a) Suppose that $V$ is the vector space , ${x}_{0}$ is a vector in $V$ and $ {y}_{0}$ is a linear functional on $V$ ,write $Ax=y_0(x)x_0$ for every $x$ in $V$ . Under what condition on $ {x}_{0} $ and $ {y}_{0}$ A is a projection .
b) if A is the projection on say M along N, characterize M and N in terms of ${x}_{0}$ and ${y}_{0}$.
For $A$ to be the projection we should have $A^2=A$ for all $x$ in $V$. We have $AAx=A(y_0(x)x_0)=y_0(y_0(x)x_0)x_0=y_0(x)y_0(x_0)x_0$ and $Ax=y_0(x)x_0$. So $A$ will be projection if $y_0(x_0)=1$. Now regarding range and null space of A, note that range is $M=span(x_0)$. For null space note that $Ax=0$ if $y_0(x)=0$ (assuming $x_0$ and $y_0$ are not zero vector because if if they are then $A=0$ is trivial projection with $M=0$ and $N=V$ ), so null space is $kernel(y_0)$.
$A$ is a projection iff $A^2=A$. It is easy to check than any of the following conditions imply that $A^2=A$:
$y_0\equiv 0$.
(This gives $A\equiv 0$, which is a projection.)
$x_0=0$.
(This gives $A\equiv 0$.)
$y_0(x_0)=1$.
(This gives $A^2x=A(y_0(x)x_0)=y_0(x)Ax_0=y_0(x)y_0(x_0)x_0=y_0(x)x_0=Ax$.)
Conversely, if conditions $1,2,3$ are false, then $A$ cannot be a projection because $$ A^2x_0 = A(y_0(x)x_0)=y_0(x_0)Ax_0 \ne Ax_0. $$ If $A$ is a projection onto $M$ along $N$, then (1) and (2) give $M=\{0\}$ and $N=V$. Finally, $(3)$ gives $M=[\{x_0\}]$ and $N=\mbox{kernel}(y_0)$.