Let $V$ be a real vector space and $E$ be an idempotent linear operator on $V$. Prove that $I + E$ is invertible.

Question

Let $V$ be a real vector space and $E$ be an idempotent linear operator on $V$, that is a projection. Prove that $I + E$ is invertible. Find $(I + E) ^{-1}$

My teacher taught me the following proof for the proposition

Since $E$ is an idempotent linear operator it is
diagonalizable. So there exists a basis of $V$
consisting of characteristics vectors of $E$ corresponding to the
characteristic values $0$ and $1$. That is, there exists a basis $\beta = \{ \beta_1, ….,\beta_n \}$ such that $E\beta_i = \beta_i$
for $i = 1, · · · , k$, and
$E\beta_i = 0$ for $i = k + 1, · · · , n$.

Then $(I + E)\beta_i = 2\beta_i$
for
$i = 1, · · · , k$ and $(I + E)\beta_i = \beta_i$
for $i = k + 1, · · · , n$, that is,

$$[I + E]\beta =
\begin{bmatrix}
2I_1 &0 \\
0 & I_2\\
\end{bmatrix}$$

where $I_1$ stands for $k × k$ identity matrix, $I_2$ is $(n − k) × (n −
k)$ identity matrix and each $0$ represents the zero matrix of
appropriate dimension. It is now easy to see that $[I + E]\beta$ is
invertible, since $det(I + E) = 2k \not=0$

To find the inverse of $(I + E)$, we note that

$$([I + E]_\beta)^{-1} =
\begin{bmatrix}
\frac{I_1}{2} &0 \\
0 & I_2\\
\end{bmatrix} = \begin{bmatrix}
I_1 &0 \\
0 & I_2\\
\end{bmatrix} + \begin{bmatrix}
\frac{-I_1}{2} &0 \\
0 & 0\\
\end{bmatrix} = I-\frac{1}{2}[E]_\beta$$

Therefore, $(I + E)
^{−1} = I −\frac{1}{2}E$

But I do not understand very well the following sentence:

So there is a basis of $ V $
consisting of characteristics vectors of $ E $ corresponding to the
characteristic values ​​$ 0 $ and $ 1 $

My question is, why the characteristic values ​​are $1$ and $0$?

Answer 1

Here is a short proof:

Let $u=I+E$. Since $E^2=E$, we have $u^2=I+2E+E^2=I+3E=3u-2I$

This can be rewritten as :

$uv=I$ where $v=\frac{1}{2}(3I-u)$

Which proves that $u$ is invertible and $u^{-1}=\frac{1}{2}(2I-E)$