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Let $A$ be an $n\times n$ symmetric matrix whose diagonal are is covered by zero blocks (square, but not of a fixed size) and all other entries are $1$ (one).

How can I find its Characteristic Polynomial and eigenvalues?

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  • $\begingroup$ Can you show us what have you tried so far? $\endgroup$ – Vlad Jul 18 '15 at 6:41
  • $\begingroup$ $$A=\begin{array}{ccc} \begin{pmatrix} b_1 & a & a\\ a & b_2 & a\\ a & a & b_3 \end{pmatrix} \end{array}.$$\\ \end{array}.$$ here A is symmetric 9x9 matrix $b_1$ is a 2x2, $b_2$ is a 3x3 and $b_3$ is a 4x4 diagonal matrix with entry 0 and all entries of a is 1 also $a_i$ are may not be sqare matrix. $\endgroup$ – L S B. user255259 Jul 18 '15 at 8:14
  • $\begingroup$ Probably the easiest way to work this out is to treat $A$ as a rank one update of a block diagonal matrix, i.e. consider $A - J$ where $J$ is the $n\times n$ matrix of all ones. $\endgroup$ – hardmath Jul 18 '15 at 13:01
  • $\begingroup$ Thank you for your suggestion. Then (A−J) will be diagonal block matrices then how to relate the Characteristic polynomial of A. $\endgroup$ – L S B. user255259 Jul 24 '15 at 10:54
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I consider a special case only leaving a (straightforward) generalization to you.

Let the number of blocks be two, i.e. we have the $(k+m)\times(k+m)$ matrix $$ \left[\matrix{0_{k\times k} & 1_{k\times m}\\1_{m\times k} & 0_{m\times m}}\right] $$ where $0_{k\times k}$ and $1_{k\times m}$ denote matrices of all zeros, resp. ones, of the subindexed size.

It is clear that the rank of this matrix is two since there are only two linearly independent rows. It gives you directly that all eigenvalues are zero except maybe two. Let's try to find those two by definition $$ \left[\matrix{0_{k\times k} & 1_{k\times m}\\1_{m\times k} & 0_{m\times m}}\right]\left[\matrix{a\cdot 1_k \\b\cdot 1_m}\right]= \left[\matrix{mb\cdot 1_k \\ka\cdot 1_m}\right]=\lambda\left[\matrix{a\cdot 1_k \\b\cdot 1_m}\right]. $$ Thus $$ \left\{ \begin{array}{l} \lambda a=mb,\\ \lambda b=ka \end{array} \right.\qquad\Rightarrow\lambda^2=mk,\quad \Rightarrow\lambda=\pm\sqrt{mk}. $$

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  • $\begingroup$ Thank you, for your answer. $\endgroup$ – L S B. user255259 Aug 4 '15 at 5:57
  • $\begingroup$ Dear A.G. can I use the same techniques to find the eigenvalue any random matrix which is not same as above with rank r> 2. $\endgroup$ – L S B. user255259 Aug 16 '15 at 16:46

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