Explicit formula for $(I+E)^{-1}$ for E is Idempotent Matrix I wanted to find formula for  $(I+E)^{-1}$
I know that this $I+E$ is invertible .
Because this is equivalent to show that -1 is not eigenvalue of E ,Idempotent matrix.
Main Problem occured when I had write power expansion of term $(I+E)^{-1}$=$(I-E+E^2-E^3.....)$ Actually which I this not converging .
How to find Inverse of I+E?
Any Help will be appreciated
 A: Assume that the base field is $K$, which is of characteristic not equal to $2$ (otherwise, $I+E=I-E$ is not invertible unless $E=0$).  You can just make a guess that $I+tE$ is the inverse of $I+E$ for some $t\in K$.  This gives $2t+1=0$, or $t=-\dfrac12$, as in Kavi Rama Murthy's answer.  
Alternatively, we study the inverse of 
$$I+xE=(1+x)\,\left(I-\frac{x}{1+x}(I-E)\right)$$
as a matrix over the field of formal Laurent power series $K(\!(x)\!)$.  From $E^2=E$, we get
$$(I-E)^2=I-2E+E^2=I-E$$
so $(I-E)^k=I-E$ for all $k=1,2,3,\ldots$.  The inverse of $I+xE$ in $K(\!(x)\!)$ is thus
$$(1+x)^{-1}\,\left(I+\frac{x}{1+x}(I-E)+\frac{x^2}{(1+x)^2}(I-E)^2+\frac{x^3}{(1+x)^3}(I-E)^3+\ldots\right)\,.$$
Simplifying the expression above, we get
$$(1+x)^{-1}\,\Biggl(I+\left({\small\sum_{k=1}^\infty\,\left(\frac{x}{1+x}\right)^k}\right)\,(I-E)\Biggr)=(1+x)^{-1}\,(I+x(I-E)\big)\,.$$
That is, the inverse of $I+xE$ in $K(\!(x)\!)$ equals
$$I-\frac{x}{1+x}E\,.$$
In particular, the evaluation at $x=1$ tells us that the inverse of $I+E$ is $I-\dfrac{1}{2}E$.
A: Or look at it this way: If $A$ is a square matrix, $p$ is a polynomial with nonzero contstant term and $p(A)=0$ then you can rearrange $p(A)=0$ to the form  $AB=I$, giving you the inverse of $A$.
Say $E$ is idempotent and $A=I+E$. It's clear how to find $p$: $$0=E^2-E=(A-I)^2-(A-I)=A^2-3A+2I,$$so $$I=\frac12A(3I-A),$$hence $A$ is invertible and $$A^{-1}=\frac12(3I-A)=I-\frac12E.$$
A: $I-\frac 1 2 E$ is the inverse of $I+E$.
A: Here is a perhaps more direct way to arrive at the answer. You were trying to use $\frac1{1+E}=1-E+E^2-\dots$. The problem is that this is only valid when $|E|<1$. However, since $E$ is idempotent, its eigenvalues may be either $0$ or $1$, so $|E|<1$ may not hold. 
However, $-\frac12+E$ will have eigenvalues in $\{-\frac12,\frac12\}$, so $|-\frac12+E|=\frac12<1$. Therefore, we can instead do
$$
\frac1{1+E}=\frac{2/3}{1-(\frac13-\frac23E)}=\frac23\Big(1+\big(\frac13-\frac23E\big)+\big(\frac13-\frac23E\big)^2+\dots\Big)\tag{*}
$$
You still have to work out what that nasty infinite summation is. The best method I could think of to evaluate the terms $\big(\frac13-\frac23E\big)^n$ was to work out small cases to find a pattern and prove it by induction. Fortunately, you can show
$$
\big(\frac13-\frac23E\big)^2=\frac19
$$
which quickly implies that
$$
\big(\frac13-\frac23E\big)^n=
\begin{cases}
\frac1{3^n} & n\text{ is even}\\
\frac1{3^n}-\frac2{3^n}E & n\text{ is odd}
\end{cases}
$$
which allows you to collect the coefficients of $E$ in $(*)$ and arrive at a simple formula for $(1+E)^{-1}$.
