# Thread: Find the inverse to (I+E) where E is a projection

1. ## Find the inverse to (I+E) where E is a projection

Hello,

I have the following problem: Let $V$ be a vector space and $E$ an idempotent linear operator on $V$. Prove that $(I+E)$ is invertible. Find $(I+E)^{-1}$.

Since the problem is for a project not an assignment I cheated a bit and found a worked solution which declares $(I+E)^{-1}=(1-\frac{1}{2}E)$. Now if you know this is the inverse then everything is very easy. Just calculate $(I+E)^{-1}(I-\frac{1}{2}E)$ and you see that it is equal to $I$ and then find $(I-\frac{1}{2}E)(I+E)^{-1}=I$ and apply a theorem ($AB=I$ and $CA=I$ implies exists $A^{-1}$).

The problem is I can't figure out how to deduce that exact form of the inverse.

I tried $(I+E)(X+Y)$ constructions but could not find a second equation so I'm stuck with 1 equation in two unknowns.

I tried setting $X=I$, which seems reasonable in general but not so intuitive, but then in the process of solving for $Y$ I get an expression which requires division.

Like, $(I+E)(I+Y)=I^2+IY+EI+EY$. Since I want this to equal $I$, set $I^2+IY+EI+EY=I$ and solve for $Y$, $(I+E)Y=-E \iff Y=\frac{-E}{I+E}$. Perhaps this expression makes sense perhaps not I don't know. I know in general that under certain conditions you can define both left and right division of linear transformations but this seems too complicated for this.

I've also tried setting $IY+EI+EY=0$ since the truth of that would cause $I^2+IY+EI+EY=I$ as $I$ is idempotent. But then this appears only to tell me $(I+E)Y=-E \iff Y=\frac{-E}{I+E}$ again.

Regards

2. ## Re: Find the inverse to (I+E) where E is a projection

in terms of polynomials,

$E=E^2$ means $E$ satisfies the polynomial $x^2-x$

similarly $A=I+E$ satisfies

$x^2-3x+2$

which means $A\left(\frac{A-3I}{-2}\right)=I$

this gives $A^{-1}$

3. ## Re: Find the inverse to (I+E) where E is a projection

Originally Posted by Idea
in terms of polynomials,

$E=E^2$ means $E$ satisfies the polynomial $x^2-x$

similarly $A=I+E$ satisfies

$x^2-3x+2$
How do you make this step? $I+E$ isn't a projection so I can't just feed it into $x^2-x=0$. Is this just a change of variable thing I don't remember how to do?

4. ## Re: Find the inverse to (I+E) where E is a projection

$A=I+E$

$A^2=(I+E)^2=I+2E+E^2=I+3E$

Eliminate $E$ from the two equations

$A^2-3A=-2I$

$\frac{A^2-3A}{-2}=I$

$\frac{A(A-3I)}{-2}=I$

$A^{-1}=\frac{A-3I}{-2}=-\frac{1}{2}A+\frac{3}{2}I=-\frac{1}{2}(I+E)+\frac{3}{2}I=I-\frac{1}{2}E$