# $W\cap U=0$ and that $\dim W+\dim U > \dim V$

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 $(*)$.

• 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 $U = \{(-t, t)\mid t\in \Bbb R\}\subseteq \operatorname{Span}C'$? Oct 2, 2017 at 12:06
• $W\cap U \supseteq \{ 0 \} \ne \emptyset$.
– mvw
Oct 2, 2017 at 12:08
• @Arthur thanks, I understood that. Is there a way to modify my proof? (I understood that I can not start with $\mathbf B$) Oct 2, 2017 at 12:14
• 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. Oct 2, 2017 at 12:16
• @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) Oct 2, 2017 at 12:33

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.

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

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 $\dim V\ge\dim(U+W)=\dim U+\dim W$.

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