0
$\begingroup$

This is a generic proof question which I thought of it in a different way different from others and needing a proof confirmation if any of the details are missing.

I use the following lemma:

Let $W_1$ and $W_2$ be subspaces of a finite-dimensional vector space $V$. Let $S$ be a basis for the subspace $W_1 \cap W_2$. There are sets of vectors $T_1$ and $T_2$ such that $S \cup T_1$ is a basis for $W_1$ and $S \cup T_2$ is a basis for $W_2$. Also $S \cup T_1 \cup T_2$ is a basis for $W_1 + W_2$.

Claim: $\dim(W_1+W_2) = \dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$

Proof:

Let $S = \{x_1, ..., x_n\}$ be a basis for $W_1 \cap W_2$.

$\implies \dim(W_1 \cap W_2)=n.$

By the lemma, $\exists \ T_1$ such that $S \cup T_1$ is a basis for $W_1$. Define $T_1 = \{y_1,...,y_a\}$.

$\implies S \cup T_1= \{x_1,...,x_n, y_1,...,y_a\}$.

$\implies \dim(W_1) = | S \cup T_1|=n+a$.

Furthermore, $\exists \ T_2$ such that $S \cup T_2$ is a basis for $W_2$. Define $T_2 = \{z_1,...,z_b\}$.

$\implies S \cup T_2= \{x_1,...,x_n, z_1,...,z_b\}$.

$\implies \dim(W_2) = | S \cup T_2|=n+b$.

Then the claim can be re-written as the following:

$\dim(W_1+W_2) = (n+a)+(n+b)-n=n+a+b$

Also, by the lemma we have that $S \cup T_1 \cup T_2$ is a basis for $W_1+W_2$.

Defined as the following, $S \cup T_1 \cup T_2 = \{x_1,...,x_n,y_1,...,y_a,z_1,...,z_b\}$.

$\implies \dim(W_1+W_2)=|S \cup T_1 \cup T_2|=n+a+b$.

Thus we can see that the claim does hold as needed.

If there is anything you can possibly point out, please do so. Will be appreciated!

$\endgroup$
  • $\begingroup$ A small detail is: If $W_1 \cap W_2 = \{0\}$ no such basis $S$ would exist as it would be (trivially) linearly dependent, but the claim holds since you'd be considering the direct sum of $W_1$ and $W_2$ $\endgroup$ – Ron Jun 21 '18 at 20:38
  • $\begingroup$ @Ron What about $S=\emptyset$, when $W_1\cap W_2=\{0\}$? What's making $\emptyset$ not a basis for the trivial subspace? $\endgroup$ – egreg Jun 21 '18 at 21:21
  • $\begingroup$ Consider $(w_1, w_2) \mapsto w_1 - w_2$ and use the dimension theorem. $\endgroup$ – Will M. Jun 21 '18 at 21:30
  • $\begingroup$ You're right @egreg. I mainly pointed that out because of when OP said "Let $S=\{x1,...,xn\}$ be a basis..." as that would look as if $S$ had to be non empty $\endgroup$ – Ron Jun 21 '18 at 21:31
  • $\begingroup$ Check @egreg's answer on this question math.stackexchange.com/questions/1637740/…. $\endgroup$ – Fareed AF Feb 11 at 14:54
0
$\begingroup$

Your proof looks good, though you may benefit from explicitly mentioning why $ S $, $ T_1 $, and $ T_2 $ are pairwise disjoint, as the result $ | S \cup T_1 \cup T_2 | = n + a + b $ fails without this. You could also shorten the proof by appealing to the Inclusion Exclusion Principle on $ S \cup T_1 $ and $ S \cup T_2 $.

$\endgroup$
  • $\begingroup$ So without having that T1∩T2=∅ which part of my proof would fail? And how would I go about proving that T1∩T2=∅. Also to clarify what you said in the last bit I wouldn't have to show that S∩T1=∅ and S∩T2=∅ right? Because as you said from the choices of T1 and T2 which I was also able to see from my proof of the lemma $\endgroup$ – javacoder Jun 22 '18 at 0:55
  • $\begingroup$ You would have to show that all three intersections are empty, but showing $ S \cap T_i $ for i = 1, 2 is easy, as the $ T_i $ by definition extend the basis $\endgroup$ – user571438 Jun 22 '18 at 1:01
  • $\begingroup$ To show T1 and T2 are disjoint, recall that they extend the basis from W1 intersect W2 to W1 and W2 respectively, so if there was something in T1 intersect T2, it would lie in W1 intersect W2, so use linear independence to conclude. $\endgroup$ – user571438 Jun 22 '18 at 1:06
0
$\begingroup$

You're missing the main point: the lemma already tells you that $S\cup T_1\cup T_2$ is a basis. What you have to show is that it has precisely $n+a+b$ elements.

Thus what you need to observe is that $S$, $T_1$ and $T_2$ are pairwise disjoint. It is sufficient to prove that $T_1\cap T_2=\emptyset$, though, because $S\cap T_1=\emptyset$ and $S\cap T_2=\emptyset$ follows from the choices of $T_1$ and $T_2$.

$\endgroup$
  • $\begingroup$ So without having that $T_1 \cap T_2 = \emptyset$ which part of my proof would fail? And how would I go about proving that $T_1 \cap T_2 = \emptyset$. Also to clarify what you said in the last bit I wouldn't have to show that $S \cap T_1 = \emptyset$ and $S \cap T_2 = \emptyset$ right? Because as you said from the choices of $T_1$ and $T_2$ which I was also able to see from my proof of the lemma. $\endgroup$ – javacoder Jun 21 '18 at 22:49
  • $\begingroup$ @javacoder $T_1$ is obtained by completing $S$ to a basis of $W_1$, so no vector in $T_1$ can be in $S$. Since $W_1\cap W_2=\{0\}$, a vector in $T_1\cap T_2$ should be… $\endgroup$ – egreg Jun 22 '18 at 7:33

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.