Hermitian operator in an orthonormal eigenbasis In page 36 of Shankar's Principles of Quantum Mechanics is given a theorem:

Theorem 10. To every Hermitian Operator $\Omega$, there exists (at least) a basis consisting of its orthonormal eigenvectors. It is diagonal in this eigenbasis and has its eigenvalues as its diagonal entries. 

There is a part of the proof that I do not understand.  Turning to page 36, one reads that, corresponding to the eigenvalue $\omega_{1}$ is a normalized eigenvector $|\omega_{1}\rangle$. Considering $|\omega_{1}\rangle$ to be a basis, $\Omega$ has a matrix form $$\begin{bmatrix}
 \omega_{1}  &  0  &. &. &. & 0\\ 
 0& \omega_{2}& . & . & . &. \\ 
. & . & \omega_{3}&. & . &. \\ 
. &.  &.  &. &.  &. \\ 
. &  .&  .&  .& . &. \\ 
 0& . &.  &  .&  .&\omega_{n}
\end{bmatrix}$$
Where the dots indicate a series of zeroes. This leads me to question. Why does $\Omega$ take that form? Also, why is the first column the image of $|\omega_{1}\rangle$ after $\Omega$ has acted on it? 
 A: $\vert\omega_1\rangle$ would not be a basis, but the set $\{\vert\omega_1\rangle,\vert\omega_2\rangle, \ldots,\vert\omega_n\rangle\}$ is a basis.  The eigenvector $\vert\omega_i\rangle$ is such that $\Omega\vert\omega_i\rangle=\omega_i\vert\omega_i\rangle$ so...
With the identification
$$
\vert\omega_1\rangle \to \left(\begin{array}{c} 1 \\ 0 \\ \vdots \\ 0\end{array}\right)\, ,\quad
\vert\omega_2\rangle \to \left(\begin{array}{c} 0 \\ 1 \\ \vdots \\ 0\end{array}\right)\, ,\ldots \, ,
\vert\omega_n\rangle \to \left(\begin{array}{c} 0 \\ \vdots \\ 0 \\
1 \end{array}\right)
$$
and the matrix representation of $\Omega$ as you suggest you find by simple matrix multiplication that
$$
\Omega \vert \omega_n\rangle = \omega_n\vert\omega_n\rangle
$$
as per the properties of eigenstates of $\Omega$.  Note that under this identification the vectors $\vert\omega_i\rangle$ are an orthonormal basis since $\langle \omega_i\vert\omega_j\rangle=\delta_{ij}$.
A: Hermitian operators are important because their eigenvectors corresponding to different eigenvalues are orthogonal to each other (and can be normalized if required), and they form a basis for the Hilbert space on which the operators act.
Take, for instance, the $\sigma_z$ operator. Its eigenvalues are $\pm 1$ and its eigenvectors are $(1,0)^T, (0,1)^T$. You can express any vector of the 2-dimensional Hilbert space as a linear combination of these eigenvectors.
