Show that $\ker(T)=\{\varphi _n\mid\lambda_n\neq 0\}^\perp $ Let $T:H \to H$ be defined as $Tx=\sum_{n=1}^{\infty} \lambda_n \langle x,\varphi _n \rangle \varphi _n$, given that $\{\varphi _n\}_{n=1}^\infty$ is an orthonormal  sequence (not necessarily a basis) and $\{\lambda_n\}_{n=1}^\infty$ is a sequence of numbers (which may be complex if the Hilbert space is complex).
Show that $\ker (T)=\{\varphi _n\mid\lambda_n\neq 0\}^\perp $.
What does this $\{\}^\perp $ notation mean? Do I need to show that $\varphi _n$ are perpendicular to each other? If so how?
 A: The notation $S^\perp$ means $\{x\in H\mid \langle x,y\rangle=0,\text{ for all }y\in S\}$.
Let $x\in\ker T$; you need to prove that, for every $m$ with $\lambda_m\ne0$, you have $\langle x,\varphi_m\rangle=0$.
You know that $\sum_n\lambda_n\langle x,\varphi_n\rangle\varphi_n=0$, so also
$$
\Bigl<\sum_n\lambda_n\langle x,\varphi_n\rangle\varphi_n,\varphi_m\Bigr>
=0
$$
Since the series converges, you can deduce that
$$
0=\sum_n\langle\lambda_n\langle x,\varphi_n\rangle\varphi_n,\varphi_m\rangle
=\sum_n\lambda_n\langle x,\varphi_n\rangle\,\langle\varphi_n,\varphi_m\rangle=
\lambda_m\langle x,\varphi_m\rangle
$$
Since $\lambda_m\ne0$, by assumption, it follows that $\langle x,\varphi_m\rangle=0$.
Conversely, you need to show that, if $\langle x,\varphi_m\rangle=0$ whenever $\lambda_m\ne0$, then $x\in\ker T$, meaning that $\sum_n\lambda_n\langle x,\varphi_n\rangle\varphi_n=0$. Can you show it?
A: HINT: You need to use the fact that $\phi_n$ is orthonormal. That is, $\left\langle \phi_n,\phi_m \right\rangle = \left\lbrace \begin{array}{cc} 
0 & n \neq m \\
1 & n = m
\end{array} \right.$. Start by supposing that $f \in \ker(T)$ so that $T(f) = \sum_{n=1}^\infty \lambda_n \left\langle f,\phi_n \right\rangle \phi_n=0$. What can you say if $f$ is one of the $\phi_m$s? Try working with the definition of orthogonal.
