help with a blowup I'm currently learning the basics of blowups and I find that a bit hard. I would like to work out the following example. Could you help me?
Let $k$ be a field and $\mathbb{A}^n_k=\operatorname{Spec}(k[x_1, \ldots, x_n])$. Consider the linear subspace generated by the $r$ first vectors, that is, 
$$
\Lambda=(x_{r+1}=\cdots=x_n=0) \subset \mathbb{A}^n 
$$
and let $X=\operatorname{Bl}_\Lambda \mathbb{A}^n_k$ be the blowup of $\mathbb{A}^n_k$ along $\Lambda$. 
Question 1: Which are the explicit equations of $X$? 
Question 2: What is the exceptional divisor in these equations? 
Now let $\Lambda'$ and $\Lambda''$ be two other linear subspaces with the property that 
$$
\Lambda \cap \Lambda', \quad \Lambda \cap \Lambda'' \quad and \quad \Lambda' \cap \Lambda''
$$ are all reduced to $\{0\}$. We can assume WLOG that
$$
\Lambda'=(x_1=\cdots=x_r=x_{r+s+1}=\cdots=x_n=0) \subset \mathbb{A}^n_k
$$ 
$$
\Lambda'=(x_1=\cdots=x_{r+s}=x_{r+s+t+1}=\cdots=x_n=0) \subset \mathbb{A}^n_k
$$
Question 3: Which are the equations of the proper transforms of $\Lambda'$ and $\Lambda''$? 
Question 4: What does the intersection of these transforms with the exceptional divisor look like? 
 A: We can play with an easy version first, which will help us get our bearings.
Let $X = \operatorname{Bl}_{\Lambda}\mathbb A^3,$ where $\mathbb A^3 = \operatorname{Spec}(k[x,y,z])$ and $\Lambda = V(y,z)$ is the $x$-axis (so $r=1$). We can describe $X$ by two charts, determined by the two equations of $\Lambda:$ there is a $y$-chart $$U_y = \operatorname{Spec}(k[x,y,z,z/y]) = \operatorname{Spec}(k[x,y,z/y])$$ and also a $z$-chart $$U_z = \operatorname{Spec}(k[x,y,z,y/z]) = \operatorname{Spec}(k[x,y/z,z]).$$ Note that each of these charts is isomorphic to $\mathbb A^3.$ Further, I haven't given any equations for $X;$ this construction is intrinsic in that we haven't yet described $X$ as an embedded quasi-projective variety.
Alternatively, we can describe $X$ as follows. Let $g: \mathbb A^3\setminus\Lambda\to\mathbb P^1$ be defined as $g(a) = [y(a):z(a)] = [a_2:a_3]$ for $a = (a_1,a_2,a_3).$ The graph $\Gamma(g)$ of $g$ is a subvariety of $(\mathbb A^3\setminus\Lambda)\times\mathbb P^1,$ and we define $X$ to be the closure of this graph in $\mathbb A^3\times\mathbb P^1.$ We see that there are two charts, coming from the two charts of $\mathbb P^1.$ Away from $\Lambda$ we have $y$ or $z$ nonzero, but $x$ is arbitrary, so $g$ collapses the $x$-variable and projectivizes the remaining $y,z$-variables. Intuitively we see that the fibre is two-dimensional. Being a little more careful, we again see the charts are actually $\mathbb A^3$ as above.
In this formulation, there is one equation of $X\subseteq\mathbb A^3\times\mathbb P^1$ given by $uz-vy=0.$ Can you see how this pops out of the definition $X = \overline{\Gamma(g)}$?
We can now see the exceptional divisor rather easily. Namely, $E = V(y,z)\subseteq\mathbb A^3\times\mathbb P^1.$ That is, if we set $y,z=0$, then our equation $uz-vy=0$ is certainly satisfied for all $u,v$, so we have a copy of $\mathbb P^1$ in $X$ living above this point on $\Lambda.$ Thus $E\cong \mathbb A^1\times\mathbb P^1.$
For $\Lambda'$ and $\Lambda''$ we can take as an example the $y$- and $z$-axis respectively. So $\Lambda' = V(x,z)\subseteq\mathbb A^3$ and $\Lambda'' = V(x,y)\subseteq\mathbb A^3.$ Again, the equations are easy. For example, $\widetilde\Lambda' = V(x,z),$ which has a unique point above any point on the $y$-axis outside the origin, but which contains a $\mathbb P^1$ in the fibre over the origin. A similar scenario occurs for $\widetilde\Lambda'',$ so that we see that $\widetilde\Lambda'\cap\widetilde\Lambda''\cong\mathbb P^1.$ In fact this intersection is also contained in $E$ so that this is the intersection $\widetilde\Lambda'\cap\widetilde\Lambda''\cap E$ as well.
Hopefully this "pet case" contains enough information to get you going on the general case.
