Let $H$ be a Hilbert space with the inner product $(\cdot,\cdot)_H$ and let $T:H\to H$ be a bounded compact and self-adjoint operator on $H$. In this case there exists an $H$-orthonormalsystem $(\varphi_j)$ of eigenvectors of $T$ and corresponding eigenvalues $(\lambda_j)$ such that $$Tf = \sum_j\lambda_j(f,\varphi_j)_H\varphi_j\quad\text{for every }f\in H$$ and $\lambda_n\to 0$ for $n\to\infty$, which is basically the spectral theorem for compact self-adjoint Operators
How can I follow from this, that $\text{span}\{\varphi_j\}_j$ is dense in $\text{Im } T$? Well in this book they say it follows from the expansion above and from $\varphi_j=\frac{1}{\lambda_j} T\varphi_j$. From this I tried to do it and came up with
$$Tf = \sum_j\lambda_j(f,\varphi_j)_H \varphi_j = \sum_j\lambda_j(f,\varphi_j)_H\frac{1}{\lambda_j}T\varphi_j = \sum_j(f,\varphi_j)_HT\varphi_j\in \text{Im }T$$
And yeah, I am not sure why this would prove the statement... So this equation would tell me that we can represent $Tf$ by linear combinations of the $T\varphi_j$ so that would mean the span of $(T\varphi_j)_j$ is dense in $\text{Im }T$.
Maybe you guys can help me with this. Thank you very for your answers!
We can immediately follow $\text{Im } T\subset\overline{\text{span}\{\varphi_j\}_j}$ from the equation \begin{equation}\label{eq1}\tag{1} Tf=\sum_j \lambda_j(f,\varphi_j)_H\varphi_j \end{equation} since $\text{Im }T=\{Tf\mid f\in H\}$. So if $g\in \text{Im } T$ there's an $f\in H$ such that $Tf = g$ and from the equation in \eqref{eq1} we have $$g = \sum_j \lambda_j(f,\varphi_j)_H\varphi_j\in\overline{\text{span}\{\varphi_j\}_j}.$$
The other direction, namely $\overline{\text{span}\{\varphi_j\}_j}\subset\text{Im } T$, follows from the fact that
\begin{equation}\label{eq2}\tag{2} T\varphi_j = \lambda_j\varphi_j\Longleftrightarrow \varphi_j=\frac{1}{\lambda_j} T\varphi_j \end{equation}
and therefore $\varphi_j\in\text{Im }T$ and since $\text{Im }T\subset H$ is a linear subspace and $T$ is compact we have $\overline{\text{span}\{\varphi_j\}_j}\subset\text{Im }T$.