,v_2,\ldots,v_m,u_1,u_2,\ldots,u_n\} = V$.

$\blacksquare$

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

1 Answer 1

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The dimension formula is needed since Atif Farooq only proves that the list spans V, not that the list is lineary independent .

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