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 $u^{-1}=\frac{1}{2}(2I-E)$

Answer 2

$$EE=E$$ $$Ex=\lambda x$$ $$EEx=E(Ex)=E(\lambda x)=\lambda^2 x$$ $$EEx=Ex$$ $$\lambda^2 x = \lambda x$$ $$(\lambda-1)\lambda x = 0$$ Since an eigenvector $x$ cannot be zero, we get $\lambda=0$ or $\lambda=1$.

Answer 3

Consider this explanation:

E is an idempotent operator on V $\space \Leftrightarrow \space \space $ E$^2$ = E $\space \space \Leftrightarrow \space$ E is a projection operator on V having range say R and null space say N.

Then we have, $\beta \in R \Leftrightarrow E\beta$ = $\beta$,and V = R $\oplus$ N ... $\mathtt result$

(clearly, $\beta \in N \Leftrightarrow E\beta =0$ )

So, if $\mathscr B_R =${ $\beta_1,...,\beta_k $} is an ordered basis of R and $\mathscr B_N =${ $\beta_{k+1},...,\beta_n $} is an ordered basis of N, then $\mathscr B $= { $\mathscr B_R, \mathscr B_N $ } = {$\beta_1,...,\beta_k, \beta_{k+1},...,\beta_n $} is an ordered basis of V = R $\oplus$ N, where, $E_{\beta_i} = \beta_i \mathit \space for\space i = 1,...,k $ and $E_{\beta_i} = 0 \mathit \space for \space i = k+1,...,n.$

It is trivial that range R is a characteristic space and it corresponds to the characteristic value 1, null space N is a characteristic space and it corresponds to the characteristic value 0.

[ Let me know if you want a short proof of the $\mathtt result$. The problem that I have from the above mentioned reasoning is that one can in a similar way derive that $\mathit EEEx = Ex \Rightarrow (\lambda^2 - 1)\lambda x = 0 \Rightarrow \lambda = 0, 1$ or $ -1 $(and so on).

Let me know what you think. ]

Thank You!