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How to find the integer eigenvalues of the $10 \times 10 $ matrix $$P=$$\begin{bmatrix} 6I&& J\\J^t&& K\end{bmatrix} where $I$ is a $4\times 4 $ matrix, $J$ is a $4\times 6$ matrix of all $1$ and $K$ denotes

$K=\begin{bmatrix} 8&1&1&0&1&1\\1&8&1&1&0&1\\1&1&8&1&1&0\\0&1&1&8&1&1\\1&0&1&1&8&1\\1&1&0&1&1&8 \end{bmatrix}$

I found that by Wolfram alpha $8$ is an eigenvalue with multiplicity $3$ and $6$ with multiplicity $5$.

I tried doing $P-6I$ where I got $4$ identical rows and hence $6$ is an eigenvalue with multiplicity at least $3$.

But I am stuck on how to do for $8$ and $6$??

Can I get some help?

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Let $E_{m\times n}$ denotes the $m\times n$ matrix of ones and let $r=1+4(x-6)^{-1}$. \begin{aligned} &\det\left[xI_{10}-\pmatrix{ 6I_4 &E_{4\times6}\\ E_{6\times4} &K}\right]\\ &=\det\pmatrix{ (x-6)I_4 &-E_{4\times6}\\ -E_{6\times4} &xI_6-K}\\ &\stackrel{(1)}{=}(x-6)^4 \det\left(xI_6-K - (x-6)^{-1}E_{6\times4}E_{4\times6}\right)\\ &=(x-6)^4 \det\left(xI_6-K - 4(x-6)^{-1}E_{6\times6}\right)\\ &=(x-6)^4 \det\pmatrix{ (x-7)I_3-rE_{3\times3} &I_3-rE_{3\times3}\\ I_3-rE_{3\times3} &(x-7)I_3-rE_{3\times3}}\\ &\stackrel{(2)}{=}(x-6)^4 \det\left(\left[(x-7)I_3-rE_{3\times3}\right]^2 -\left[I_3-rE_{3\times3}\right]^2\right)\\ &=(x-6)^4 \det\left(\left[(x-7)^2I_3-2(x-7)rE_{3\times3}+r^2E_{3\times3}^2\right] -\left[I_3-2rE_{3\times3}+r^2E_{3\times3}^2\right]\right)\\ &=(x-6)^4 \det\left((x-14x+48)I_3-2(x-8)rE_{3\times3}\right)\\ &=(x-6)^4 \det\left((x-6)(x-8)I_3-2(x-8)rE_{3\times3}\right)\\ &=(x-6)^4(x-8)^3 \det\left((x-6)I_3-2rE_{3\times3}\right)\\ &=(x-6)(x-8)^3 \det\left((x-6)^2I_3-2(x-6)rE_{3\times3}\right)\\ &=(x-6)(x-8)^3 \det\left((x-6)^2I_3-2(x-2)E_{3\times3}\right)\\ &\stackrel{(3)}{=}(x-6)(x-8)^3 \det\left((x-6)^2I_3-2(x-2)\operatorname{diag}(3,0,0)\right)\\ &=(x-6)(x-8)^3 \left[(x-6)^2-6(x-2)\right](x-6)^4\\ &=(x-6)^5(x-8)^3 (x^2-18x+48)\\ &=(x-6)^5(x-8)^3 \left[x-\left(9+\sqrt{33}\right)\right]\left[x-\left(9-\sqrt{33}\right)\right]. \end{aligned} Remarks:

  1. Use Schur complement here.
  2. Let $M=\pmatrix{A&B\\ C&D}$ where $A,B,C,D$ are square sub-blocks of equal sizes. If $CD=DC$, then $\det(M)=\det(AD-BC)$.
  3. $E_{3\times3}$ is similar to $\operatorname{diag}(3,0,0)$.
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Hint: The row-reduced echelon form of $\ P-6I\ $ is $$ \pmatrix{1&1&1&1&0&0&0&0&0&0\\ 0&0&0&0&1&0&0&0&1&1\\ 0&0&0&0&0&1&0&0&-1&0\\ 0&0&0&0&0&0&1&0&0&-1\\ 0&0&0&0&0&0&0&1&1&1\\ 0&0&0&0&0&0&0&0&0&0&\\ 0&0&0&0&0&0&0&0&0&0&\\ 0&0&0&0&0&0&0&0&0&0&\\ 0&0&0&0&0&0&0&0&0&0&\\ 0&0&0&0&0&0&0&0&0&0}\ , $$ which has a nullspace of dimension $5$, and the row-reduced echelon form of $\ P-8I\ $ similarly has a nullspace of dimension $3$. The characteristic polynomial of $\ P\ $ is $\ \left(x-6\right)^5\left(x-8\right)^3\left(x^2-18x+48\right)\ $. What does that tell you about the remaining eigenvalues of $\ P\ $?

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