Proving any vector in $\Bbb R^n$ can be written on the form $x = u + v$ I'm having a hard time understanding the solution of this exercise. The exercise says:
Let A be an $n\times n$ matrix so that $$A^2 = A$$ Show that every vector $x$ in $\Bbb R^n$ can be written as $$x = u + v$$ where $Au = u$ and $Av = 0$.
First of all, when they're asking me to use that $Au = u$ and $Av = 0$, I have very little understanding of what this actually means, please explain this (in an intuitive way).
The solution goes like this: let $x$ be a vector in $\Bbb R^n$. Let $u = Ax$ and $v = x-u$. We then get $$Au = A(Ax) = A^2x = Ax = u$$ because $A^2 = A$. Then $$Av = Ax –Au = u – u = 0$$
 A: Just follow your nose: 
Since $x=Ax+x-Ax$, take $u=Ax$ and $v=x-Ax$, and you only need to see that $v$ satisfies the required property: $Av=A(x-Ax)=Ax-A^{2}x=Ax-Ax=0$.
Have you studied projections yet?
A: Hint: For any vector $x$ we can write
$$x=Ax+(x-Ax) $$
A: $A$ represents projection $E$. ($E(x)=Ax$)
If we look at $$E(x)=x+(E(x)-x)$$
It applies for all $x$.
Now try to understand where $E(x)-x$ "lives" (maybe by operating $E$ on it).
A: we will show that $R^n$ is the direct sum of the kernel and the image of $A.$ 
suppose $\dim(\ker(A) = dim(ker(A^2) = n-r.$ the dimension theorem tells you $rank(A) = rank(A^2) = r$  all that remains is to show that there is no overlap between the $\ker(A)$ and the $image(A).$  
suppose $x \in \ker(A) \cap image(A).$ then there is a $y$  such that $$Ay = x, Ax = 0.$$  multiplying $Ay = x$ on the left by $A$ gives $x = Ay = A^2y = Ax = 0$  that is $x = 0.$  
we have now shown that $$(a) dim(\ker(A) + dim(image(A) = n, \ker(A) \cap image(A) = 0 $$ that means the space is the direct sum of the kernel and the image of $A.$ in particular every element can be written as the sum from each in a unique way.
A: Suppose the problem has been solved for the vector $x$, that is, you are able to write $x=u+v$ with $u=Au$ and $Av=0$.
Then you have
$$
Ax=A(u+v)=Au+Av=u+0=u
$$
so $u=Ax$ and therefore $v=x-Ax$. So these are the vectors you're looking for and what's needed now is to show that they indeed solve the problem.
So, given $x$, consider $u=Ax$ and $v=x-Ax$. It's true that $x=u+v$ and, moreover,
\begin{align}
Au&=A(Ax)=A^2x=Ax=u\\
Av&=A(x-Ax)=Ax-A^2x=Ax-Ax=0
\end{align}
so you're done.
