I want to find the non-zero eigen-values of a matrix of the following form

\begin{align} M = \begin{bmatrix} \mathbf{0}_{n_1 \times n_1} & \mathbf{0}_{n_1 \times n_2} & \mathbf{0}_{n_1 \times n_3} \\ \mathbf{A}_{n_2 \times n_1} & \mathbf{B}_{n_2 \times n_2} & \mathbf{C}_{n_2 \times n_3} \\ \mathbf{0}_{n_3 \times n_1} & \mathbf{0}_{n_3 \times n_2} & \mathbf{0}_{n_3 \times n_3} \\ \end{bmatrix} \end{align} where the subscripts indicate the size of the corresponding block matrix, that is $\mathbf{A}_{n_2 \times n_1}$ is a $n_2 \times n_1$ matrix. My guess is that eigen-values of $M$ include $n_1 + n_3$ zeros and eigenvalues of $\mathbf{B}$. For simple examples, this intuition is correct, but I want to know whether this is correct for the block matrix $M$ described above? Any suggestion?

  • 1
    $\begingroup$ The problem is that these matrices are not square. For square block matrices you can use this question. So you could change the first and second "block-row" of your matrix and form square matrices. Maybe these formed square matrices have nice properties. $\endgroup$ – P. Siehr Nov 10 '17 at 7:55
  • $\begingroup$ @P.Siehr Thanks for your suggestion. I think the question you linked actually can solve my problem. Why do we need all matrices to be square? I think we can apply determinant to block matrices and conclude my conjecture. $\endgroup$ – m0_as Nov 10 '17 at 8:02

We work over the algebraic closure of the base field. Suppose $\lambda\neq 0$ is an eigenvalue of $M$. Then, $$\mathbf{M}-\lambda \mathbf{I}=\left[ \begin{array}{ccc} -\lambda \mathbf{I} & \textbf{0} &\textbf{0} \\ \mathbf{A}&\mathbf{B}-\lambda \mathbf{I}&\mathbf{C} \\ \textbf{0}&\textbf{0}&-\lambda\mathbf{I} \end{array} \right]$$ Using a sequence of row operations, we can reduce $\mathbf{M}-\lambda \mathbf{I}$ to a form $$\left[ \begin{array}{ccc} -\lambda \mathbf{I} & \textbf{0} &\textbf{0} \\ \textbf{0}&\mathbf{B}-\lambda \mathbf{I}&\mathbf{C} \\ \textbf{0}&\textbf{0}&-\lambda\mathbf{I} \end{array} \right]\,.$$ This shows that the characteristic polynomial of $\mathbf{M}$ is indeed $$\chi_\mathbf{M}(t)=t^{n_1+n_3}\,\chi_{\mathbf{B}}(t)\,,$$ where $\chi_\mathbf{X}$ indicates the characteristic polynomial of a square matrix $\mathbf{X}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.