Suppose $W$ is a subspace of the finite-dimensional vector space $V$ and $\dim W = \dim V=n$
Let $\{\overrightarrow w_1,\ldots,\overrightarrow w_n\}$ be a basis for $W$ and $\{\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ be a basis for $V$.
The set $\{\overrightarrow w_1,\ldots,\overrightarrow w_n,\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ still spans $W$. Since $W$ is a subspace of $V$, $W\subset V$ and all the $w_i$'s can be written in terms of $\{ \overrightarrow v_{1},\ldots,\overrightarrow v_n\}$.
$\therefore$ We can remove one by one all the $w_i$'s from the set $\{\overrightarrow w_{1},\ldots,\overrightarrow w_n,\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ while still having it span $W$, which leaves us with the set $\{\overrightarrow v_1,\ldots,\overrightarrow v_n\}$. We know it to be linearly independent since it is a basis for $V$. Therefore the basis of $V$ is also a basis of $W$. From there on, it is trivial to prove double inclusion $W\subset V$ and $V\subset W$ which gives us $W=V$.
Is my approach correct?
Best Answer
Your approach is incorrect: the set $\{\overrightarrow w_1,\ldots,\overrightarrow w_n,\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ spans $V$ and saying that it spans $W$ is the same as assuming $W=V$.
You should rather consider a basis $\{\overrightarrow w_1,\ldots,\overrightarrow w_n\}$ of $W$ and assume, by contradiction, that $W\ne V$. Then there exists $\vec{v}\in V$ with $\vec{v}\notin W$. It's easy to prove that $\{\overrightarrow w_1,\ldots,\overrightarrow w_n,\overrightarrow v\}$ is linearly independent and now this is a contradiction, because we found a linearly independent set in $V$ with more elements than $\dim V$.