Finding $T^{\perp}$!
Solution 1:
If $\lambda_1\ne0$, then $\lambda_n=\frac{\lambda_1}{\sqrt n}$. But $\left(\frac{\lambda_1}{\sqrt n}\right)_{n\in\Bbb N}=\lambda_1\left(\frac1{\sqrt n}\right)_{n\in\Bbb N}\notin\ell^2$. So, $T^\perp=\{0\}$.
Solution 2:
You have proven that $\lambda_n = \frac{\lambda_1}{\sqrt{n}}$.
Now, use the fact that $x\in\ell^2$, which means that $$\sum_{n=1}^\infty |\lambda_n|^2$$ must converge.