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?
Best Answer
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 $A$.