Rank and Pseudo Inverse of a Matrix Let $T$ be a linear transformation   $L(x) = Ax$ from $ℝ^m → ℝ^n$. 
The pseudo inverse $L^+$ of $L$ is the transformation from $ℝ^n → ℝ^m$ given by minimal least-squares solution of the system $L(x) = y$
Question: 
Suppose $\operatorname{rank}(A) = n$ 
Find the matrix for $L^+$.
proof: $\operatorname{rank}(A) = n$  implies that $\operatorname{image}(A)$ spans $ℝ^n$. From this, $\ker(A^T) = \operatorname{im}(A)^\perp = \{0\}$.
this is where I get stuck. Don't know how to proceed. 

Hint: The least squares solution solves the equation
$$
A^TA x = A^Ty
$$
In this case, we can note that $A^TA$ is a square matrix with full rank.
