Q:

Eigen values of a principal sub-matrix of a symmetric matrix

I am stuck on the following problem:
By Courant-Fischer Theorem : If $A$ is a symmetric $n\times n$ matrix and if $\lambda_1\le \lambda_2\le \cdots \le\lambda_n$ be the eigenvalues of $A$ then 
$$\lambda_k=\min_{U\subset \Bbb R^n,\ \dim U=k} \left\{\max_{x\in U}\dfrac{x^TAx}{x^Tx}\right\}.$$
Using this theorem :
Show that if $B$ is a principal sub-matrix of $A$ and if $ \mu_1\le \mu_2\le \cdots \le\mu_{n-1}$ be the eigenvalues of $B$ then prove that:
$$\lambda_1\le \mu_1\le \lambda_2\le \mu_2\le \cdots\le\mu _{n-1}\le \lambda_n$$
How should I do this this ? Please give some way. 
I am not getting how to start  it.

A:

As with any proof involving the Courant-Fischer inequality, this will require some comfort with manipulating inequalities.  In particular, we want to relate the eigenvalues of $B$ to the eigenvalues of $A$ in such a way that we can see that one is less than the other.
Now, let $I \subset \{1,\dots,n\}$ be such that $B = (a_{ij})_{i,j \in I}$.  Let $S_I \subset \Bbb R^n$ denote the set of vectors $x$ satisfying $x_i = 0$ for all $i \notin I$. We note that
$$
\lambda_k(B)=\min_{U\subset S_I,\ \dim U=k} \left\{\max_{x\in U}\dfrac{x^TAx}{x^Tx}\right\} \geq 
\lambda_k(A)=\min_{U\subset \Bbb R^n,\ \dim U=k} \left\{\max_{x\in U}\dfrac{x^TAx}{x^Tx}\right\}
$$
Notice the change in the subscript to the minimum!
Perhaps you can use similar reasoning on the max-min formulation to get the other inequality.  Or, simply note that $\lambda_k(-B) \geq \lambda_k(-A)$.

