How does the Hermiticity of an operator imply that functions have an expansion in in multiple bases? In Shankar QM it is stated that since the $\boldsymbol K$ operator is Hermitian, vectors, which are expanded in the $\boldsymbol X$ basis with components $f(x) = \langle x | f \rangle$, must have an expansion in the $\boldsymbol K$ basis (i.e. the eigenbasis of $\boldsymbol K$).
Why is this?
 A: Theorem. If $\hat{K}$ is Hermitian then the eigenvectors of $\hat{K}$ can be chosen so as to form an orthonormal basis $|\xi_i\rangle$. 
(The "can be chosen" wording completely handles the cases of degenerate eigenvalues, and the fact that an eigenvector times a constant is still an eigenvector)
(the converse is not true. In fact replace the word "Hermitian" with "Normal" and the theorem still holds. See: the wikipedia page on the spectral theorem. Also, I'm mainly considering the finite dimensional case)
Theorem. If $\{|\xi_i\rangle\}$ forms an orthonormal basis, then $I=\sum |\xi_i\rangle \langle \xi_i|$, where $I$ is the identity operator.
So, if your vector is $|\varphi\rangle$, you can also write it as $|\varphi\rangle=I|\varphi\rangle=\sum |\xi_i\rangle \langle \xi_i|\varphi\rangle$
if you're in a basis given by $\{ |n\rangle \}$, then $\langle n|\varphi\rangle$ are your components and $I=\sum |\xi_i\rangle \langle \xi_i|=\sum |n\rangle \langle n|$, so:
$|\varphi\rangle=II|\varphi\rangle=\sum_{i,n}|\xi_i\rangle \langle \xi_i|n\rangle \langle n|\varphi\rangle$
which is a change of basis represented by matrix multiplication, with $\langle \xi_i|n\rangle$ being the matrix elements.
My point in all this is that those two theorems when combined with Dirac notation are very nice!
Bonus:
Consider two particularly interesting bases:


*

*The basis of delta functions $\{|x\rangle\}$ where $\langle x | y\rangle =\delta(x - y)$. The $|x\rangle$ vectors are vectors with specific position in space.

*The basis of exponentials $\{|\nu \rangle\}$ where $\langle x | \nu \rangle = e^{i 2 \pi \nu x}$. These have specific spatial frequency. Note that $\langle \nu | \mu \rangle = \delta(\nu - \mu)$.


Now consider the equation we had at the end of the previous section:
\begin{align}
|f\rangle
&= \int dx d\nu | \nu \rangle \langle \nu | x \rangle \langle x | f \rangle \\
\langle \nu' | f \rangle &= \int dx d\nu \langle \nu'| \nu \rangle e^{-i 2 \pi \nu x} f(x)\\
&= \int dx d\nu \delta(\nu - \nu') e^{-i 2 \pi \nu px} f(x) \\
\tilde{f}(\nu') &= \int dx \, e^{-i 2 \pi \nu' x} f(x) \, .
\end{align}
So yeah, the Fourier transform is a change of basis.
