Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I come across an interesting problem on my journey of cracking open some old math books and cracking down on problems from boredom. I cannot seem to wrap my head around this problem of subspaces. The problem is,

Let $W_1$ and $W_2$ be subspaces of a finite dimensional vector space $V$. Can the following be proved?

(a) $W_1+W_2=\{w_1+w_2:w_1 \in W_1,w_2 \in W_2\}$ is a subspace of $V$.

(b) $W_1 \cap W_2$ is a subspace of $V$.

(c) $\dim(W_1)+ \dim(W_2)= \dim(W_1+W_2)+ \dim(W_1 \cap W_2)$.

Any ideas on how to go about solving this?

Thank you in advance.

share|cite|improve this question
up vote 1 down vote accepted

(a) and (b) are relatively easy - you just have to verify the definition of the subspace.

For (c): Choose any basis $a_1,\ldots,a_k$ of $W_1\cap W_2$. Then this basis can be extended to a basis $a_1,\ldots,a_k,b_1,\ldots,b_l$ of $W_1$ and to a basis $a_1,\ldots,a_k,c_1,\ldots,c_m$ of $W_2$. (A consequence of Steinitz exchange lemma.)

If you are able to prove that $a_1,\ldots,a_k,b_1,\ldots,b_l,c_1,\ldots,c_m$ is a basis of $W_1+W_2$, then you are done. (The dimensions of $W_1\cap W_2$, $W_1$, $W_2$ and $W_1+W_2$ are $k$, $k+l$, $k+m$ and $k+l+m=(k+l)+(k+m)-k$, respectively.

share|cite|improve this answer
Very helpful, this will shall do the trick. Thanks. – night owl Jun 5 '11 at 10:36

To show that $U \subset V$ is a subspace you have to prove that

(a) $0 \in U$

(b) $\forall a,b \in U, \lambda \in K: a+\lambda \cdot b \in U$

That your first 2 expressions are subspaces directly follows from the fact that $W_1$ and $W_2$ are already subspaces. For the dimension you can look at a base of $W_1$ and $W_2$ and see what the operations change about the number of vectors in your base or do this.

share|cite|improve this answer
Thanks for the hint & link provided. – night owl Jun 5 '11 at 10:37
No problem, feel free to ask if something is still not clear. – Listing Jun 5 '11 at 12:38

For c), let $U=W_1\times W_2$, that is, $U$ is the set of all ordered pairs $(w_1,w_2)$ with $w_1$ in $W_1$ and $w_2$ in $W_2$. Observe that $U$ is a vector space of dimension ${\rm dim}(W_1)+{\rm dim}(W_2)$. Consider the map $T:U\to W_1+W_2$ given by $T(w_1,w_2)=w_1+w_2$. Observe that $T$ is linear and onto. Now study the kernel of $T$ and show that the nullity of $T$ is ${\rm dim}(W_1\cap W_2)$, and then use the rank plus nullity theorem to get the desired result.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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