# How to find the eigen values of the following matrix:

Is there any way to find the eigen values of the following matrix:

$$A_{2n\times 2n}=$$ $$\begin{bmatrix}\textbf{0} & E_{n\times n}\\E^T&\textbf{0}\end{bmatrix}$$

where $$E=$$ $$\begin{bmatrix}1&1&1&\ldots1\\2&2&2&\ldots 2\\2&2&2&\ldots 2\\\ldots&\ldots&\ldots&\ldots\\\ldots&\ldots&\ldots&\ldots\\\ldots&\ldots&\ldots&\ldots \\2&2&2&\ldots 2\\\end{bmatrix}$$

My try:

I find that rows of $$E$$ are linearly dependent. Also every row of $$E$$ is just a scalar multiple of the first row.

So I was guessing may be $$0$$ may occur as its eigen value many number of times.

What are some methods to calculate the characteristic polynomial?

Can someone kindly help?

Let $$e$$ denotes the all-one vector and write $$E=ve^T$$. Then $$\det(xI_{2n}-A)=\det(x^2I_n-EE^T)=\det(x^2I_n-ve^Tev^T)=\det(x^2I_n-nvv^T).$$ Hence the eigenvalues of $$A$$ are given by $$2n-2$$ copies of $$0$$ and $$\pm\sqrt{n}\|v\|=\pm\sqrt{n(4n-3)}$$.
• Can you kindly tell what is $v$? – Math_Freak May 17 at 11:56
• And how is $vv^t=\sqrt{4n-3}$ – Math_Freak May 17 at 13:11
• Also i don't follow how you obtained values of $x$ from $\det(x^2I-nvv^T)=0$ – Math_Freak May 17 at 13:13