Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

If $U$ and $W$ are subspaces of $V$, a vector space, and $A$ is the spanning set of $U$ and $B$ is the spanning set of $W$. Find the spanning set of $U+W$, in terms of $A$ and $B$, and prove that this is the spanning set. Note that we have seen that $U+W$ is a subspace of $V$.

So from the definition of $\operatorname{span}$, I have that $\operatorname{span}(A)=U$ and $\operatorname{span}(B)=W$. Then $U+W=\operatorname{span}(A)+\operatorname{span}(B)$. This should be true since all elements in $A$ and $B$ are in $U$ and $W$, and by the sum of subspaces. I'm not sure whether I am thinking about that right. Any help is appreciated and thanks in advance.

share|cite|improve this question
This really seems ok to me. – Tunococ Sep 11 '12 at 4:46
Also, $U+W = \mathbb{sp} ( A+B)$. – copper.hat Sep 11 '12 at 5:02
Is that different than what I wrote? i.e. does span(A+B)=span(A)+span(B)? – tk2 Sep 11 '12 at 5:06
tkrm, you haven't actually found a spanning set, have you? A set $C$, such that the span of $C$ is $U+W$? Think of it this way: everything in $U+W$ is of the form $u+w$ with $u$ in $U$ and $w$ in $W$; then $u$ is a linear combination of elements of $A$, and $w$ is a linear combination of elements of $B$, so $u+w$ is a linear combination of...what? – Gerry Myerson Sep 11 '12 at 6:19
By the way, there's no such thing as "$A$ is the spanning set of $U$;" rather, $A$ is a spanning set of $U$. – Gerry Myerson Sep 11 '12 at 6:20
up vote 0 down vote accepted

Try to prove that if $U$ and $W$ are subspaces of $V$, and $A$ and $B$ are spanning sets for $U$ and $W$, respectively, then the union of $A$ and $B$ is a spanning set for $U+W$.

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.