Let $A$ be a matrix with $Nul(A) = 0$. For a given vector $b$, there is a unique least squares "solution" $x = T(b)$ of the equation $Ax = b$. Show that the map $T: b \rightarrow T(b) $ is linear. Give the matrix that describes it.
So from what I understand this means I have to show $T(\alpha b) = \alpha T(b)$ and $T(a+b) = T(a) + T(b)$
we have $A^T \alpha b = \alpha A^T b$ and $A^T(a+b) = A^Ta + A^Tb$. That should prove the map is linear. I guess the matrix that describes the map is $A^T$?
Since $x$ is the least-squares solution to $A x = b$, one can employ the normal equations to reformulate this condition as $(A^TA) x = A^T b$. Because $Null(A)=\emptyset$, $A^TA$ is invertible, thus $T = (A^TA)^{-1}A^T$, which basically gives the pseudo-inverse of $A$.