0
$\begingroup$

Let $W$ be a subspace of a vector space $V$. Prove that there is no subspace $U$ such that $W\cap U=0$ and that $\dim W+\dim U > \dim V$.

I am using the fact that a subset of a linearly independent set is linearly independent.$\tag{$*$}$

Let, $\mathbf B=\{v_1,v_2,\dots, v_n\}$ is a basis set of $V$. Choose a subset $\mathbf C\subset\mathbf B$ and $S_1=span \mathbf C$ and $S_2=span \mathbf C'$ such that $W\subset S_1$ and $U\subset S_2$. By construction $W\cap U=0$. Note that $\dim W\leq \dim S_1$ and $\dim U\leq \dim S_2$. Then, $\dim W+\dim U\leq \dim S_1+\dim S_2=\dim V$ (using $(*)$). Hence, the statement is true.

Is my proof writing is ok, or I am missing something. Thanks.

Also, is there a method without using $(*)$.

$\endgroup$
6
  • $\begingroup$ Given an arbitrary basis, you can't be certain that $U$ and $W$ are spanned by subsets of that basis. For instance, for $V = \Bbb R^2$, given the basis $\{(1,0),(0,1)\}$, how will you choose a $C$ such that $W = \{(t, t)\mid t\in \Bbb R\}\subseteq \operatorname{Span}C$ while at the same time $U = \{(-t, t)\mid t\in \Bbb R\}\subseteq \operatorname{Span}C'$? $\endgroup$
    – Arthur
    Oct 2, 2017 at 12:06
  • $\begingroup$ $W\cap U \supseteq \{ 0 \} \ne \emptyset$. $\endgroup$
    – mvw
    Oct 2, 2017 at 12:08
  • $\begingroup$ @Arthur thanks, I understood that. Is there a way to modify my proof? (I understood that I can not start with $\mathbf B$) $\endgroup$ Oct 2, 2017 at 12:14
  • $\begingroup$ Start with a basis of each of $U$ and $W$ instead, and show that $U\cap W = \{0\}$ implies that the union of the two bases is a set of $\dim U + \dim W$ linearly independent vectors. $\endgroup$
    – Arthur
    Oct 2, 2017 at 12:16
  • $\begingroup$ @Arthur Is this argument ok? : Since $W\cap U=0$, any element of $B_U$ can not be written as linear combination of elements of $B_W$. Hence, $|B_U\cup B_W|=|B_U|+|B_W|$. (Notations are taken from the answer) $\endgroup$ Oct 2, 2017 at 12:33

2 Answers 2

1
$\begingroup$

Let $B_U = \{u_1, \ldots, u_m\}$ and $B_W = \{w_1,\ldots,w_n\}$ be bases of $U$ and $W$ respectively. We want to show that $B_U\cup B_W = \{u_1, \ldots, u_m,w_1,\ldots,w_n\}$ is a linearly independent set of vectors (because if it is, then the dimension of $V$, where all of this takes place, is at least $m+n$).

Now, let's take some linear combination of those vectors that add to $0$: $$ r_1v_1 + \cdots + r_mv_m + s_1w_1 + \cdots + s_nw_n = 0 $$ where $r_i, s_i \in \Bbb R$ (I assume we are working over the real numbers; if not, you can replace $\Bbb R$ with your field of choice, it doesn't matter). We want to show that all the $r_i, s_i$ must be $0$.

Now, take that linear combination, and reorder it: $$ r_1v_1 + \cdots + r_mv_m = -s_1w_1 - \cdots - s_nw_n $$ We see that the left-hand side add up to a vector in $U$, and the right-hand side add up to a vector in $W$. But they are the same vector, which means that it must be some vector in $U\cap W = \{0\}$. Thus we have $$ r_1v_1 + \cdots + r_mv_m =0\\ -s_1w_1 - \cdots - s_nw_n = 0 $$ Now we use the linear independence of $B_U$ and $B_W$ separately, and we see that all the $r_i$ and all the $s_i$ must be $0$, and we are done.

$\endgroup$
1
  • $\begingroup$ sorry, I had to accept first, then upvote (because of my reputation was less than 15) $\endgroup$ Oct 2, 2017 at 12:46
1
$\begingroup$

Let $B_W$ be a basis of $W$ and $B_U$ a basis of $U$. Since $U\cap W=0$, $B_U\cup B_W$ is a basis of $U+W$, and $\dim V\ge\dim(U+W)=\dim U+\dim W$.

$\endgroup$
3
  • $\begingroup$ Thanks, so I need to prove $\dim(U+W)=\dim U+\dim W$ $\endgroup$ Oct 2, 2017 at 12:16
  • $\begingroup$ Since $B_U$ and $B_W$ are disjoint, $\dim(U+W)=|B_U\cup B_W|=|B_U|+|B_W|=\dim U+\dim W$. $\endgroup$
    – ajotatxe
    Oct 2, 2017 at 12:19
  • $\begingroup$ Yeah! Since $W\cap U=0$, any element of $B_U$ can not be written as linear combination of elements of $B_W$. Hence, $|B_U\cup B_W|=|B_U|+|B_W|$. $\endgroup$ Oct 2, 2017 at 12:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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