Prove that dim(U + V ) ≤ dim U + dim V .

Question

I'm not sure how to prove part a); can someone please provide a proof (not too complex please)

Better show that $\dim(U+V) + \dim(U\cap V) = \dim U + \dim V$. Then it's obvious. And this has been answered multiple times, like here https://math.stackexchange.com/questions/289971/given-two-subspaces-u-w-of-vector-space-v-how-to-show-that-textdim?lq=1 or here https://math.stackexchange.com/questions/1637740/prove-that-dimuw-dimu-cap-w-dim-u-dim-w?rq=1. — Stefan Perko, Apr 19 '16 at 13:29

score 3 · Accepted Answer · answered Apr 19 '16 at 13:31

3

Dimension here is defined as the number of basis elements that span the subspace. Assume that $\{u_1,...,u_n\}$ and $\{v_1,...,v_k\}$ are bases for $U$ and $V$ respectively. If you can show that $U+V$ can be spanned with at most $n+k$ many vectors then you are done. Can you think of a way to get it from here?

answered Apr 19 '16 at 13:31

T. Eskin

8,303

hmm..as U can be spanned by n vectors and V can be spanned by k vectors, these components of U+V can be spanned by at most k+n vectors. If U or V contain any of the same basis vectors, the dimension of U+V will be lower than k+n. Is this correct? are there other circumstances (apart from sharing a basis vector) where the dimension of U+V will be lower? – Jake Apr 19 '16 at 13:41
@user6156388. Correct. And there is no other case when it would occur. In that instance, obviously U and V don't have to share a basis vector given an arbitrary basis for both, but it must be the case that you can find a vector that is a basis vector for both U and V. – T. Eskin Apr 19 '16 at 14:05
thanks; can you provide an example for part b? I'm not sure how to apply w=u+v – Jake Apr 21 '16 at 14:54
@user6156388. Take $U=V$ for example. Then $U+V=U$ so $dim(U+V)=dim(U)<2dim(U)=dim(U)+dim(V)$. And $U$ can be any non-empty subspace of any non-empty vector space. As another non-trivial example, take $W=\Bbb{R}^{3}$ and $U$ a line that is inside a plane $V$. E.g. $V={(x,z,y):z=0}$ and $U={(x,y,z):y=0,;z=0}$. Then $U+V=V$, so $dim(U+V)=dim(V)=2<3=dim(U)+dim(V)$. – T. Eskin Apr 21 '16 at 21:18

score 0 · Answer 2 · answered Apr 19 '16 at 17:21

Let $\{w_1,w_2,\ldots,w_k\}$ be the basis for $U\cap V$, extend it to the bases $\beta=\{w_1,w_2,\ldots,w_k,u_1,u_2,\ldots,u_m\}$ and $\gamma=\{w_1,w_2,\ldots,w_k,v_1,v_2,\ldots,v_n\}$ for $U$ and $V$, respectively. Then we claim that $$ \beta\cup\gamma=\{w_1,w_2,\ldots,w_k,u_1,u_2,\ldots,u_m,v_1,v_2,\ldots,v_n\} $$ is the subset of $U+V$ that generates $U+V$. First, it is clear that $\beta\cup\gamma\subseteq U+V$. To show the second statement, given $x\in U+V$, write $x=u+v$ for some $u\in V$ and $v\in V$. Because $\beta$ is the basis for $U$ and $\gamma$ is the basis for $V$, we may write $u=\sum_{i=1}^ka_iw_i+\sum_{i=1}^mb_iu_i$ and $v=\sum_{i=1}^kc_iw_i+\sum_{i=1}^nd_iv_i$ for some scalars $a_i$, $b_i$, $c_i$, and $d_i$. Thus \begin{align} x=u+v &=\sum_{i=1}^ka_iw_i+\sum_{i=1}^mb_iu_i +\sum_{i=1}^kc_iw_i+\sum_{i=1}^nd_iv_i\\ &=\sum_{i=1}^k(a_i+c_i)w_i+\sum_{i=1}^mb_iu_i+\sum_{i=1}^nd_iv_i \end{align} is evidently a linear combination of vectors of $\beta\cup\gamma$, so the claim is proved. Therefore we have $${\rm span}(\beta\cup\gamma)=U+V$$ and hence $$\dim(U+V)\le \#(\beta\cup\gamma)=m+n+k\leq(m+k)+(n+k)=\dim(U)+\dim(V).$$

Prove that dim(U + V ) ≤ dim U + dim V .

2 Answers2