I have the following question, which I need for a research work:
Suppose that $A$ is an $n \times n$ symmetric, positive definite matrix, and let $m$ be a positive integer less than $n$. Let $B$ be the $m \times m$ top principle submatrix of $A$, i.e. $B := (A_{ij})_{i\le m, j\le m}$. Then, is the spectral norm (largest magnitude eigenvalue) of $B$ less than the spectral norm of $A$? If not true, can a counterexample be given?
This is true for any complex square matrix, not just for the real symmetric ones. It follows directly from the characterisation of the spectral norm that $\|A\|_2=\max_{\|x\|_2=1}\|Ax\|_2$. More specifically, let $y\in\mathbb C^m$ be a unit vector such that $\|By\|_2=\|B\|_2$. Append zeroes to $y$ to form a unit vector $x\in\mathbb C^n$. Then $\|B\|_2=\|By\|_2\le\|Ax\|_2\le\|A\|_2$.
As a remark, note that unless the matrix is Hermitian, the spectral radius of a principal submatrix can be larger than the spectral radius matrix of the parent matrix. E.g. in the companion matrix $$ A=\pmatrix{0&-1\\ 1&2}, $$ the spectral radius of the trailing principal submatrix $2$ is greater than $1$, the spectral radius of $A$.