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搬运
We use an overlooked characterization of interlacing to give a two sentence proof of Cauchy's interlace theorem\cite{hwang}. Recall that
if polynomials $f(x)$ and $g(x)$ have all real roots $r_1\le
r_2\le\dots\le r_n$ and $s_1\le s_2\le\dots\le s_{n-1}$ then we say that $f$ and $g$ interlace if and only if
$$ r_1 \le s_1 \le r_2 \le s_2 \cdots \le s_{n-1} \le r_n.$$
The following can be found in \cite{rahman}, along with a discussion of its history back to Hermite.
Theorem.
The roots of polynomials $f,g$ interlace if and only if the linear combinations $f+\alpha g$ have all real roots for all $\alpha\in{{\mathbb R}}$.
Corollary.
If $A$ is a Hermitian matrix, and $B$ is a principle submatrix of $A$, then the eigenvalues of $B$ interlace the eigenvalues of $A$.
Proof.
Choose $\alpha\in\mathbb R$, partition $
A =\left(\begin{array}{c|c}
B & c \\\hline c^\ast & d
\end{array}\right)
$
and consider the following equation that follows from linearity of the determinant:
$$
\left|\begin{array}{c|c}
B - xI & c \\\hline c^\ast & d-x+\alpha
\end{array}\right|
=
\left|\begin{array}{c|c}
B - xI & c \\\hline c^\ast & d-x
\end{array}\right|
+
\left|\begin{array}{c|c}
B - xI & c \\\hline 0 & \alpha
\end{array}\right|
$$Since the matrix on the left hand side is the characteristic polynomial of a Hermitian matrix, $|A-xI| + \alpha|B-xI|$ has all real roots for any $\alpha$, and hence the eigenvalues interlace. $\square$ |
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