# Prove that $\dim(W_1+W_2) = \dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$ proof confirm

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!

• 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 $W_2$ – Ron Jun 21 '18 at 20:38
• @Ron What about $S=\emptyset$, when $W_1\cap W_2=\{0\}$? What's making $\emptyset$ not a basis for the trivial subspace? – egreg Jun 21 '18 at 21:21
• Consider $(w_1, w_2) \mapsto w_1 - w_2$ and use the dimension theorem. – Will M. Jun 21 '18 at 21:30
• 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 – Ron Jun 21 '18 at 21:31
• Check @egreg's answer on this question math.stackexchange.com/questions/1637740/…. – Fareed AF Feb 11 at 14:54

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 $S \cup T_2$.
• 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 – user571438 Jun 22 '18 at 1:01
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 $T_2$.
• 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. – javacoder Jun 21 '18 at 22:49
• @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… – egreg Jun 22 '18 at 7:33