Let $$T:\ell^2 \to \ell^2$$ is unilateral shift operator, defined by $$T(x_1,x_2,x_3......)=(0,x_1,x_2,x_3.....),$$ then show that $T$ has no eigenvalue. But every $\lambda \in \mathbb{C}$ such that $|\lambda|<1$ is an eigenvalue of $T^{*}$ with multiplicity one.
show that none of eigen vectors of $T^{*}$ are orthogonal to each other.
I have found
$$T^{*}(x_1,x_2,x_3......)=(x_2,x_3,x_4.....)$$ and eigenspace$\ E_{\lambda}$ corresponding to $\lambda$ equal to linear span of vector $$(1,\lambda,\lambda^2,\lambda^3....)$$
but I am stuck in the last part.how to show no two eigenvectors are orthogonal to each other.
Any hint please?