# How to find the eigenvalues?

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?

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:
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 $$\operatorname{diag}(3,0,0)$$.
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\$$?