I have a real matrix
$$A = \begin{bmatrix} 0 & A_{12} \\ 0 & A_{22} \\ \end{bmatrix}$$
where $A$ is a square matrix with dimension $N \times N$ and $A_{22}$ is also a square matrix with dimension $N'\times N'$.
I already know the spectral radius for matrix $A$ is $\lambda$. I wonder if we can have an estimated upper bound of the spectral radius for submatrix $A_{22}$. It would be great if we can prove that the spectral radius of $A_{22}$ is more or less at the same magnitude with the $\lambda$.
Suppose that we have a right eigenvector $v$ associated with an eigenvalue $\lambda \ne 0$. Write it as $v = \begin{pmatrix} v_1 v_2 \end{pmatrix}^T$. Then $$ \lambda \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \lambda v = A v = \begin{pmatrix} 0 & A_{12} \\ 0 & A_{22} \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} A_{12} v_2 \\ A_{22} v_2 \end{pmatrix} $$ Therefore, $v_2$ has to be an eigenvector of $A_{22}$ with eigenvalue $\lambda$. Conversely, for any $v_2$ that is an eigenvector of $A_{22}$ associated with an eigenvalue $\lambda \ne 0$, we can choose $v_1 = A_{12} v_2 / \lambda$ to make the eigenvalue equation above true. So there is a bijection between eigenvectors of $A$ and eigenvectors of $A_{22}$ with the same eigenvalue. Therefore, the spectral radius of $A$ has to be the same as the spectral radius of $A_{22}$.