Any linear subspace $A\subset\mathbb{P}^n(\mathbb{C})$, where $A=\bigcap^m_{i=1} Z(f_i)$ with $f_i\in\mathbb{C}[x_0,...,x_n]$ being some linear forms, is a closed sub variety and there exists uniquely a $b$ where $b=dim(A)$, such that $A\simeq\mathbb{P}^b(\mathbb{C})$.
Any advice/guidance on how to prove this would be appreciated.
Thank you in advance!
Hint : this is a closed variety by definition. Let's show it's linear. Look in $\mathbb C^{n+1}$. Your $f_i$ lift to linear form $\tilde f_i \in (\mathbb C^{n+1})^{\vee}$, i.e hyperplane of $\mathbb C^{n+1}$. Now it is clear that the intersection of these hyperplane is a linear subspace $\tilde A \subset \mathbb C^{n+1}$ of dimension $a+1$. I claim that $\pi(\tilde A) = A$, and so $b = a$.