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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.

3 Answers 3

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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.

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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.

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    No problem, $f$eel $f$ree to ask if somethi$n$g is still $n$ot clear.2011-06-05
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(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.

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    Very helpful, this will shall do the trick. Thanks.2011-06-05