When a matrix, $M$ equals $M^2$, why is this property true? While we were looking at idempotent matrices, my teacher brought up some properties about them, specifically when the square of a matrix equals itself, the matrix's null space equals the column space of the matrix minus its respective identity matrix.
Is this something that should be trivial, I've been trying to convince myself this is true but I can't seem to reason this to myself.
 A: You've already had two answers, but here's a more intuitive one.
Matrices represent linear transformations. Start with $\mathbb{R}^3$, and give it coordinate axes x, y, z. Consider what happens if you apply the transformation $\alpha = $ "project onto the (x,y)-plane".


*

*In coordinates, $\alpha(a,b,c) = (a,b,0)$.


Then consider what happens if you apply $\alpha$ twice.


*

*In coordinates, $\alpha(\alpha(a,b,c)) = \alpha(a,b,0) = (a,b,0)$.


In other words, applying $\alpha$ twice is the same as applying it once, so $\alpha^2 = \alpha$.
Likewise, matrices $M$ such that $M^2 = M$ usually look like "projections". After all, if $M^2 = M$, and $x$ is in the image of $M$, then $Mx = x$.


*

*Proof: $x$ is in the image of $M$, so say $x = My$. Then $Mx = M^2 y = My = x$.


In other words, everything that's "perpendicular" to the image of $M$ gets killed, and everything that's in the image of $M$ gets left alone.
A: Let $(M-I)y$ be an arbitrary element of the column space of $M-I$.  Then $M(M-I)y=M^2y - My = My - My = 0$.
Conversely, suppose that $Mx=0$.  Then $x=-(M-I)x$, so $x$ is in the column space of $M-I$.
A: Another viewpoint:
$$M=M^2\Longleftrightarrow M(M-\operatorname{Id})=0$$
from which it follows that $$\operatorname{Col}(M-\operatorname{Id})\subset \operatorname{Ker}M$$
The equality follows from the fact that they have the same dimension:
 $$\operatorname{rk}(M-\operatorname{Id})= \dim\operatorname{Ker}M$$
 which you can see e.g. from a triangular form of $M$, since its only non-zero eigenvalue is $1$.
