Understanding a proof that a Hermitian matrix has an orthonormal basis whose diagonal contains its eigenvalues Context. Accoring to pg. 36 of Principles of Quantum Mechanics:



My confusion surrounds the nature of the portions of the proof highlighted in red.
Notation. It seems to me the proof potentially conflates two distinct matrices and kets.

*

*First, we have the $n$-dimensional matrix $\Omega$ vs. the $n-1$-dimensional matrix $\Omega_{n-1}$, which is the lower right subset of $\Omega$ (as outlined in the proof).

*Next we have the $n-1$-dimensional $|\omega_2 \rangle_{n-1}$, which can be expressed as $(\omega_2, 0, \ldots, 0)$ against the basis $\{| \omega_2 \rangle_{n-1}, \ldots \}$. This should be contrasted with $|\omega_2 \rangle_n$, which the author implicitly asserts (but unless I'm missing something: doesn't establish) can be expressed as $(0, \omega_2, 0, \ldots, 0)$ against the basis $\{|\omega_1 \rangle, |\omega_2\rangle_n, \ldots \}$.

Problem. It seems to me that the author has established only that
$$
\Omega_{n-1} | \omega_2 \rangle_{n-1} = \omega_2 |\omega_2 \rangle_{n-1}
$$
But why does that imply that
$$
\Omega_{n} | \omega_2 \rangle_{n} = \omega_2 |\omega_2 \rangle_{n}
$$
..which is what it seems the author actually needs to show in order for his proof to work?
 A: I agree with your misgivings about the proof presented. I believe that the proof can be clarified with a more careful consideration of the relationship between an operator and its matrix relative to some basis.
It is not specified in the statement or proof, but I'll assume that $\Omega$ is an operator on the space $\Bbb C^n$. In the first part of the proof, the author constructs the orthonormal basis
$$
\mathcal B_1^n = \{|\omega_1\rangle, V^1_{\perp 1}, \dots, V^{n-1}_{\perp1}\}.
$$
The superscript $n$ merely indicates that this is a basis for the $n$-dimensional space $\Bbb C^n$. Relative to this basis, $[\Omega]_{\mathcal B_1^n}$ has the form indicated in Equation (1.8.12). This implies that the set
$$
U_1 = |\omega_1\rangle^\perp = \operatorname{span}\{V^1_{\perp 1}, \dots, V^{n-1}_{\perp1}\} \subset \Bbb C^n
$$
is an invariant subspace of $\Omega$. In other words, for any vector $|v\rangle \in U_1$, $\Omega |v\rangle$ will also be an element of $U_1$. With that in mind, we may consider the restriction of $\Omega$ to this subspace, which I will denote as $\Omega|_{U_1}: U_1 \to U_1$. This simply refers to the map defined by
$$
\Omega|_{U_1} |v \rangle = \Omega|v\rangle,
$$
with the caveat that $|v\rangle$ is required to be an element of $U_1$. From $\mathcal B_1^n$, we "inherit" a basis for $U_1$:
$$
\mathcal B_1^{n-1} = \{V^1_{\perp 1}, \dots, V^{n-1}_{\perp1}\}.
$$
I claim that the matrix of $[\Omega|_{U_1}]_{\mathcal B_1^{n-1}}$ is equal to the "boxed submatrix" indicated in the proof. I would suggest that you try to prove that this is the case for yourself. Because $\Omega|_{U_1}$ is also Hermitian, we consider an eigenvalue $\omega_2$ of $\Omega|_{U_1}$ to construct a new orthonormal basis of $U_1$,
$$
\mathcal B_{2}^{n-1} = \{|\omega_2 \rangle, V_{\perp1,2}^1,\dots,V^{n-2}_{\perp1,2}\}.
$$
Note that each of these vectors from the basis are elements of $U_1$ and hence elements of $\Bbb C^n$. In particular, this eigenvector $|\omega_2\rangle$ satisfies
$$
\Omega|_{U_1} |\omega_2 \rangle = \Omega|\omega_2 \rangle = \omega_2 |\omega_2 \rangle.
$$
That is, $|\omega_2\rangle$ is not only an eigenvector of the restricted operator $\Omega|_{U_1}$, but also of the full operator $\Omega$ over $\Bbb C^n$.
The matrix of $\Omega|_{U_1}$ relative to this basis has the block-diagonal form
$$
[\Omega|_{U_1}]_{\mathcal B^{n-1}_{2}} = 
\pmatrix{
\omega_2 & 0 & \cdots & 0\\
0 & \star & \cdots & \star\\
\vdots & \vdots & \ddots & \vdots\\
0 & \star & \cdots & \star }.
$$
Now, this basis can be extended to the orthonormal basis
$$
\mathcal B_2^{n} = \{|\omega_1\rangle,|\omega_2 \rangle, V_{\perp1,2}^1,\dots,V^{n-2}_{\perp1,2}\}.
$$
I claim that the matrix $[\Omega]_{\mathcal B_2^n}$ has the above-mentioned matrix $[\Omega|_{U_1}]_{\mathcal B_2^{n-1}}$ as a submatrix. Thus, the matrix of $\Omega$ relative to $\mathcal B_2^n$ has the form
$$
[\Omega]_{\mathcal B_2^n} = 
\pmatrix{
\omega_1 & 0 & 0 & \cdots & 0\\
0 & \omega_2 & 0 & \cdots & 0\\
0 & 0 & \star & \cdots & \star\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & \star & \cdots & \star }.
$$
