Spectrum of the operator $(Ax)(t) = x(0) + t x(1)$ for $A: C[0,1] \to C[0,1]$. I have a question:
How can we find the spectrum and resolvent of the operator $A: C[0,1] \to C[0,1]$ which defines as  $(Ax)(t) = x(0) + t x(1)$? 
$\textbf{Some definitions and facts:}$
The resolvent set of $A$ is:
$$\rho(A) = \{ \lambda \in \mathbb{C} \mid A - \lambda I ~ \text{is invertible} \} = \{ \lambda \in \mathbb{C} \mid A - \lambda I ~ \text{is bijective} \} $$
And the spectrum of $A$ is:
$$\sigma(A) = \mathbb{C}\ \rho(A) = \{ \lambda \in \mathbb{C} \mid A - \lambda I ~ \text{is not invertible} \} = \{ \lambda \in \mathbb{C} \mid A - \lambda I ~ \text{is not bijective} \} $$.
And we know that $A$ is a positive and a compact operator. If needed I will prove it.
Can you please give me the idea on how to find its spectrum?
Thanks!
 A: Obviously, the image of $A$ is the 2-dimensional subspace $S=\{\alpha+\beta t\}$. Let $P$ be the projector onto that subspace, $(Px)(t)=x(0)+(x(1)-x(0))t$. For  $y=(I-P)x$, it follows that $y(0)=y(1)=0$, so $Ay=A(I-P)x=0$. That means $0$ is certainly in the spectrum of $A$, and $AP=A$. Since $PA=A=AP$, $A$ and $P$ commute.
For $x\in S$, we can check easily that $(A^2-2A+I)x=0$, since that's true for the basis elements $e_1(t)=1$ and $e_2(t)=t$. In the latter case, we have $Ae_2=e_2$, so $1$ is in the spectrum, too. It follows that $(A^2-2A+I)P=0$. From those identities, one can easily check that
$$(A-\lambda I)\left(\frac{2-\lambda}{(\lambda-1)^2}I-\frac1{(\lambda-1)^2}A\right)P=P,$$ so we get eventually
$$(A-\lambda I)^{-1}=\left(\frac{2-\lambda}{(\lambda-1)^2}I-\frac1{(\lambda-1)^2}A\right)P-\frac1{\lambda}(I-P),$$ showing that the spectrum of $A$ is really just $\sigma(A)=\{0,1\}$.
A: Looks like only $\lambda=0$ and $\lambda=1$ is in the spectrum.


*

*For $\lambda=0$: take $x (t)=t (1-t) $ and $x\ne 0$ but $Ax=0$.

*For $\lambda=1$: take $x (t)=t $. Note $Ax=x $ i.e. $ (A-I)x=0$ but $x\ne 0$.

*For $\lambda\ne 0, 1$: denote$B=A-\lambda I $. For $x\in C [0,1] $, let $y =Bx$, so $y (t)=(Bx)(t)=(A-\lambda I)(x)(t)=x (0)+tx (1)-\lambda x (t) $. By substituting $t=0$ and $t=1$: $y (0)=(1-\lambda)x (0) $, $y (1)=x (0)+x (1)(1-\lambda) $, you can solve for $x (0), x (1) $ (as $\lambda\ne 1$) and then $x (t)=\frac {1}{\lambda}(y (t)-x (0)-tx (1)) $ as $\lambda\ne 0$. You can then rigorously prove that this mapping $y\mapsto x $ is the inverse of $A-\lambda I $.
