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
$I-\frac 1 2 E$ is the inverse of $I+E$.
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 $|E|&lt;1$. However, since $E$ is idempotent, its eigenvalues may be either $0$ or $1$, so $|E|&lt;1$ may not hold. 

However, $-\frac12+E$ will have eigenvalues in $\{-\frac12,\frac12\}$, so $|-\frac12+E|=\frac12&lt;1$. 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
$$
\big(\frac13-\frac23E\big)^n=
\begin{cases}
\frac1{3^n} &amp; n\text{ is even}\\
\frac1{3^n}-\frac2{3^n}E &amp; n\text{ is odd}
\end{cases}
$$
which allows you to collect the coefficients of $E$ in $(*)$ and arrive at a simple formula for $(1+E)^{-1}$.