Let A be an $n*n$ matrix. Prove that if $Ax=x$ for all $x\in{\mbox{Im A}}$, then $x-Ax\in{\mbox{Ker A}}$ for all $x\in{R^n}$

Question

Let $A\in{R^{n*n}}$. Let $Ax=x$ for all $x\in{\mbox{Im A}}$.

Prove: $x-Ax\in{\mbox{Ker A}}$ for all $x\in{R^n}$

Preferably, I want to show this without using $A^2=A$. Alternatively, if that is not possible/way harder, I am also struckling with showing that $Ax=x$ for all $x\in{\mbox{Im A}}\Rightarrow A^2 = A$

So far, I have it for all $x\in{\mbox{Im A}}$ by using $Ax = x\Leftrightarrow x-Ax = 0$ thus, $A(x-Ax)=A(0)=0$. But I am not sure how to show it for $x\notin{\mbox{Im A}}$

Answer 1

Let us first show that $A^{2}=A$. Let $e_{i}$ be the $i$-th standard basis vector of $\mathbb{R}^{n}$, for $i\in\{1,\ldots,n\}$. Since $Ae_{i}\in Im(A)$, we have $A(Ae_{i})=A^{2}e_{i}=Ae_{i}$. Note that $A^{2}e_{i}$ is the $i$-th column of the matrix $A^{2}$, and similarly $Ae_{i}$ is the $i$-th column of the matrix $A$. Hence the $i$-th column of $A^{2}$ and the $i$-th column of $A$ are equal. This holds for any $i\in\{1,\ldots,n\}$, so $A^{2}=A$.

Using that $A^{2}=A$, we get for any $x\in\mathbb{R}^{n}$:

$$A(x-Ax)=Ax-A^{2}x=Ax-Ax=0.$$ So $x-Ax$ lies in the kernel of $A$.

Answer 2

We see that $A$ is a projection operator and thus $A^2=A$ as you mentioned. So we have: $$Ax=A(Ax)\\0=A(Ax)-Ax\\0=A(Ax-x)$$ Since $A$ maps $Ax-x$ to $0$, we see that $Ax-x$ is in the kernel of $A$ (which works for any $x\in\Bbb R^n$). So in this solution you do use the fact that $A^2=A$, otherwise I don't know how to solve this (and don't think it is true otherwise). Note that this is equivalent to showing that $x-Ax\in\ker (A)$.