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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?

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    $\begingroup$ 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
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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 $\mathbf{X}$.

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