Suppose that $W$ is a subspace of a finite-dimensional vector space $V$.
Prove that there exists a subspace $W'$ and a function $T:V\rightarrow V$ such that $T$ is a projection on $W$ along $W'$.
MY ATTEMPT
Since $W$ admits a basis $\mathcal{B}_{W}$, we have that $W = \text{span}(\mathcal{B}_{W})$. Let us suppose that $\dim W < \dim V$. Therefore there exists a vector $w'_{1}\in V$ such that $w'_{1}\not\in W$. Consequently, the set $\mathcal{B}_{W}\cup\{w'_{1}\}$ is LI. If $\dim\text{span}(\mathcal{B}_{W}\cup\{w'_{1}\}) = \dim V$, we have that $W' = \text{span}(\{w'_{1}\})$.
Otherwise there exists a vector $w'_{2}\in V$ such that $w'_{2}\not\in\text{span}(\mathcal{B}_{W}\cup\{w'_{1}\})$. Consequently, the set $\mathcal{B}_{W}\cup\{w'_{1},w'_{2}\}$ is LI. If $\dim\text{span}(\mathcal{B}_{W}\cup\{w'_{1},w'_{2}\}) = \dim V$, we have that $W' = \text{span}\{w'_{1},w'_{2}\}$.
Otherwise, we can proceed as before until we obtain a basis for $V$. In such case, we conclude that $W' = \{w'_{1},w'_{2},\ldots,w'_{n}\}$, where $n = \dim V - \dim W$.
We can define the function $T:V\rightarrow V$ such that $T$ is a projection on $W$ along $W'$ as next \begin{align*} x & = a_{1}w_{1} + a_{2}w_{2} + \ldots + a_{m}w_{m} + b_{1}w'_{1} + b_{2}w'_{2} + \ldots + b_{n}w'_{n} \Longrightarrow\\\\ T(x) & = a_{1}w_{1} + a_{2}w_{2} + \ldots + a_{m}w_{m} \end{align*}
where $\mathcal{B}_{W} = \{w_{1},w_{2},\ldots,w_{m}\}$.
Can someone verify if I am reasoning rightly?