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

Question

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 $(*)$.

Answer 1

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.

Answer 2

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