The below theorem I am to prove. Perhaps you have a critique...

Theorem 2.4 Let $W_1$ and $W_2$ be two subspaces of a vector space $V$. Then $\dim(W_1 \cap W_2)=\dim(W_1)$ if and only if $W_1 \subset W_2$.

My—perhaps lazy—attempt to prove it... I say lazy in the sense that I don't show $\subseteq$ then $\supseteq$.

Let $W_1$ and $W_2$ be finite subspaces of the vector space $V$. Let \begin{eqnarray} \{\alpha_1,\alpha_2,\dots,\alpha_l\} \end{eqnarray} be a basis for $W_1 \cap W_2$. Because $W_1 \cap W_2 \subset W_1$, then we have \begin{eqnarray} \{\alpha_1,\alpha_2,\dots,\alpha_l,\beta_1,\beta_2,\dots,\beta_m\} \end{eqnarray} as a basis for $W_1$, and similarly we have \begin{eqnarray} \{\alpha_1,\alpha_2,\dots,\alpha_l,\gamma_1,\gamma_2,\dots,\gamma_n\} \end{eqnarray} as a basis for $W_2$, so $W_1+W_2$ is spanned by the linearly independent set of vectors \begin{eqnarray} \alpha_1,\dots,\alpha_l,\beta_1,\dots,\beta_m,\gamma_1,\dots,\gamma_n \end{eqnarray} and has dimension $l+m+n$. Thus, from (14) and (15) we have that $\dim W_1 = l+m$ and $\dim W_2 = l+n$, so \begin{eqnarray} \dim W_1 + \dim W_2 & = & (l+m)+(l+n) \\ &=&l+(l+m+n) \\ &=&\dim(W_1 \cap W_2) + \dim(W_1+W_2). \end{eqnarray} This being the case, then the second statement in Theorem 2.4 implies that $\dim W_2 = \dim (W_1+W_2)$, or $l+n=l+m+n$. This means that $m=0$, which implies that the basis for $W_1$ has the form \begin{eqnarray} \{\alpha_1,\dots,\alpha_l\}, \end{eqnarray} so since $W_2$ has the form \begin{eqnarray} \{\alpha_1,\dots,\alpha_l,\gamma_1,\dots,\gamma_n\}, \end{eqnarray} then it is clear that this can only be the case when $W_1 \subset W_2$.

  • 1
    $\begingroup$ I really don't see why you keep tagging questions with [elementary-set-theory]. $\endgroup$ – Asaf Karagila Apr 16 '13 at 22:26
  • $\begingroup$ @AsafKaragila Sets are involved. I want non-topic-specific insight. Ideas from other regions of thought slosh over to others. Insight through these other regions are necessary. Have you heard of the Ketengban tribe? Perhaps understanding their methods of ornithological identification could help simple-minded people have a deeper mode of classification. I don't know that I'm really getting my point across here... $\endgroup$ – Trancot Apr 16 '13 at 22:33
  • 1
    $\begingroup$ Then perhaps tag it using [english-written-problems] as well, seeing how English is involved? $\endgroup$ – Asaf Karagila Apr 16 '13 at 22:43
  • $\begingroup$ I can't yet. My "reputation" isn't high enough. ^_^ $\endgroup$ – Trancot Apr 16 '13 at 22:46
  • $\begingroup$ Someone with a more purely set-theoretic background would perhaps have a stronger insight into the problem than one focused primarily with the other tags... That's all... $\endgroup$ – Trancot Apr 16 '13 at 22:52

Depending on how well is one acquainted with these things I'd prove this much shorter and using basic set theory:

$$W_1\subset W_2\iff W_1\cap W_2=W_1$$

and thus for finite dimensional vector subspaces:

$$W_1\subset W_2\iff \dim(W_1\cap W_2)=\dim W_1$$

  • $\begingroup$ I'm fairly unacquainted with "these things," and I think that will always be the case... Even for everyone else on the planet too, don't you think? I mean in a Feynman kind of way... $\endgroup$ – Trancot Apr 16 '13 at 22:21
  • $\begingroup$ It is for an honors class, so I assume it must be detailed... $\endgroup$ – Trancot Apr 16 '13 at 22:22
  • $\begingroup$ Well, it is not the most usual thing to see someone trying some basic exercises in linear algebra without some basic set theory... $\endgroup$ – DonAntonio Apr 16 '13 at 22:22
  • $\begingroup$ Oh, and the above is very detailed...and honor class in high school, college, university...? $\endgroup$ – DonAntonio Apr 16 '13 at 22:23
  • $\begingroup$ UC... Tell me about it! This Honors Linear Algebra course is "supposed" to be our introduction to proofs... $\endgroup$ – Trancot Apr 16 '13 at 22:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.