Positive definite matrix submatrix eigenvalues For a positive definite matrix $n\times n$ prove that for its any $(n-1)\times (n-1)$ submatrix all eigen-values lie within the range of the eigenvalues of the original matrix.
Please provide some intuition as of why this is the case.
 A: It seems to me that this is not true.  The identity matrix 
$$\left( \begin{array}{ccc} 
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right)$$
has eigenvalue $1$.  However the submatrix 
$$\left( \begin{array}{cc} 
 0 & 1 \\
0  & 0\end{array} \right)$$
is nilpotent and so only has eigenvalue $0$.
A: As the other answer shows, this is not true (and the eigenvalues of an arbitrary submatrix of a positive definite matrix are not necessarily real in the first place). The correct statement should be that if $A$ is a positive definite matrix of size $n$ and $B$ is a principal submatrix of $A$ of size $n-1$, then $\lambda_\min(A)\le\lambda_\min(B)\le\lambda_\max(B)\le\lambda_\max(A)$. This is simply because the maximum eigenvalue of a Hermitian matrix $H$ is given by $\max_{\|x\|=1}x^\ast Hx$, and $\{y^\ast Ky: y\in\mathbb C^{n-1},\,\|y\|=1\}$ is a subset of $\{x^\ast Hx: x\in\mathbb C^n,\,\|x\|=1\}$ whenever $K$ is a principal submatrix of $H$.
We actually have a stronger result, known as Cauchy's interlacing inequality, which says that if $A$ is a Hermitian matrix of size $n$ and $B$ is a principal submatrix of $A$ of size $n-1$, then $\lambda_j(A)\le\lambda_j(B)\le\lambda_{j+1}(A)$ for each $j$, when the eigenvalues of each matrix are arranged in ascending order. This is a consequence of Courant-Fischer minimax principle.
