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Is the following proof correct?

Theorem. Given that $U$ and $W$ are subspaces of the vector space $V$ such that $V=U\oplus W$ and $u_1,u_2,...,u_m$ and $w_1,w_2,...,w_n$ are bases for $U$ and $W$ respectively, then $$u_1,u_2,...,u_m,w_1,w_2,...,w_n$$ is a bases for $V$.

Proof. We construct the proof by proving a series of Lemma.

Lemma(1). $span(u_1,u_2,...,u_m,w_1,w_2,...,w_n)=V$

Proof. Consider an arbitrary element of $span(u_1,u_2,...,u_m,w_1,w_2,...,w_n)$ say $$v=\sum_{i=1}^{m}a_iu_i+\sum_{j=1}^{n}b_jw_j\tag{1}$$ evidently $u=\sum_{i=1}^{m}a_iu_i\in U$ and $w=\sum_{j=1}^{m}b_jw_j\in W$ but since $U\oplus W=V$, It follows that $v\in V$. Since $v$ was arbitrary it follows that $span(u_1,u_2,...,u_m,w_1,w_2,...,w_n)\subseteq V$.

Assume now that $v\in V=U\oplus W$ it then immediately follows that for some $u=\sum_{i=1}^{m}a_iu_i\in U$ and $u=\sum_{j=1}^{n}b_jw_j\in W$ it is the case that $v=u+w$ implying that $v\in span(u_1,u_2,...,u_m,w_1,w_2,...,w_n)$.

We may now conclude that $span(u_1,u_2,...,u_m,w_1,w_2,...,w_n)\subseteq V$.

consequently $span(u_1,u_2,...,u_m,w_1,w_2,...,w_n)=V$.

$\square$

Lemma(2). The list $u_1,u_2,...,u_m,w_1,w_2,...,w_n$ is linearly independent.

Proof. Consider an arbitrary $v\in V$ and assume that $$v=\sum_{i=1}^{m}\alpha_iu_i+\sum_{j=1}^{n}\beta_jw_j=\sum_{i=1}^{m}\phi_iu_i+\sum_{j=1}^{n}\psi_jw_j\tag{2}$$ After a little algebraic manipulation we have $$\sum_{i=1}^{m}(\alpha_i-\phi_i)u_i+\sum_{j=1}^{n}(\beta_j-\psi_j)w_j=0\tag{3}$$ but $u=\sum_{i=1}^{m}(\alpha_i-\phi_i)u_i\in U$ and $w=\sum_{j=1}^{n}(\beta_j-\psi_j)w_j\in W$ and since $U+W=U\oplus W$ it follows that $u=w=0$ i.e. $$u=\sum_{i=1}^{m}(\alpha_i-\phi_i)u_i=0\tag{4}$$ $$w=\sum_{j=1}^{n}(\beta_i-\psi_i)w_i=0\tag{5}$$ But $u_1,u_2,...,u_m$ and $w_1,w_2,...,w_m$ are linearly independent lists in $U$ and $W$ and therefore in $V$ consequently $$\forall i\in\{1,2,...,m\}(\alpha_i=\phi_i)\tag{6}$$ $$\forall j\in\{1,2,...,m\}(\beta_j=\psi_j)\tag{7}$$ Therefore $v$ has only one unique representation in $V$ as linear combination of $u_1,u_2,...,u_m,w_1,w_2,...,w_m$.

$\square$

Taking the above two Lemma together we have the required result.

$\blacksquare$

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Besides some typos it is ok, but I think it is too long.

Proof of Lemma 1.

Since $u_1,\ldots,u_m,w_1,\ldots,w_n \in V$ then $span(u_1,\ldots,u_m,w_1,\ldots,w_n)\subset V.$ If $v \in V=U \oplus W$ then $v=u+w$ for some $u \in U,$ $w\in W.$ Then $u=\sum_i\alpha_i u_i$ and $w=\sum_j \beta_j w_j$ for scalars $\alpha_i, \beta_j,$ so $v \in span(u_1,\ldots,u_m,w_1,\ldots,w_n)$ and hence $V \subset span(u_1,\ldots,u_m,w_1,\ldots,w_n).$

Proof of Lemma 2.

Suppose that $$ \alpha_1u_1+\cdots\alpha_mu_m+\beta_1w_1+\cdots \beta_nw_n=0$$ for some scalars $\alpha_i, \beta_j.$ Then $$ \alpha_1u_1+\cdots\alpha_mu_m=-(\beta_1w_1+\cdots \beta_nw_n) \in U \cap W=0.$$ Since the $u_i$ are linear independent then $\alpha_i=0$ for all $i.$ The same argument shows that $\beta_j=0$ for all $j.$ Therefore the list $u_1,\ldots,u_m,w_1,\ldots w_n$ is linear independent.

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  • $\begingroup$ Wow i like your proof much better any advice against writing long winded proofs? $\endgroup$ – Atif Farooq Jul 27 '17 at 20:02
  • $\begingroup$ Write lots of proofs and do lots of exercises, no matter if proofs are long, and you'll keep improving and beginning to write shorter proofs. $\endgroup$ – positrón0802 Jul 27 '17 at 20:07
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Your thinking is correct but I would suggest some improvements.

  • For the fist lemma you don't need to show in such detail that $$ \operatorname{span}\{u_1,\dots,u_m,w_1,\dots,w_n\} \subseteq V. $$ We know that given any subset $S \subseteq V$ that $\operatorname{span}(S) \subseteq V$ because where else would the linear combinations land? If you are working inside the vector space $V$ then any linear combination is in $V$.

  • When you are showing that $V \subseteq \operatorname{span}\{u_1,\dots,u_m,w_1,\dots,w_n\}$ you write $u = \sum_{i=1}^m a_iu_i$ without further comment. I would suggest you say that this can be done because $\{u_1,\dots,u_m\}$ span $U$ just as for the second lemma you appeal explicitly to the linear independence of $\{u_1,\dots,u_m\}$.

  • For the second lemma, from $u=\sum_{i=1}^{m}(\alpha_i-\phi_i)u_i\in U$ and $w=\sum_{j=1}^{n}(\beta_j-\psi_j)w_j\in W$, I would say $u = -w \in U \cap W = \{0\}$ which makes it more clear what $U + W = U \oplus W$ is supposed to imply.

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