Eigenvalues and eigenvectors of $T^*T$ I am working on the following problem:

Define a bounded linear operator $T$ on $L^2[0,1]$ as $$Tf(x)=\int_x^1 f(t)dt, \; f \in L^2[0,1], \; x \in [0,1].$$ Find the range of $T^*+T$ and the eigenvalues and eigenvectors of $T^*T$.

I started by noticing that
$$
T^*g(x)= \int_{0}^{x}g(t)dt.
$$
So I got $$(T^*+T)f=\int_0^1f(t)dt.$$ At this point, by the Cauchy-Schwarz inequelity I have $$(T^*+T)f \leq \|f\|_2,$$ and so I want to say that $\text{range}(T^*+T)=\{c \in \mathbb{R} \text{ such that } c=\|f\|_2 \text{ for some } f \in L^2[0,1] \}$.
Finally, I think $T^*T$ has no eigenvalues, since neither $T$ nor $T^*$ have some. Am I correct? Any help is appreciated.
Edit: as Aaron pointed out, we have $(T^*T)^*=T^*T^{**}=T^*T$. So the operator is self-adjoint and therefore it will have non-negative eigenvalues.
 A: As you found, $T+T^*$ is a rank-one operator, and its range are the constant functions.
As $T$ is compact, so is $T^*T$. So its spectrum will consist of zero and positive eigenvalues.
If $\lambda>0$ and $T^*Tf=\lambda f$, we have
$$\tag1
\lambda f=\int_0^x\int_t^1f(s)\,ds\,dt.
$$
Since $f$ is in $L^2$, it is also in $L^1$ (use Cauchy-Schwarz) and so $Tf$ is continuous. It follows that $\lambda f$ is differentiable, and going back and forth we get that $f$ is C$^\infty$. Differentiating $(1)$ we obtain
$$\tag2
\lambda f'=\int_x^1f(s)\,ds,
$$
and differentiating once more
$$\tag3
\lambda f''=-f.
$$
The solutions to $(3)$ are of the form
$$
f(t)=\alpha\,\cos\frac t{\sqrt\lambda}+\beta\,\sin\frac t{\sqrt\lambda}.
$$
From $(1)$ we have the initial condition $f(0)=0$, which gives $\alpha=0$. And from $(2)$ we have $f'(1)=0$, which gives
$$
0=\beta\,\cos\frac1{\sqrt\lambda}.
$$
As the sine is zero on $\frac\pi2+k\pi$, $k\in\mathbb Z$, we get
$$
\lambda_k=\frac4{(2k+1)^2\pi^2},\qquad k\in\mathbb N,
$$
with eigenvectors
$$
f_k(t)=\sin 2k\pi t.
$$
