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 $\pm\sqrt{n}\|v\|=\pm\sqrt{n(4n-3)}$.

  • $\begingroup$ Can you kindly tell what is $v$? $\endgroup$ – Math_Freak May 17 at 11:56
  • $\begingroup$ And how is $vv^t=\sqrt{4n-3}$ $\endgroup$ – Math_Freak May 17 at 13:11
  • $\begingroup$ Also i don't follow how you obtained values of $x$ from $\det(x^2I-nvv^T)=0$ $\endgroup$ – Math_Freak May 17 at 13:13

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