# Spectrum relationship between non-symmetric matrix's principle sub-matrices

I'm considering this: Given a non-symmetric $$\{0,1\}$$ matrix $$A$$, define $$A_{(i)}$$ to be the principal submatrix which deletes $$i$$th row and column. If we denote $$\rho(A)$$ as spectral radius of the matrix $$A$$, then is there a criteria that helps to easily compare $$\rho(A_{(i)})$$ and $$\rho(A_{(j)})$$?

Intuitively, the sub matrix that have been deleted the rows and columns with more non-zero entries has lower spectrum radius, but I can not prove this, could anyone help me with it?

• Let $A$ be $2\times 2$ and nilpotent. What kinds of problems does this create? Commented Feb 21, 2023 at 18:54
• Thank you for comment, but I seems to can not understand what you mean, could you be more specifier? Commented Feb 21, 2023 at 20:35
• @Duber Consider the specific matrix $$A = \pmatrix{1&1\\-1&-1}$$ Commented Feb 21, 2023 at 20:43
• @BenGrossmann Thank you, I think for this matrix, it is clear that $\rho(A_{(1)})<\rho(A_{(2)})$, but the matrix $A$ should be a $\{0,1\}$ matrix, and if the dimension increases, I think it is still quite hard to analyse. I think I may not totally get your idea, could you explain a little bit more? Commented Feb 21, 2023 at 20:51
• This isn't true. Consider $$A=\pmatrix{1&0&0&0\\ 0&0&1&1\\ 0&0&0&1\\ 0&0&0&0}.$$ Then $A_{(1)}$ has fewer nonzero elements of $A$ deleted than $A_{(4)}$ does, but $\rho(A_{(1)})=0<1=\rho(A_{(4)})$. Commented Feb 22, 2023 at 0:53