number of elements of basis of subspaces May I ask if my proof is correct for the following?
Let $V$ be a vector space and dim(V) = n, and let $W$ be a subspace of $V$ and dim (W) = n-1. If $U$ is a subspace of $V$ such that $U \not\subset$ W show that $dim (W \cap U) = dim (U) - 1.$
Proof. Since $dim (W) = n-1,$ then $ n-1 < dim (U) \leq n.$ We should have $dim (V) = dim (U)$. Hence, $U = V$ So,
\begin{align*}
 dim (W \cap U) = dim (W \cap V) = dim (W) = n-1.
\end{align*}
 A: As I point out in the comments, your proof is not correct.
Hint: Let $k = \dim(W \cap U)$ and $d = \dim(U)$. Let $\{\alpha_1,\dots,\alpha_k\}$ be a basis of $W \cap U$. Note that this set can be extended both to a basis $\{\alpha_1,\dots,\alpha_k,\dots,\alpha_{n-1}\}$ of $W$ and a basis $\{\alpha_1,\dots,\alpha_k,\beta_{k+1},\dots,\beta_d\}$ of $U$.
A: We are given that
(i) the vector space $V$ has dimension $n$,
and also that
(ii) $U$ and $W$ are subspaces of $V$ such that $\dim W = n-1$ and $U \not\subset W$.
As $U \not\subset W$, so either $W \subsetneqq U$ or $W \not\subset U$.
Case 1. Suppose that $W \subsetneqq U$. Then $W$ is a proper subspace of $U$ and thus we have
$$
n-1 = \dim W < \dim U \leq \dim V = n
$$
from which it follows that $\dim U = n$, and so we must have $U = V$. Then the subspace $U+W$ of $V$ must equal $U=V$ also. Therefore we obtain
$$
\dim V = \dim (U+W) = \dim U + \dim V - \dim (U \cap W) = \dim V + \dim W - \dim (U \cap W),
$$
which implies that
$$
\dim (U \cap W) = \dim W = n-1 = \dim U - 1. 
$$
Case 2. Suppose that $W \not\subset U$. Thus we have $U \not\subset W$ and $W \not\subset U$. So the subspace $U+W$ of $V$ satisfies
$$
U+W \supset U \cup W \supsetneqq W,
$$
which implies that
$$
n= \dim V \geq \dim (U+W) > \dim W = n-1,
$$
and hence
$$
\dim (U+W) = n,
$$
from which it follows that
$$
U+W = V. 
$$
Therefore
\begin{align}
 \dim (U \cap W) &= \dim U + \dim W - \dim (U+W) \\
&= \dim U + (n-1) - n \\
&= \dim U - 1,
\end{align}
as required.
Hope this helps.
